Chapter 7 Integrals

(i) For the function $f(x) = \cos 2x$, we need to find a function whose derivative is $\cos 2x$.

We know that $\frac{d}{dx}(\sin x) = \cos x$.

Using the chain rule, $\frac{d}{dx}(\sin(ax)) = a \cos(ax)$.

We want the derivative to be $\cos 2x$. Let's consider the derivative of $\sin 2x$.

$\frac{d}{dx}(\sin 2x) = 2 \cos 2x$.

This is twice the function we want. So, we should consider $\frac{1}{2} \sin 2x$.

$\frac{d}{dx}\left(\frac{1}{2} \sin 2x\right) = \frac{1}{2} \cdot (2 \cos 2x) = \cos 2x$.

Thus, an anti derivative of $\cos 2x$ is $\mathbf{\frac{1}{2} \sin 2x}$.

(ii) For the function $f(x) = 3x^2 + 4x^3$, we need to find a function whose derivative is $3x^2 + 4x^3$. We can find the anti derivative of each term separately.

For the term $3x^2$, we know that $\frac{d}{dx}(x^n) = nx^{n-1}$. We want the derivative to be $3x^2$, which suggests the original function involved $x^3$.

$\frac{d}{dx}(x^3) = 3x^2$.

So, an anti derivative of $3x^2$ is $x^3$.

For the term $4x^3$, we want the derivative to be $4x^3$. This suggests the original function involved $x^4$.

$\frac{d}{dx}(x^4) = 4x^3$.

So, an anti derivative of $4x^3$ is $x^4$.

By the linearity of differentiation, the derivative of a sum is the sum of the derivatives. Therefore, the anti derivative of the sum is the sum of the anti derivatives.

An anti derivative of $3x^2 + 4x^3$ is $x^3 + x^4$.

Check: $\frac{d}{dx}(x^3 + x^4) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^4) = 3x^2 + 4x^3$.

Thus, an anti derivative of $3x^2 + 4x^3$ is $\mathbf{x^3 + x^4}$.

(iii) For the function $f(x) = \frac{1}{x}$, $x \neq 0$, we need to find a function whose derivative is $\frac{1}{x}$.

We know that $\frac{d}{dx}(\log|x|) = \frac{1}{x}$ for $x \neq 0$. The condition $x \neq 0$ is given, which means we should use the absolute value in the logarithm.

Thus, an anti derivative of $\frac{1}{x}$ is $\mathbf{\log|x|}$.

Solution:

(i) We need to find $\int \frac{x^3 − 1}{x^2} \;dx$.

We can rewrite the integrand by splitting the terms in the numerator:

$\frac{x^3 − 1}{x^2} = \frac{x^3}{x^2} - \frac{1}{x^2} = x - x^{-2}$

Now, we integrate term by term using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), and the rule $\int dx = x + C$:

$\int \frac{x^3 − 1}{x^2} \;dx = \int (x - x^{-2}) \;dx$

$= \int x \;dx - \int x^{-2} \;dx$

$= \frac{x^{1+1}}{1+1} - \frac{x^{-2+1}}{-2+1} + C$

$= \frac{x^2}{2} - \frac{x^{-1}}{-1} + C$

$= \frac{x^2}{2} - (- \frac{1}{x}) + C$

$= \frac{x^2}{2} + \frac{1}{x} + C$

(ii) We need to find $\int (x^{\frac{2}{3}} + 1) \;dx$.

We integrate term by term using the power rule, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, and $\int dx = x + C$:

$\int (x^{\frac{2}{3}} + 1) \;dx = \int x^{\frac{2}{3}} \;dx + \int 1 \;dx$

$= \frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1} + x + C$

$= \frac{x^{\frac{5}{3}}}{\frac{5}{3}} + x + C$

$= \frac{3}{5} x^{\frac{5}{3}} + x + C$

(iii) We need to find $\int (x^{\frac{3}{2}} + 2e^x − \frac{1}{x}) \;dx$.

We integrate term by term using the power rule, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, the rule for exponential functions, $\int e^x dx = e^x + C$, and the rule for the reciprocal function, $\int \frac{1}{x} dx = \log |x| + C$. We also use the linearity of the integral, $\int c f(x) dx = c \int f(x) dx$:

$\int (x^{\frac{3}{2}} + 2e^x − \frac{1}{x}) \;dx = \int x^{\frac{3}{2}} \;dx + \int 2e^x \;dx - \int \frac{1}{x} \;dx$

$= \int x^{\frac{3}{2}} \;dx + 2 \int e^x \;dx - \int \frac{1}{x} \;dx$

$= \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} + 2e^x - \log |x| + C$

$= \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + 2e^x - \log |x| + C$

$= \frac{2}{5} x^{\frac{5}{2}} + 2e^x - \log |x| + C$

Solution:

(i) We need to find $\int (\sin x + \cos x ) \;dx$.

Using the linearity of the integral, we can integrate each term separately:

$\int (\sin x + \cos x ) \;dx = \int \sin x \;dx + \int \cos x \;dx$

We know the standard integrals $\int \sin x \;dx = -\cos x + C_1$ and $\int \cos x \;dx = \sin x + C_2$.

So, $\int (\sin x + \cos x ) \;dx = -\cos x + \sin x + C$, where $C = C_1 + C_2$ is the constant of integration.

$= \sin x - \cos x + C$

(ii) We need to find $\int \text{cosec}\; x (\text{cosec}\; x + \cot x) \;dx$.

First, expand the integrand:

$\text{cosec}\; x (\text{cosec}\; x + \cot x) = \text{cosec}^2 x + \text{cosec}\; x \cot x$

Now, integrate the expanded form term by term:

$\int (\text{cosec}^2 x + \text{cosec}\; x \cot x) \;dx = \int \text{cosec}^2 x \;dx + \int \text{cosec}\; x \cot x \;dx$

We know the standard integrals $\int \text{cosec}^2 x \;dx = -\cot x + C_1$ and $\int \text{cosec}\; x \cot x \;dx = -\text{cosec}\; x + C_2$.

So, $\int \text{cosec}\; x (\text{cosec}\; x + \cot x) \;dx = -\cot x - \text{cosec}\; x + C$, where $C = C_1 + C_2$ is the constant of integration.

$= -\cot x - \text{cosec}\; x + C$

(iii) We need to find $\int \frac{1 − \sin x}{\cos^2 x} \;dx$.

We can rewrite the integrand by splitting the fraction:

$\frac{1 − \sin x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x}$

Using trigonometric identities, $\frac{1}{\cos^2 x} = \sec^2 x$ and $\frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x$.

So the integral becomes:

$\int (\sec^2 x - \sec x \tan x) \;dx$

Using the linearity of the integral, integrate each term separately:

$= \int \sec^2 x \;dx - \int \sec x \tan x \;dx$

We know the standard integrals $\int \sec^2 x \;dx = \tan x + C_1$ and $\int \sec x \tan x \;dx = \sec x + C_2$.

So, $\int \frac{1 − \sin x}{\cos^2 x} \;dx = \tan x - \sec x + C$, where $C = C_1 - C_2$ is the constant of integration.

$= \tan x - \sec x + C$

Solution:

We are asked to find the antiderivative F(x) of the function $f(x) = 4x^3 - 6$, such that $F(0) = 3$.

By definition, the antiderivative F(x) is the indefinite integral of f(x). So, we have:

$F(x) = \int f(x) \;dx$

$F(x) = \int (4x^3 - 6) \;dx$

Using the properties of integration (linearity) and the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$) and $\int c \;dx = cx + C$:

$F(x) = \int 4x^3 \;dx - \int 6 \;dx$

$F(x) = 4 \int x^3 \;dx - 6 \int dx$

$F(x) = 4 \cdot \frac{x^{3+1}}{3+1} - 6x + C$

$F(x) = 4 \cdot \frac{x^4}{4} - 6x + C$

$F(x) = x^4 - 6x + C$

Here, C is the constant of integration.

We are given the condition that $F(0) = 3$. We use this to find the value of C.

Substitute $x=0$ into the expression for F(x):

$F(0) = (0)^4 - 6(0) + C$

Since $F(0) = 3$, we have:

$3 = 0 - 0 + C$

$3 = C$

Thus, the constant of integration is $C = 3$.

Substituting the value of C back into the expression for F(x), we get the specific antiderivative:

$F(x) = x^4 - 6x + 3$

Solution:

We need to find an antiderivative of $f(x) = \sin(2x)$ by the method of inspection.

The method of inspection involves finding a function whose derivative is the given function.

We know that the derivative of $\cos x$ is $-\sin x$. Therefore, we can guess that the antiderivative of $\sin(2x)$ might involve $\cos(2x)$.

Let's find the derivative of $\cos(2x)$:

$\frac{d}{dx}(\cos(2x)) = -\sin(2x) \cdot \frac{d}{dx}(2x) = -\sin(2x) \cdot 2 = -2\sin(2x)$

We want the derivative to be $\sin(2x)$, but we got $-2\sin(2x)$. To adjust this, we need to multiply our initial guess, $\cos(2x)$, by a constant such that its derivative is $\sin(2x)$.

Consider the function $-\frac{1}{2}\cos(2x)$. Let's find its derivative:

$\frac{d}{dx}\left(-\frac{1}{2}\cos(2x)\right) = -\frac{1}{2} \cdot \frac{d}{dx}(\cos(2x))$

$= -\frac{1}{2} \cdot (-2\sin(2x))$

$= \sin(2x)$

The derivative of $-\frac{1}{2}\cos(2x)$ is indeed $\sin(2x)$.

Therefore, an antiderivative of $\sin(2x)$ is $-\frac{1}{2}\cos(2x)$.

The general antiderivative (or integral) is obtained by adding the constant of integration C.

So, the antiderivative of $\sin(2x)$ is $-\frac{1}{2}\cos(2x) + C$.

Solution:

We need to find an antiderivative of $f(x) = \cos(3x)$ by the method of inspection.

The method of inspection involves finding a function whose derivative is the given function.

We know that the derivative of $\sin x$ is $\cos x$. Therefore, we can guess that the antiderivative of $\cos(3x)$ might involve $\sin(3x)$.

Let's find the derivative of $\sin(3x)$:

$\frac{d}{dx}(\sin(3x)) = \cos(3x) \cdot \frac{d}{dx}(3x)$

$= \cos(3x) \cdot 3$

$= 3\cos(3x)$

We want the derivative to be $\cos(3x)$, but we got $3\cos(3x)$. To adjust this, we need to multiply our initial guess, $\sin(3x)$, by a constant such that its derivative is $\cos(3x)$. The constant should be $\frac{1}{3}$.

Consider the function $\frac{1}{3}\sin(3x)$. Let's find its derivative:

$\frac{d}{dx}\left(\frac{1}{3}\sin(3x)\right) = \frac{1}{3} \cdot \frac{d}{dx}(\sin(3x))$

$= \frac{1}{3} \cdot (3\cos(3x))$

$= \cos(3x)$

The derivative of $\frac{1}{3}\sin(3x)$ is indeed $\cos(3x)$.

Therefore, an antiderivative of $\cos(3x)$ is $\frac{1}{3}\sin(3x)$.

The general antiderivative (or integral) is obtained by adding the constant of integration C.

So, the antiderivative of $\cos(3x)$ is $\frac{1}{3}\sin(3x) + C$.

Solution:

We need to find an antiderivative of $f(x) = e^{2x}$ by the method of inspection.

The method of inspection involves finding a function whose derivative is the given function.

We know that the derivative of $e^x$ is $e^x$. Therefore, we can guess that the antiderivative of $e^{2x}$ might involve $e^{2x}$.

Let's find the derivative of $e^{2x}$:

$\frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x)$

$= e^{2x} \cdot 2$

$= 2e^{2x}$

We want the derivative to be $e^{2x}$, but we got $2e^{2x}$. To adjust this, we need to multiply our initial guess, $e^{2x}$, by a constant such that its derivative is $e^{2x}$. The constant should be $\frac{1}{2}$.

Consider the function $\frac{1}{2}e^{2x}$. Let's find its derivative:

$\frac{d}{dx}\left(\frac{1}{2}e^{2x}\right) = \frac{1}{2} \cdot \frac{d}{dx}(e^{2x})$

$= \frac{1}{2} \cdot (2e^{2x})$

$= e^{2x}$

The derivative of $\frac{1}{2}e^{2x}$ is indeed $e^{2x}$.

Therefore, an antiderivative of $e^{2x}$ is $\frac{1}{2}e^{2x}$.

The general antiderivative (or integral) is obtained by adding the constant of integration C.

So, the antiderivative of $e^{2x}$ is $\frac{1}{2}e^{2x} + C$.

Solution:

We need to find an antiderivative of $f(x) = (ax + b)^2$ by the method of inspection.

The method of inspection involves finding a function whose derivative is the given function.

We know that the derivative of a function of the form $(u(x))^n$ is $n(u(x))^{n-1} \cdot u'(x)$. In our case, $u(x) = ax + b$ and the power is 2. This suggests that the original function might be of the form $(ax+b)^3$.

Let's find the derivative of $(ax + b)^3$ using the chain rule:

$\frac{d}{dx}((ax + b)^3) = 3(ax + b)^{3-1} \cdot \frac{d}{dx}(ax + b)$

$= 3(ax + b)^2 \cdot (a \cdot 1 + 0)$

$= 3(ax + b)^2 \cdot a$

$= 3a(ax + b)^2$

We want the derivative to be $(ax + b)^2$, but we obtained $3a(ax + b)^2$. To get rid of the factor $3a$, we need to multiply our initial guess, $(ax + b)^3$, by $\frac{1}{3a}$ (assuming $a \neq 0$).

Consider the function $\frac{1}{3a}(ax + b)^3$. Let's find its derivative:

$\frac{d}{dx}\left(\frac{1}{3a}(ax + b)^3\right) = \frac{1}{3a} \cdot \frac{d}{dx}((ax + b)^3)$

$= \frac{1}{3a} \cdot (3a(ax + b)^2)$

$= (ax + b)^2$

The derivative of $\frac{1}{3a}(ax + b)^3$ is indeed $(ax + b)^2$.

Therefore, an antiderivative of $(ax + b)^2$ is $\frac{1}{3a}(ax + b)^3$.

The general antiderivative (or integral) is obtained by adding the constant of integration C.

So, the antiderivative of $(ax + b)^2$ is $\frac{1}{3a}(ax + b)^3 + C$.

(Note: If $a=0$, the function is $f(x) = b^2$, which is a constant. The antiderivative of a constant $k$ is $kx + C$. So if $a=0$, $\int b^2 dx = b^2 x + C$. The formula $\frac{1}{3a}(ax + b)^3 + C$ is for $a \neq 0$.)

Solution:

We need to find an antiderivative of $f(x) = \sin(2x) - 4e^{3x}$ by the method of inspection.

The method of inspection involves finding a function whose derivative is the given function. We can find the antiderivative of each term separately.

First, consider the term $\sin(2x)$. We know that the derivative of $\cos(ax)$ is $-a\sin(ax)$.

So, $\frac{d}{dx}(\cos(2x)) = -2\sin(2x)$.

To get $\sin(2x)$, we need to multiply by $-\frac{1}{2}$.

$\frac{d}{dx}\left(-\frac{1}{2}\cos(2x)\right) = -\frac{1}{2} \cdot (-2\sin(2x)) = \sin(2x)$.

So, an antiderivative of $\sin(2x)$ is $-\frac{1}{2}\cos(2x)$.

Next, consider the term $4e^{3x}$. We know that the derivative of $e^{ax}$ is $ae^{ax}$.

So, $\frac{d}{dx}(e^{3x}) = 3e^{3x}$.

We have $4e^{3x}$. We can guess a function of the form $A e^{3x}$ such that its derivative is $4e^{3x}$.

$\frac{d}{dx}(A e^{3x}) = A \cdot 3e^{3x} = 3A e^{3x}$.

We want $3A e^{3x} = 4e^{3x}$, which means $3A = 4$, so $A = \frac{4}{3}$.

Thus, $\frac{d}{dx}\left(\frac{4}{3}e^{3x}\right) = \frac{4}{3} \cdot 3e^{3x} = 4e^{3x}$.

So, an antiderivative of $4e^{3x}$ is $\frac{4}{3}e^{3x}$.

Combining the antiderivatives, an antiderivative of $\sin(2x) - 4e^{3x}$ is the difference of the individual antiderivatives:

$-\frac{1}{2}\cos(2x) - \frac{4}{3}e^{3x}$.

The general antiderivative includes the constant of integration C.

Therefore, the antiderivative of $\sin(2x) - 4e^{3x}$ is $-\frac{1}{2}\cos(2x) - \frac{4}{3}e^{3x} + C$.

Solution:

We need to find the integral $\int (4e^{3x} + 1)\;dx$.

Using the linearity property of integration, we can split the integral into two parts:

$\int (4e^{3x} + 1)\;dx = \int 4e^{3x} \;dx + \int 1 \;dx$

We can take the constant factor out of the integral:

$= 4 \int e^{3x} \;dx + \int dx$

Now, we use the standard integral formulas: $\int e^{ax} \;dx = \frac{1}{a}e^{ax} + C_1$ and $\int dx = x + C_2$.

For the first integral, $a=3$. So, $\int e^{3x} \;dx = \frac{1}{3}e^{3x} + C_1$.

For the second integral, $\int dx = x + C_2$.

Substituting these back into the expression:

$= 4 \left(\frac{1}{3}e^{3x}\right) + x + C$

where $C = 4C_1 + C_2$ is the combined constant of integration.

Simplifying the expression:

$= \frac{4}{3}e^{3x} + x + C$

Solution:

We need to find the integral $\int x^2 (1 − \frac{1}{x^2} ) \;dx$.

First, simplify the integrand by expanding the expression:

$x^2 \left(1 − \frac{1}{x^2}\right) = x^2 \cdot 1 - x^2 \cdot \frac{1}{x^2}$

$= x^2 - \frac{x^2}{x^2}$

$= x^2 - 1$

Now, we integrate the simplified expression:

$\int (x^2 - 1) \;dx$

Using the linearity property of integration, we can integrate each term separately:

$= \int x^2 \;dx - \int 1 \;dx$

We use the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$, and the rule $\int dx = x + C$.

For the first term, $n=2$:

$\int x^2 \;dx = \frac{x^{2+1}}{2+1} + C_1 = \frac{x^3}{3} + C_1$

For the second term:

$\int 1 \;dx = x + C_2$

Combining the results, we get:

$\int (x^2 - 1) \;dx = \left(\frac{x^3}{3} + C_1\right) - (x + C_2)$

$= \frac{x^3}{3} - x + (C_1 - C_2)$

Let $C = C_1 - C_2$ be the constant of integration.

So, the integral is:

$\int x^2 (1 − \frac{1}{x^2} ) \;dx = \frac{x^3}{3} - x + C$

Solution:

We need to find the integral $\int (ax^2 + bx + c) \;dx$.

Using the linearity property of integration, we can integrate each term separately:

$\int (ax^2 + bx + c) \;dx = \int ax^2 \;dx + \int bx \;dx + \int c \;dx$

We can take the constant factors out of the integrals:

$= a \int x^2 \;dx + b \int x \;dx + c \int dx$

Now, we use the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), and the rule $\int dx = x + C$.

For the first term, $n=2$:

$\int x^2 \;dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$ (plus constant)

For the second term, $n=1$:

$\int x \;dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2}$ (plus constant)

For the third term, which is a constant integral:

$\int c \;dx = cx$ (plus constant)

Combining the results and adding a single constant of integration C:

$= a \left(\frac{x^3}{3}\right) + b \left(\frac{x^2}{2}\right) + c(x) + C$

So, the integral is:

$\int (ax^2 + bx + c) \;dx = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + C$

Solution:

We need to find the integral $\int (2x^2 + e^x) \; dx$.

Using the linearity property of integration, we can split the integral into two parts:

$\int (2x^2 + e^x) \; dx = \int 2x^2 \; dx + \int e^x \; dx$

We can take the constant factor out of the first integral:

$= 2 \int x^2 \; dx + \int e^x \; dx$

Now, we use the standard integral formulas: the power rule $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), and $\int e^x \; dx = e^x + C$.

For the first integral, $n=2$. So, $\int x^2 \; dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$ (plus constant).

For the second integral, $\int e^x \; dx = e^x$ (plus constant).

Combining the results and adding a single constant of integration C:

$= 2 \left(\frac{x^3}{3}\right) + e^x + C$

So, the integral is:

$\int (2x^2 + e^x) \; dx = \frac{2}{3}x^3 + e^x + C$

Solution:

We need to find the integral $\int \left( \sqrt{x} − \frac{1}{\sqrt{x}} \right)^2\;dx$.

First, we expand the expression inside the integral:

$\left( \sqrt{x} − \frac{1}{\sqrt{x}} \right)^2 = (\sqrt{x})^2 - 2 \cdot \sqrt{x} \cdot \frac{1}{\sqrt{x}} + \left(\frac{1}{\sqrt{x}}\right)^2$

$= x - 2 \cdot 1 + \frac{1}{x}$

$= x - 2 + \frac{1}{x}$

Now, we integrate the expanded expression:

$\int \left( x - 2 + \frac{1}{x} \right) \;dx$

Using the linearity property of integration, we integrate each term separately:

$= \int x \;dx - \int 2 \;dx + \int \frac{1}{x} \;dx$

We use the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), the integral of a constant, $\int c \;dx = cx + C$, and the integral of $\frac{1}{x}$, $\int \frac{1}{x} \;dx = \log|x| + C$.

For the first term ($\int x \;dx$), $n=1$: $\frac{x^{1+1}}{1+1} = \frac{x^2}{2}$.

For the second term ($\int 2 \;dx$): $2x$.

For the third term ($\int \frac{1}{x} \;dx$): $\log|x|$.

Combining these results and adding the constant of integration C:

$\int \left( \sqrt{x} − \frac{1}{\sqrt{x}} \right)^2\;dx = \frac{x^2}{2} - 2x + \log|x| + C$

Solution:

We need to find the integral $\int \frac{x^3 + 5x^2 − 4}{x^2} \; dx$.

First, we simplify the integrand by dividing each term in the numerator by the denominator:

$\frac{x^3 + 5x^2 − 4}{x^2} = \frac{x^3}{x^2} + \frac{5x^2}{x^2} - \frac{4}{x^2}$

$= x + 5 - 4x^{-2}$

Now, we integrate the simplified expression:

$\int (x + 5 - 4x^{-2}) \; dx$

Using the linearity property of integration, we integrate each term separately:

$= \int x \; dx + \int 5 \; dx - \int 4x^{-2} \; dx$

$= \int x^1 \; dx + \int 5 \; dx - 4 \int x^{-2} \; dx$

We use the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), and the integral of a constant, $\int c \;dx = cx + C$.

For $\int x^1 \; dx$ ($n=1$): $\frac{x^{1+1}}{1+1} = \frac{x^2}{2}$.

For $\int 5 \; dx$: $5x$.

For $\int x^{-2} \; dx$ ($n=-2$): $\frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -x^{-1} = -\frac{1}{x}$.

Combining these results with the constant factors and adding the constant of integration C:

$= \frac{x^2}{2} + 5x - 4\left(-\frac{1}{x}\right) + C$

$= \frac{x^2}{2} + 5x + \frac{4}{x} + C$

So, the integral is:

$\int \frac{x^3 + 5x^2 − 4}{x^2} \; dx = \frac{x^2}{2} + 5x + \frac{4}{x} + C$

Solution:

We need to find the integral $\int \frac{x^3 + 3x + 4}{\sqrt{x}}\; dx$.

First, we simplify the integrand by dividing each term in the numerator by $\sqrt{x} = x^{1/2}$:

$\frac{x^3 + 3x + 4}{\sqrt{x}} = \frac{x^3}{x^{1/2}} + \frac{3x}{x^{1/2}} + \frac{4}{x^{1/2}}$

Using the rule $\frac{x^a}{x^b} = x^{a-b}$:

$= x^{3 - 1/2} + 3x^{1 - 1/2} + 4x^{0 - 1/2}$

$= x^{6/2 - 1/2} + 3x^{2/2 - 1/2} + 4x^{-1/2}$

$= x^{5/2} + 3x^{1/2} + 4x^{-1/2}$

Now, we integrate the simplified expression term by term:

$\int \left( x^{5/2} + 3x^{1/2} + 4x^{-1/2} \right) \; dx$

Using the linearity property and the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

$= \int x^{5/2} \; dx + \int 3x^{1/2} \; dx + \int 4x^{-1/2} \; dx$

$= \int x^{5/2} \; dx + 3 \int x^{1/2} \; dx + 4 \int x^{-1/2} \; dx$

For $\int x^{5/2} \; dx$ ($n=5/2$): $\frac{x^{5/2+1}}{5/2+1} = \frac{x^{7/2}}{7/2} = \frac{2}{7}x^{7/2}$.

For $\int x^{1/2} \; dx$ ($n=1/2$): $\frac{x^{1/2+1}}{1/2+1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$.

For $\int x^{-1/2} \; dx$ ($n=-1/2$): $\frac{x^{-1/2+1}}{-1/2+1} = \frac{x^{1/2}}{1/2} = 2x^{1/2}$.

Combining these results with the constant factors and adding the constant of integration C:

$= \frac{2}{7}x^{7/2} + 3 \left(\frac{2}{3}x^{3/2}\right) + 4 \left(2x^{1/2}\right) + C$

$= \frac{2}{7}x^{7/2} + 2x^{3/2} + 8x^{1/2} + C$

So, the integral is:

$\int \frac{x^3 + 3x + 4}{\sqrt{x}}\; dx = \frac{2}{7}x^{7/2} + 2x^{3/2} + 8\sqrt{x} + C$

Solution:

We need to find the integral $\int \frac{x^3 − x^2 + x − 1}{x − 1}\; dx$.

First, we simplify the integrand by factoring the numerator. We can group terms in the numerator:

$x^3 − x^2 + x − 1 = (x^3 − x^2) + (x − 1)$

$= x^2(x − 1) + 1(x − 1)$

$= (x^2 + 1)(x − 1)$

So the integrand becomes:

$\frac{x^3 − x^2 + x − 1}{x − 1} = \frac{(x^2 + 1)(x − 1)}{x − 1}$

For $x \neq 1$, we can cancel the $(x-1)$ term:

$= x^2 + 1$

Now, we integrate the simplified expression:

$\int (x^2 + 1) \; dx$

Using the linearity property of integration, we integrate each term separately:

$= \int x^2 \; dx + \int 1 \; dx$

We use the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), and the rule for the integral of a constant, $\int c \;dx = cx + C$.

For $\int x^2 \; dx$ ($n=2$): $\frac{x^{2+1}}{2+1} = \frac{x^3}{3}$ (plus constant).

For $\int 1 \; dx$: $x$ (plus constant).

Combining these results and adding a single constant of integration C:

$= \frac{x^3}{3} + x + C$

So, the integral is:

$\int \frac{x^3 − x^2 + x − 1}{x − 1}\; dx = \frac{x^3}{3} + x + C$

Solution:

We need to find the integral $\int (1 − x) \sqrt{x}\; dx$.

First, we simplify the integrand by multiplying $(1-x)$ by $\sqrt{x} = x^{1/2}$:

$(1 − x) \sqrt{x} = 1 \cdot x^{1/2} - x \cdot x^{1/2}$

Using the rule $x^a \cdot x^b = x^{a+b}$:

$= x^{1/2} - x^{1 + 1/2}$

$= x^{1/2} - x^{3/2}$

Now, we integrate the simplified expression term by term:

$\int \left( x^{1/2} - x^{3/2} \right) \; dx$

Using the linearity property and the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

$= \int x^{1/2} \; dx - \int x^{3/2} \; dx$

For $\int x^{1/2} \; dx$ ($n=1/2$): $\frac{x^{1/2+1}}{1/2+1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$.

For $\int x^{3/2} \; dx$ ($n=3/2$): $\frac{x^{3/2+1}}{3/2+1} = \frac{x^{5/2}}{5/2} = \frac{2}{5}x^{5/2}$.

Combining these results and adding the constant of integration C:

$= \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C$

So, the integral is:

$\int (1 − x) \sqrt{x}\; dx = \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C$

Solution:

We need to find the integral $\int \sqrt{x}(3x^2 + 2x + 3)\; dx$.

First, we simplify the integrand by multiplying $(3x^2 + 2x + 3)$ by $\sqrt{x} = x^{1/2}$:

$\sqrt{x}(3x^2 + 2x + 3) = x^{1/2}(3x^2) + x^{1/2}(2x) + x^{1/2}(3)$

Using the rule $x^a \cdot x^b = x^{a+b}$:

$= 3x^{1/2 + 2} + 2x^{1/2 + 1} + 3x^{1/2}$

$= 3x^{1/2 + 4/2} + 2x^{1/2 + 2/2} + 3x^{1/2}$

$= 3x^{5/2} + 2x^{3/2} + 3x^{1/2}$

Now, we integrate the simplified expression term by term:

$\int \left( 3x^{5/2} + 2x^{3/2} + 3x^{1/2} \right) \; dx$

Using the linearity property and the power rule for integration, $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

$= \int 3x^{5/2} \; dx + \int 2x^{3/2} \; dx + \int 3x^{1/2} \; dx$

$= 3 \int x^{5/2} \; dx + 2 \int x^{3/2} \; dx + 3 \int x^{1/2} \; dx$

For $\int x^{5/2} \; dx$ ($n=5/2$): $\frac{x^{5/2+1}}{5/2+1} = \frac{x^{7/2}}{7/2} = \frac{2}{7}x^{7/2}$.

For $\int x^{3/2} \; dx$ ($n=3/2$): $\frac{x^{3/2+1}}{3/2+1} = \frac{x^{5/2}}{5/2} = \frac{2}{5}x^{5/2}$.

For $\int x^{1/2} \; dx$ ($n=1/2$): $\frac{x^{1/2+1}}{1/2+1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$.

Combining these results with the constant factors and adding the constant of integration C:

$= 3 \left(\frac{2}{7}x^{7/2}\right) + 2 \left(\frac{2}{5}x^{5/2}\right) + 3 \left(\frac{2}{3}x^{3/2}\right) + C$

$= \frac{6}{7}x^{7/2} + \frac{4}{5}x^{5/2} + 2x^{3/2} + C$

So, the integral is:

$\int \sqrt{x}(3x^2 + 2x + 3)\; dx = \frac{6}{7}x^{7/2} + \frac{4}{5}x^{5/2} + 2x^{3/2} + C$

Solution:

We need to find the integral $\int (2x − 3 \cos x + e^x)\; dx$.

Using the linearity property of integration, we can split the integral into separate integrals for each term:

$\int (2x − 3 \cos x + e^x)\; dx = \int 2x \; dx - \int 3 \cos x \; dx + \int e^x \; dx$

Take the constant factors out of the first two integrals:

$= 2 \int x \; dx - 3 \int \cos x \; dx + \int e^x \; dx$

Now, use the standard integral formulas: $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$), $\int \cos x \; dx = \sin x + C$, and $\int e^x \; dx = e^x + C$.

For $\int x \; dx$ ($n=1$): $\frac{x^{1+1}}{1+1} = \frac{x^2}{2}$ (plus constant).

For $\int \cos x \; dx$: $\sin x$ (plus constant).

For $\int e^x \; dx$: $e^x$ (plus constant).

Combine these results with the constant factors and add a single constant of integration C:

$= 2 \left(\frac{x^2}{2}\right) - 3 (\sin x) + e^x + C$

Simplify the expression:

$= x^2 - 3 \sin x + e^x + C$

So, the integral is:

$\int (2x − 3 \cos x + e^x)\; dx = x^2 - 3 \sin x + e^x + C$

Solution:

We need to find the integral $\int (2x^2 − 3 \sin x + 5 \sqrt{x}) \;dx$.

First, rewrite $\sqrt{x}$ as $x^{1/2}$:

$\int (2x^2 − 3 \sin x + 5 x^{1/2}) \;dx$

Using the linearity property of integration, we split the integral into separate integrals for each term:

$= \int 2x^2 \; dx - \int 3 \sin x \; dx + \int 5 x^{1/2} \; dx$

Take the constant factors out of each integral:

$= 2 \int x^2 \; dx - 3 \int \sin x \; dx + 5 \int x^{1/2} \; dx$

Now, use the standard integral formulas: $\int x^n \;dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$) and $\int \sin x \; dx = -\cos x + C$.

For $\int x^2 \; dx$ ($n=2$): $\frac{x^{2+1}}{2+1} = \frac{x^3}{3}$ (plus constant).

For $\int \sin x \; dx$: $-\cos x$ (plus constant).

For $\int x^{1/2} \; dx$ ($n=1/2$): $\frac{x^{1/2+1}}{1/2+1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$ (plus constant).

Combine these results with the constant factors and add a single constant of integration C:

$= 2 \left(\frac{x^3}{3}\right) - 3 (-\cos x) + 5 \left(\frac{2}{3}x^{3/2}\right) + C$

Simplify the expression:

$= \frac{2}{3}x^3 + 3 \cos x + \frac{10}{3}x^{3/2} + C$

So, the integral is:

$\int (2x^2 − 3 \sin x + 5 \sqrt{x}) \;dx = \frac{2}{3}x^3 + 3 \cos x + \frac{10}{3}x^{3/2} + C$

Solution:

We need to find the integral $\int \sec x (\sec x + \tan x)\; dx$.

First, we simplify the integrand by multiplying $\sec x$ through the parenthesis:

$\sec x (\sec x + \tan x) = \sec^2 x + \sec x \tan x$

Now, we integrate the simplified expression:

$\int (\sec^2 x + \sec x \tan x)\; dx$

Using the linearity property of integration, we can split the integral into separate integrals for each term:

$= \int \sec^2 x \; dx + \int \sec x \tan x \; dx$

Now, we use the standard integral formulas: $\int \sec^2 x \; dx = \tan x + C_1$ and $\int \sec x \tan x \; dx = \sec x + C_2$.

Combine these results and add a single constant of integration C ($C = C_1 + C_2$):

$= \tan x + \sec x + C$

So, the integral is:

$\int \sec x (\sec x + \tan x)\; dx = \tan x + \sec x + C$

Solution:

We need to find the integral $\int \frac{\sec^2 x}{\text{cosec}^2 \;x}\; dx$.

First, we simplify the integrand using the identities $\sec x = \frac{1}{\cos x}$ and $\text{cosec}\; x = \frac{1}{\sin x}$:

$\frac{\sec^2 x}{\text{cosec}^2 \;x} = \frac{\frac{1}{\cos^2 x}}{\frac{1}{\sin^2 x}}$

$= \frac{1}{\cos^2 x} \cdot \frac{\sin^2 x}{1}$

$= \frac{\sin^2 x}{\cos^2 x}$

Using the identity $\tan x = \frac{\sin x}{\cos x}$:

$= \tan^2 x$

Now, we need to integrate $\tan^2 x$. We use the trigonometric identity $\tan^2 x + 1 = \sec^2 x$, which implies $\tan^2 x = \sec^2 x - 1$.

So the integral becomes:

$\int (\sec^2 x - 1) \; dx$

Using the linearity property of integration, we integrate each term separately:

$= \int \sec^2 x \; dx - \int 1 \; dx$

Now, use the standard integral formulas: $\int \sec^2 x \; dx = \tan x + C_1$ and $\int 1 \; dx = x + C_2$.

Combine these results and add a single constant of integration C ($C = C_1 - C_2$):

$= \tan x - x + C$

So, the integral is:

$\int \frac{\sec^2 x}{\text{cosec}^2 \;x}\; dx = \tan x - x + C$

Solution:

We need to find the integral $\int \frac{2 − 3 \sin x}{\cos^2 x}\; dx$.

First, we simplify the integrand by splitting the fraction:

$\frac{2 − 3 \sin x}{\cos^2 x} = \frac{2}{\cos^2 x} - \frac{3 \sin x}{\cos^2 x}$

Using trigonometric identities, $\frac{1}{\cos^2 x} = \sec^2 x$ and $\frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x$:

$= 2 \sec^2 x - 3 \frac{\sin x}{\cos x} \frac{1}{\cos x}$

$= 2 \sec^2 x - 3 \tan x \sec x$

Now, we integrate the simplified expression:

$\int (2 \sec^2 x - 3 \sec x \tan x)\; dx$

Using the linearity property of integration, we split the integral into separate integrals for each term:

$= \int 2 \sec^2 x \; dx - \int 3 \sec x \tan x \; dx$

Take the constant factors out of each integral:

$= 2 \int \sec^2 x \; dx - 3 \int \sec x \tan x \; dx$

Now, use the standard integral formulas: $\int \sec^2 x \; dx = \tan x + C_1$ and $\int \sec x \tan x \; dx = \sec x + C_2$.

Combine these results with the constant factors and add a single constant of integration C ($C = 2C_1 - 3C_2$):

$= 2 (\tan x) - 3 (\sec x) + C$

$= 2 \tan x - 3 \sec x + C$

So, the integral is:

$\int \frac{2 − 3 \sin x}{\cos^2 x}\; dx = 2 \tan x - 3 \sec x + C$

Solution:

We need to find the antiderivative of the function $f(x) = \sqrt{x} + \frac{1}{\sqrt{x}}$.

The antiderivative is given by the indefinite integral $\int f(x) \; dx$.

Rewrite the function using exponents: $f(x) = x^{1/2} + x^{-1/2}$.

Now, integrate the function:

$\int \left( x^{1/2} + x^{-1/2} \right) \; dx$

Using the linearity of the integral, we integrate term by term:

$= \int x^{1/2} \; dx + \int x^{-1/2} \; dx$

Apply the power rule for integration, $\int x^n \; dx = \frac{x^{n+1}}{n+1} + C$, for $n \neq -1$:

For the first term, $n = 1/2$:

$\int x^{1/2} \; dx = \frac{x^{1/2 + 1}}{1/2 + 1} + C_1 = \frac{x^{3/2}}{3/2} + C_1 = \frac{2}{3}x^{3/2} + C_1$

For the second term, $n = -1/2$:

$\int x^{-1/2} \; dx = \frac{x^{-1/2 + 1}}{-1/2 + 1} + C_2 = \frac{x^{1/2}}{1/2} + C_2 = 2x^{1/2} + C_2$

Combining the results and letting $C = C_1 + C_2$ be the constant of integration:

$\int \left( x^{1/2} + x^{-1/2} \right) \; dx = \frac{2}{3}x^{3/2} + 2x^{1/2} + C$

Comparing this result with the given options, we see that it matches option (C).

The correct answer is (C) $\frac{2}{3} x^{\frac{3}{2}} + 2x^{\frac{1}{2}} + C$.

Solution:

We are given that the derivative of a function $f(x)$ is $\frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}$.

To find the function $f(x)$, we need to find the antiderivative (integral) of the given derivative:

$f(x) = \int \left( 4x^3 - \frac{3}{x^4} \right) \; dx$

Rewrite the term $\frac{3}{x^4}$ as $3x^{-4}$:

$f(x) = \int (4x^3 - 3x^{-4}) \; dx$

Using the linearity property of integration, we integrate each term separately:

$f(x) = \int 4x^3 \; dx - \int 3x^{-4} \; dx$

Take the constant factors out of the integrals:

$f(x) = 4 \int x^3 \; dx - 3 \int x^{-4} \; dx$

Apply the power rule for integration, $\int x^n \; dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$):

For $\int x^3 \; dx$ ($n=3$): $\frac{x^{3+1}}{3+1} = \frac{x^4}{4}$ (plus constant).

For $\int x^{-4} \; dx$ ($n=-4$): $\frac{x^{-4+1}}{-4+1} = \frac{x^{-3}}{-3}$ (plus constant).

Combine these results with the constant factors and add a single constant of integration C:

$f(x) = 4 \left(\frac{x^4}{4}\right) - 3 \left(\frac{x^{-3}}{-3}\right) + C$

Simplify the expression:

$f(x) = x^4 - (-\frac{3}{3}x^{-3}) + C$

$f(x) = x^4 - (-x^{-3}) + C$

$f(x) = x^4 + x^{-3} + C$

We can write $x^{-3}$ as $\frac{1}{x^3}$:

$f(x) = x^4 + \frac{1}{x^3} + C$

We are given the condition that $f(2) = 0$. Use this to find the value of C.

Substitute $x=2$ into the expression for $f(x)$:

$f(2) = (2)^4 + \frac{1}{(2)^3} + C$

Since $f(2) = 0$, we have:

$0 = 16 + \frac{1}{8} + C$

To find C, isolate C:

$C = -16 - \frac{1}{8}$

Find a common denominator:

$C = -\frac{16 \cdot 8}{8} - \frac{1}{8}$

$C = -\frac{128}{8} - \frac{1}{8}$

$C = -\frac{128 + 1}{8}$

$C = -\frac{129}{8}$

Substitute the value of C back into the expression for $f(x)$:

$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$

Compare this result with the given options. It matches option (A).

The correct answer is (A) $x^4 + \frac{1}{x^3} - \frac{129}{8}$.

Solution:

(i) We need to integrate $\sin(mx)$ with respect to x, i.e., find $\int \sin(mx) \; dx$.

We use the method of substitution. Let $u = mx$.

Differentiating both sides with respect to x, we get $\frac{du}{dx} = m$.

This implies $du = m \; dx$, or $dx = \frac{1}{m} du$.

Substitute $u = mx$ and $dx = \frac{1}{m} du$ into the integral:

$\int \sin(mx) \; dx = \int \sin(u) \cdot \frac{1}{m} \; du$

We can take the constant $\frac{1}{m}$ out of the integral:

$= \frac{1}{m} \int \sin(u) \; du$

We know the standard integral $\int \sin(u) \; du = -\cos(u) + C'$, where $C'$ is the constant of integration.

$= \frac{1}{m} (-\cos(u)) + C$

$= -\frac{1}{m} \cos(u) + C$

Now, substitute back $u = mx$:

$= -\frac{1}{m} \cos(mx) + C$

Thus, $\int \sin(mx) \; dx = -\frac{1}{m} \cos(mx) + C$.

(ii) We need to integrate $2x \sin(x^2 + 1)$ with respect to x, i.e., find $\int 2x \sin(x^2 + 1) \; dx$.

We use the method of substitution. Let $u = x^2 + 1$.

Differentiating both sides with respect to x, we get $\frac{du}{dx} = \frac{d}{dx}(x^2 + 1) = 2x$.

This implies $du = 2x \; dx$.

Substitute $u = x^2 + 1$ and $du = 2x \; dx$ into the integral:

$\int \sin(x^2 + 1) \cdot (2x \; dx) = \int \sin(u) \; du$

We know the standard integral $\int \sin(u) \; du = -\cos(u) + C$, where $C$ is the constant of integration.

$= -\cos(u) + C$

Now, substitute back $u = x^2 + 1$:

$= -\cos(x^2 + 1) + C$

Thus, $\int 2x \sin(x^2 + 1) \; dx = -\cos(x^2 + 1) + C$.

(iii) We need to integrate $\frac{\tan^4 \sqrt{x} \sec^2 \sqrt{x}}{\sqrt{x}}$ with respect to x, i.e., find $\int \frac{\tan^4 \sqrt{x} \sec^2 \sqrt{x}}{\sqrt{x}} \; dx$.

We use the method of substitution. Let $v = \tan(\sqrt{x})$.

Differentiating both sides with respect to x, we use the chain rule:

$\frac{dv}{dx} = \frac{d}{dx}(\tan(\sqrt{x})) = \sec^2(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x})$

We know that $\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

So, $\frac{dv}{dx} = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\sec^2 \sqrt{x}}{2\sqrt{x}}$.

This implies $dv = \frac{\sec^2 \sqrt{x}}{2\sqrt{x}} \; dx$, or $2 \; dv = \frac{\sec^2 \sqrt{x}}{\sqrt{x}} \; dx$.

Rewrite the integral as $\int (\tan(\sqrt{x}))^4 \cdot \frac{\sec^2 \sqrt{x}}{\sqrt{x}} \; dx$.

Substitute $v = \tan(\sqrt{x})$ and $2 \; dv = \frac{\sec^2 \sqrt{x}}{\sqrt{x}} \; dx$ into the integral:

$\int v^4 \cdot (2 \; dv) = 2 \int v^4 \; dv$

Apply the power rule for integration, $\int v^n \; dv = \frac{v^{n+1}}{n+1} + C'$:

$= 2 \cdot \frac{v^{4+1}}{4+1} + C$

$= 2 \cdot \frac{v^5}{5} + C$

$= \frac{2}{5} v^5 + C$

Now, substitute back $v = \tan(\sqrt{x})$:

$= \frac{2}{5} (\tan(\sqrt{x}))^5 + C$

$= \frac{2}{5} \tan^5 \sqrt{x} + C$

Thus, $\int \frac{\tan^4 \sqrt{x} \sec^2 \sqrt{x}}{\sqrt{x}} \; dx = \frac{2}{5} \tan^5 \sqrt{x} + C$.

(iv) We need to integrate $\frac{\sin (\tan^{−1} x)}{1 + x^2}$ with respect to x, i.e., find $\int \frac{\sin (\tan^{−1} x)}{1 + x^2} \; dx$.

We use the method of substitution. Let $u = \tan^{-1} x$.

Differentiating both sides with respect to x, we get $\frac{du}{dx} = \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2}$.

This implies $du = \frac{1}{1 + x^2} \; dx$.

Rewrite the integral as $\int \sin(\tan^{-1} x) \cdot \frac{1}{1 + x^2} \; dx$.

Substitute $u = \tan^{-1} x$ and $du = \frac{1}{1 + x^2} \; dx$ into the integral:

$\int \sin(u) \; du$

We know the standard integral $\int \sin(u) \; du = -\cos(u) + C$, where $C$ is the constant of integration.

$= -\cos(u) + C$

Now, substitute back $u = \tan^{-1} x$:

$= -\cos(\tan^{-1} x) + C$

Thus, $\int \frac{\sin (\tan^{−1} x)}{1 + x^2} \; dx = -\cos(\tan^{-1} x) + C$.

Solution:

(i) We need to find $\int \sin^3 x \cos^2 x \;dx$.

We can rewrite $\sin^3 x$ as $\sin^2 x \sin x$. Using the identity $\sin^2 x = 1 - \cos^2 x$, the integral becomes:

$\int (1 - \cos^2 x) \cos^2 x \sin x \;dx$

Let $u = \cos x$. Then $du = -\sin x \; dx$. So, $\sin x \; dx = -du$.

Substitute $u$ and $du$ into the integral:

$\int (1 - u^2) u^2 (-du)$

$= -\int (u^2 - u^4) \;du$

$= \int (u^4 - u^2) \;du$

Now, integrate with respect to $u$ using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + C'$:

$= \frac{u^{4+1}}{4+1} - \frac{u^{2+1}}{2+1} + C$

$= \frac{u^5}{5} - \frac{u^3}{3} + C$

Substitute back $u = \cos x$:

$= \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$

Thus, $\int \sin^3 x \cos^2 x \;dx = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C$.

(ii) We need to find $\int \frac{\sin x}{\sin (x + a)} \;dx$.

Let $u = x + a$. Then $du = dx$. Also, $x = u - a$, so $\sin x = \sin(u - a)$.

Substitute $u$ and $dx = du$ into the integral:

$\int \frac{\sin (u - a)}{\sin u} \;du$

Use the trigonometric identity for $\sin(A - B) = \sin A \cos B - \cos A \sin B$:

$\sin(u - a) = \sin u \cos a - \cos u \sin a$

So the integral becomes:

$\int \frac{\sin u \cos a - \cos u \sin a}{\sin u} \;du$

Split the fraction into two terms:

$= \int \left( \frac{\sin u \cos a}{\sin u} - \frac{\cos u \sin a}{\sin u} \right) \;du$

For $u$ where $\sin u \neq 0$, this simplifies to:

$= \int (\cos a - \cot u \sin a) \;du$

Since $a$ is a constant, $\cos a$ and $\sin a$ are constants. Use linearity of integration:

$= \int \cos a \;du - \int \cot u \sin a \;du$

$= \cos a \int du - \sin a \int \cot u \;du$

We know $\int du = u + C_1$ and $\int \cot u \;du = \log |\sin u| + C_2$.

$= \cos a (u) - \sin a (\log |\sin u|) + C$

Substitute back $u = x + a$:

$= (x + a) \cos a - \sin a \log |\sin (x + a)| + C$

Thus, $\int \frac{\sin x}{\sin (x + a)} \;dx = (x + a) \cos a - \sin a \log |\sin (x + a)| + C$.

(iii) We need to find $\int \frac{1}{1 + \tan x} \;dx$.

Rewrite $\tan x$ in terms of $\sin x$ and $\cos x$:

$\int \frac{1}{1 + \frac{\sin x}{\cos x}} \;dx = \int \frac{1}{\frac{\cos x + \sin x}{\cos x}} \;dx = \int \frac{\cos x}{\cos x + \sin x} \;dx$

We can write the numerator $\cos x$ as a linear combination of the denominator $(\cos x + \sin x)$ and its derivative $(\cos x - \sin x)$.

Let $\cos x = A(\cos x + \sin x) + B(\cos x - \sin x)$.

$\cos x = (A+B)\cos x + (A-B)\sin x$

Comparing coefficients of $\cos x$ and $\sin x$ on both sides:

$A + B = 1$

$A - B = 0$

Adding the two equations, $2A = 1 \implies A = \frac{1}{2}$.

Substituting $A = \frac{1}{2}$ into $A - B = 0$, we get $\frac{1}{2} - B = 0 \implies B = \frac{1}{2}$.

So, $\cos x = \frac{1}{2}(\cos x + \sin x) + \frac{1}{2}(\cos x - \sin x)$.

Substitute this back into the integrand:

$\frac{\cos x}{\cos x + \sin x} = \frac{\frac{1}{2}(\cos x + \sin x) + \frac{1}{2}(\cos x - \sin x)}{\cos x + \sin x}$

$= \frac{1}{2} \frac{\cos x + \sin x}{\cos x + \sin x} + \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x}$

$= \frac{1}{2} + \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x}$

Now integrate term by term:

$\int \left( \frac{1}{2} + \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x} \right) \;dx$

$= \int \frac{1}{2} \;dx + \int \frac{1}{2} \frac{\cos x - \sin x}{\cos x + \sin x} \;dx$

$= \frac{1}{2} \int dx + \frac{1}{2} \int \frac{\cos x - \sin x}{\cos x + \sin x} \;dx$

The first integral is $\frac{1}{2}x$. For the second integral, let $v = \cos x + \sin x$. Then $dv = (-\sin x + \cos x) \;dx = (\cos x - \sin x) \;dx$.

The second integral becomes $\int \frac{1}{2} \frac{1}{v} \;dv = \frac{1}{2} \int \frac{1}{v} \;dv = \frac{1}{2} \log |v| + C_1$.

Substitute back $v = \cos x + \sin x$:

$= \frac{1}{2} \log |\cos x + \sin x| + C_1$

Combining the integrals and adding the constant of integration C:

$= \frac{1}{2}x + \frac{1}{2} \log |\cos x + \sin x| + C$

Thus, $\int \frac{1}{1 + \tan x} \;dx = \frac{1}{2}x + \frac{1}{2} \log |\cos x + \sin x| + C$.

Let the given integral be $I$.

$I = \int \frac{2x}{1 + x^2} dx$

We use the method of substitution.

Let $u = 1 + x^2$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(1 + x^2)$

$\frac{du}{dx} = 0 + 2x$

$\frac{du}{dx} = 2x$

Rearranging the terms, we get:

$du = 2x \, dx$

Now, substitute $u = 1 + x^2$ and $du = 2x \, dx$ into the integral $I$:

$I = \int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log_e |u| + C$, where $C$ is the constant of integration.

$I = \log_e |u| + C$

Substitute back $u = 1 + x^2$:

$I = \log_e |1 + x^2| + C$

Since $x^2 \geq 0$ for any real number $x$, $1 + x^2 \geq 1$. Therefore, $1 + x^2$ is always positive, and the absolute value is not necessary.

$I = \log_e (1 + x^2) + C$

Thus, the integral of the given function is $\log_e (1 + x^2) + C$.

To evaluate the integral $\int \frac{(\log x)^2}{x} dx$, we can use the method of substitution.

Let $u = \log |x|$.

Differentiating $u$ with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\log |x|)$

$\frac{du}{dx} = \frac{1}{|x|} \times \text{sign}(x)$

$\frac{du}{dx} = \frac{1}{x}$ (for $x \neq 0$)

This means $du = \frac{1}{x} dx$.

Now, substitute $u = \log |x|$ and $du = \frac{1}{x} dx$ into the integral $\int \frac{(\log x)^2}{x} dx$. Note that $(\log x)^2 = (\log |x|)^2$ for $x \neq 0$. So the integral becomes:

$\int u^2 du$

Now, we integrate this expression with respect to $u$ using the power rule for integration, which states that $\int y^n dy = \frac{y^{n+1}}{n+1} + C$ (for $n \neq -1$).

$\int u^2 du = \frac{u^{2+1}}{2+1} + C$

$\int u^2 du = \frac{u^3}{3} + C$

Finally, substitute back $u = \log |x|$ to express the result in terms of $x$.

The integral is equal to:

$\frac{(\log |x|)^3}{3} + C$

Where $C$ is the constant of integration.

So, the solution is:

$\int \frac{(\log x)^2}{x} dx = \frac{1}{3} (\log |x|)^3 + C$

Let the given integral be $I$.

$I = \int \frac{1}{x + x \log x} dx$

First, simplify the denominator by factoring out $x$:

$x + x \log x = x(1 + \log x)$

So the integral becomes:

$I = \int \frac{1}{x(1 + \log x)} dx$

$I = \int \frac{1}{(1 + \log x)} \cdot \frac{1}{x} dx$

We use the method of substitution.

Let $u = 1 + \log x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(1 + \log x)$

$\frac{du}{dx} = 0 + \frac{1}{x}$

$\frac{du}{dx} = \frac{1}{x}$

Rearranging the terms, we get:

$du = \frac{1}{x} dx$

Now, substitute $u = 1 + \log x$ and $du = \frac{1}{x} dx$ into the integral $I$:

$I = \int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log_e |u| + C$, where $C$ is the constant of integration.

$I = \log_e |u| + C$

Substitute back $u = 1 + \log x$:

$I = \log_e |1 + \log x| + C$

Thus, the integral of the given function is $\log_e |1 + \log x| + C$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \sin x \;\sin \;(\cos x) \;dx$

We use the method of substitution.

Let $u = \cos x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\cos x)$

$\frac{du}{dx} = -\sin x$

Rearranging the terms, we get:

$du = -\sin x \;dx$

So, $\sin x \;dx = -du$.

Now, substitute $u = \cos x$ and $\sin x \;dx = -du$ into the integral $I$:

$I = \int \sin(u) (-du)$

$I = - \int \sin(u) \;du$

The integral of $\sin(u)$ with respect to $u$ is $-\cos(u) + C$, where $C$ is the constant of integration.

$I = - (-\cos(u)) + C$

$I = \cos(u) + C$

Substitute back $u = \cos x$:

$I = \cos(\cos x) + C$

Thus, the integral of the given function is $\cos(\cos x) + C$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \sin(ax + b) \; \cos(ax + b) \;dx$

We can use the trigonometric identity $\sin(2\theta) = 2 \sin\theta \cos\theta$.

Let $\theta = ax + b$. Then the integrand is $\frac{1}{2} \sin(2(ax + b)) = \frac{1}{2} \sin(2ax + 2b)$.

The integral becomes:

$I = \int \frac{1}{2} \sin(2ax + 2b) \;dx$

$I = \frac{1}{2} \int \sin(2ax + 2b) \;dx$

Now, we integrate $\sin(2ax + 2b)$ with respect to $x$. We can use substitution for the argument of sine, let $u = 2ax + 2b$. Then $du = \frac{d}{dx}(2ax + 2b) \;dx = 2a \;dx$. This implies $dx = \frac{du}{2a}$.

So the integral $\int \sin(2ax + 2b) \;dx$ becomes $\int \sin(u) \frac{du}{2a} = \frac{1}{2a} \int \sin(u) \;du$.

The integral of $\sin(u)$ is $-\cos(u)$.

$\frac{1}{2a} \int \sin(u) \;du = \frac{1}{2a} (-\cos(u)) + C' = -\frac{1}{2a} \cos(u) + C'$.

Substitute back $u = 2ax + 2b$:

$-\frac{1}{2a} \cos(2ax + 2b) + C'$.

Now, substitute this back into the expression for $I$:

$I = \frac{1}{2} \left( -\frac{1}{2a} \cos(2ax + 2b) \right) + C$

$I = -\frac{1}{4a} \cos(2ax + 2b) + C$, where $C = \frac{C'}{2}$ is the constant of integration.

Thus, the integral of the given function is $\mathbf{-\frac{1}{4a} \cos(2ax + 2b) + C}$.

Alternate Solution (using substitution):

Let the given integral be $I$.

$I = \int \sin(ax + b) \; \cos(ax + b) \;dx$

We use the method of substitution.

Let $u = \sin(ax + b)$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\sin(ax + b))$

Using the chain rule, $\frac{du}{dx} = \cos(ax + b) \cdot \frac{d}{dx}(ax + b) = \cos(ax + b) \cdot a$

$\frac{du}{dx} = a \cos(ax + b)$

Rearranging the terms, we get:

$du = a \cos(ax + b) \;dx$

So, $\cos(ax + b) \;dx = \frac{du}{a}$.

Now, substitute $u = \sin(ax + b)$ and $\cos(ax + b) \;dx = \frac{du}{a}$ into the integral $I$:

$I = \int u \cdot \frac{du}{a}$

$I = \frac{1}{a} \int u \;du$

Integrate $u$ with respect to $u$ using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C''$):

$I = \frac{1}{a} \left( \frac{u^{1+1}}{1+1} \right) + C$

$I = \frac{1}{a} \left( \frac{u^2}{2} \right) + C$

$I = \frac{u^2}{2a} + C$

Substitute back $u = \sin(ax + b)$:

$I = \frac{(\sin(ax + b))^2}{2a} + C$

$I = \frac{\sin^2(ax + b)}{2a} + C$, where $C$ is the constant of integration.

Both results obtained by Method 1 and Method 2 are equivalent, differing only by a constant.

Let the given integral be $I$.

$I = \int \sqrt{ax + b} \;dx$

$I = \int (ax + b)^{1/2} \;dx$

We use the method of substitution.

Let $u = ax + b$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(ax + b)$

$\frac{du}{dx} = a + 0$

$\frac{du}{dx} = a$

Rearranging the terms, we get:

$du = a \;dx$

So, $dx = \frac{du}{a}$.

Now, substitute $u = ax + b$ and $dx = \frac{du}{a}$ into the integral $I$:

$I = \int u^{1/2} \cdot \frac{du}{a}$

$I = \frac{1}{a} \int u^{1/2} \;du$

Integrate $u^{1/2}$ with respect to $u$ using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C'$):

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C'$

$\int u^{1/2} \;du = \frac{u^{3/2}}{3/2} + C'$

$\int u^{1/2} \;du = \frac{2}{3} u^{3/2} + C'$

Now, substitute this back into the expression for $I$:

$I = \frac{1}{a} \left( \frac{2}{3} u^{3/2} \right) + C$

$I = \frac{2}{3a} u^{3/2} + C$, where $C = \frac{C'}{a}$ is the constant of integration.

Substitute back $u = ax + b$:

$I = \frac{2}{3a} (ax + b)^{3/2} + C$

Thus, the integral of the given function is $\mathbf{\frac{2}{3a} (ax + b)^{3/2} + C}$.

Let the given integral be $I$.

$I = \int x \sqrt{x + 2} \;dx$

We use the method of substitution.

Let $u = x + 2$.

From the substitution, we can express $x$ in terms of $u$:

$x = u - 2$

Differentiating the substitution $u = x + 2$ with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x + 2)$

$\frac{du}{dx} = 1 + 0$

$\frac{du}{dx} = 1$

Rearranging the terms, we get:

$du = dx$

Now, substitute $u = x + 2$, $x = u - 2$, and $dx = du$ into the integral $I$:

$I = \int (u - 2) \sqrt{u} \;du$

$I = \int (u - 2) u^{1/2} \;du$

Distribute $u^{1/2}$ inside the parenthesis:

$I = \int (u \cdot u^{1/2} - 2 \cdot u^{1/2}) \;du$

$I = \int (u^{1 + 1/2} - 2u^{1/2}) \;du$

$I = \int (u^{3/2} - 2u^{1/2}) \;du$

Integrate term by term using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C'$):

$I = \int u^{3/2} \;du - 2 \int u^{1/2} \;du$

$\int u^{3/2} \;du = \frac{u^{3/2 + 1}}{3/2 + 1} = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}$

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$

Substitute the results back into the expression for $I$:

$I = \frac{2}{5} u^{5/2} - 2 \left( \frac{2}{3} u^{3/2} \right) + C$

$I = \frac{2}{5} u^{5/2} - \frac{4}{3} u^{3/2} + C$, where $C$ is the constant of integration.

Substitute back $u = x + 2$:

$I = \frac{2}{5} (x + 2)^{5/2} - \frac{4}{3} (x + 2)^{3/2} + C$

Thus, the integral of the given function is $\mathbf{\frac{2}{5} (x + 2)^{5/2} - \frac{4}{3} (x + 2)^{3/2} + C}$.

Let the given integral be $I$.

$I = \int x \sqrt{1 + 2x^2} \;dx$

$I = \int (1 + 2x^2)^{1/2} \cdot x \;dx$

We use the method of substitution.

Let $u = 1 + 2x^2$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(1 + 2x^2)$

$\frac{du}{dx} = 0 + 2(2x)$

$\frac{du}{dx} = 4x$

Rearranging the terms, we get:

$du = 4x \;dx$

We have $x \;dx$ in the integrand, so we can write:

$x \;dx = \frac{du}{4}$

Now, substitute $u = 1 + 2x^2$ and $x \;dx = \frac{du}{4}$ into the integral $I$:

$I = \int u^{1/2} \cdot \frac{du}{4}$

$I = \frac{1}{4} \int u^{1/2} \;du$

Integrate $u^{1/2}$ with respect to $u$ using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C'$):

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C'$

$\int u^{1/2} \;du = \frac{u^{3/2}}{3/2} + C'$

$\int u^{1/2} \;du = \frac{2}{3} u^{3/2} + C'$

Now, substitute this back into the expression for $I$:

$I = \frac{1}{4} \left( \frac{2}{3} u^{3/2} \right) + C$

$I = \frac{2}{12} u^{3/2} + C$

$I = \frac{1}{6} u^{3/2} + C$, where $C = \frac{C'}{4}$ is the constant of integration.

Substitute back $u = 1 + 2x^2$:

$I = \frac{1}{6} (1 + 2x^2)^{3/2} + C$

Thus, the integral of the given function is $\mathbf{\frac{1}{6} (1 + 2x^2)^{3/2} + C}$.

Let the given integral be $I$.

$I = \int (4x + 2) \sqrt{x^2 + x + 1} \;dx$

We can factor out a 2 from the term $(4x + 2)$: $4x + 2 = 2(2x + 1)$.

$I = \int 2(2x + 1) \sqrt{x^2 + x + 1} \;dx$

$I = 2 \int (2x + 1) \sqrt{x^2 + x + 1} \;dx$

We use the method of substitution.

Let $u = x^2 + x + 1$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x^2 + x + 1)$

$\frac{du}{dx} = 2x + 1 + 0$

$\frac{du}{dx} = 2x + 1$

Rearranging the terms, we get:

$du = (2x + 1) \;dx$

Now, substitute $u = x^2 + x + 1$ and $(2x + 1) \;dx = du$ into the integral $I$:

$I = 2 \int \sqrt{u} \;du$

$I = 2 \int u^{1/2} \;du$

Integrate $u^{1/2}$ with respect to $u$ using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C'$):

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C'$

$\int u^{1/2} \;du = \frac{u^{3/2}}{3/2} + C'$

$\int u^{1/2} \;du = \frac{2}{3} u^{3/2} + C'$

Now, substitute this back into the expression for $I$:

$I = 2 \left( \frac{2}{3} u^{3/2} \right) + C$

$I = \frac{4}{3} u^{3/2} + C$, where $C = 2C'$ is the constant of integration.

Substitute back $u = x^2 + x + 1$:

$I = \frac{4}{3} (x^2 + x + 1)^{3/2} + C$

Thus, the integral of the given function is $\mathbf{\frac{4}{3} (x^2 + x + 1)^{3/2} + C}$.

Let the given integral be $I$.

$I = \int \frac{1}{x - \sqrt{x}} dx$

Factor out $\sqrt{x}$ from the denominator:

$x - \sqrt{x} = \sqrt{x} \cdot \sqrt{x} - \sqrt{x} = \sqrt{x}(\sqrt{x} - 1)$

The integral becomes:

$I = \int \frac{1}{\sqrt{x}(\sqrt{x} - 1)} dx$

$I = \int \frac{1}{(\sqrt{x} - 1)} \cdot \frac{1}{\sqrt{x}} dx$

We use the method of substitution.

Let $u = \sqrt{x} - 1$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\sqrt{x} - 1)$

$\frac{du}{dx} = \frac{d}{dx}(x^{1/2} - 1)$

$\frac{du}{dx} = \frac{1}{2} x^{1/2 - 1} - 0$

$\frac{du}{dx} = \frac{1}{2} x^{-1/2}$

$\frac{du}{dx} = \frac{1}{2\sqrt{x}}$

Rearranging the terms, we get:

$du = \frac{1}{2\sqrt{x}} dx$

$2 \, du = \frac{1}{\sqrt{x}} dx$

Now, substitute $u = \sqrt{x} - 1$ and $\frac{1}{\sqrt{x}} dx = 2 \, du$ into the integral $I$:

$I = \int \frac{1}{u} \cdot 2 \, du$

$I = 2 \int \frac{1}{u} \, du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log_e |u| + C'$, where $C'$ is the constant of integration.

$I = 2 (\log_e |u| + C')$

$I = 2 \log_e |u| + 2C'$

Let $C = 2C'$ be the constant of integration.

$I = 2 \log_e |u| + C$

Substitute back $u = \sqrt{x} - 1$:

$I = 2 \log_e |\sqrt{x} - 1| + C$

Thus, the integral of the given function is $\mathbf{2 \log_e |\sqrt{x} - 1| + C}$.

Let the given integral be $I$.

$I = \int \frac{x}{\sqrt{x + 4}} \;dx$

We use the method of substitution.

Let $u = x + 4$.

From the substitution, we can express $x$ in terms of $u$:

$x = u - 4$

Differentiating the substitution $u = x + 4$ with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x + 4)$

$\frac{du}{dx} = 1 + 0$

$\frac{du}{dx} = 1$

Rearranging the terms, we get:

$du = dx$

Now, substitute $u = x + 4$, $x = u - 4$, and $dx = du$ into the integral $I$:

$I = \int \frac{u - 4}{\sqrt{u}} \;du$

$I = \int \frac{u - 4}{u^{1/2}} \;du$

Simplify the integrand by dividing each term in the numerator by $u^{1/2}$:

$I = \int \left( \frac{u}{u^{1/2}} - \frac{4}{u^{1/2}} \right) \;du$

$I = \int \left( u^{1 - 1/2} - 4u^{-1/2} \right) \;du$

$I = \int \left( u^{1/2} - 4u^{-1/2} \right) \;du$

Integrate term by term using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C'$):

$I = \int u^{1/2} \;du - 4 \int u^{-1/2} \;du$

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$

$\int u^{-1/2} \;du = \frac{u^{-1/2 + 1}}{-1/2 + 1} = \frac{u^{1/2}}{1/2} = 2 u^{1/2}$

Substitute the results back into the expression for $I$:

$I = \frac{2}{3} u^{3/2} - 4 (2 u^{1/2}) + C$

$I = \frac{2}{3} u^{3/2} - 8 u^{1/2} + C$, where $C$ is the constant of integration.

Substitute back $u = x + 4$:

$I = \frac{2}{3} (x + 4)^{3/2} - 8 (x + 4)^{1/2} + C$

$I = \frac{2}{3} (x + 4)\sqrt{x + 4} - 8 \sqrt{x + 4} + C$

This can also be written by factoring $\sqrt{x+4}$:

$I = \sqrt{x + 4} \left( \frac{2}{3}(x + 4) - 8 \right) + C$

$I = \sqrt{x + 4} \left( \frac{2x}{3} + \frac{8}{3} - \frac{24}{3} \right) + C$

$I = \sqrt{x + 4} \left( \frac{2x - 16}{3} \right) + C$

$I = \frac{2}{3} (x - 8) \sqrt{x + 4} + C$

Thus, the integral of the given function is $\mathbf{\frac{2}{3} (x + 4)^{3/2} - 8 (x + 4)^{1/2} + C}$ or equivalently $\mathbf{\frac{2}{3} (x - 8) \sqrt{x + 4} + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int (x^3 - 1)^{1/3} x^5 \;dx$

We can rewrite $x^5$ as $x^3 \cdot x^2$.

$I = \int (x^3 - 1)^{1/3} x^3 \cdot x^2 \;dx$

We use the method of substitution.

Let $u = x^3 - 1$.

From the substitution, we can express $x^3$ in terms of $u$:

$x^3 = u + 1$

Differentiating the substitution $u = x^3 - 1$ with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x^3 - 1)$

$\frac{du}{dx} = 3x^2 - 0$

$\frac{du}{dx} = 3x^2$

Rearranging the terms, we get:

$du = 3x^2 \;dx$

So, $x^2 \;dx = \frac{du}{3}$.

Now, substitute $u = x^3 - 1$, $x^3 = u + 1$, and $x^2 \;dx = \frac{du}{3}$ into the integral $I$:

$I = \int u^{1/3} \cdot (u + 1) \cdot \frac{du}{3}$

$I = \frac{1}{3} \int u^{1/3} (u + 1) \;du$

Expand the integrand:

$u^{1/3} (u + 1) = u^{1/3} \cdot u^1 + u^{1/3} \cdot 1$

$u^{1/3} (u + 1) = u^{1/3 + 1} + u^{1/3}$

$u^{1/3} (u + 1) = u^{4/3} + u^{1/3}$

So the integral becomes:

$I = \frac{1}{3} \int (u^{4/3} + u^{1/3}) \;du$

Integrate term by term using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + C'$:

$I = \frac{1}{3} \left( \int u^{4/3} \;du + \int u^{1/3} \;du \right)$

$\int u^{4/3} \;du = \frac{u^{4/3 + 1}}{4/3 + 1} = \frac{u^{7/3}}{7/3} = \frac{3}{7} u^{7/3}$

$\int u^{1/3} \;du = \frac{u^{1/3 + 1}}{1/3 + 1} = \frac{u^{4/3}}{4/3} = \frac{3}{4} u^{4/3}$

Substitute the integrated terms back into the expression for $I$:

$I = \frac{1}{3} \left( \frac{3}{7} u^{7/3} + \frac{3}{4} u^{4/3} \right) + C$

$I = \frac{1}{3} \cdot \frac{3}{7} u^{7/3} + \frac{1}{3} \cdot \frac{3}{4} u^{4/3} + C$

$I = \frac{1}{7} u^{7/3} + \frac{1}{4} u^{4/3} + C$, where $C$ is the constant of integration.

Substitute back $u = x^3 - 1$:

$I = \frac{1}{7} (x^3 - 1)^{7/3} + \frac{1}{4} (x^3 - 1)^{4/3} + C$

Thus, the integral of the given function is $\mathbf{\frac{1}{7} (x^3 - 1)^{7/3} + \frac{1}{4} (x^3 - 1)^{4/3} + C}$.

Let the given integral be $I$.

$I = \int \frac{x^2}{(2 + 3x^3)^3} dx$

We use the method of substitution.

Let $u = 2 + 3x^3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(2 + 3x^3)$

$\frac{du}{dx} = 0 + 3(3x^2)$

$\frac{du}{dx} = 9x^2$

Rearranging the terms, we get:

$du = 9x^2 \;dx$

We have $x^2 \;dx$ in the numerator of the integrand, so we can write:

$x^2 \;dx = \frac{du}{9}$

Now, substitute $u = 2 + 3x^3$ and $x^2 \;dx = \frac{du}{9}$ into the integral $I$:

$I = \int \frac{1}{u^3} \cdot \frac{du}{9}$

$I = \frac{1}{9} \int \frac{1}{u^3} du$

$I = \frac{1}{9} \int u^{-3} du$

Integrate $u^{-3}$ with respect to $u$ using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C'$, where $n \neq -1$:

$\int u^{-3} du = \frac{u^{-3 + 1}}{-3 + 1} + C'$

$\int u^{-3} du = \frac{u^{-2}}{-2} + C'$

$\int u^{-3} du = -\frac{1}{2} u^{-2} + C'$

$\int u^{-3} du = -\frac{1}{2u^2} + C'$

Now, substitute this back into the expression for $I$:

$I = \frac{1}{9} \left( -\frac{1}{2u^2} \right) + C$

$I = -\frac{1}{18u^2} + C$, where $C = \frac{C'}{9}$ is the constant of integration.

Substitute back $u = 2 + 3x^3$:

$I = -\frac{1}{18(2 + 3x^3)^2} + C$

Thus, the integral of the given function is $\mathbf{-\frac{1}{18(2 + 3x^3)^2} + C}$.

Let the given integral be $I$.

$I = \int \frac{1}{x(\log x)^m} dx$

We can rewrite the integrand as:

$\frac{1}{x(\log x)^m} = \frac{1}{(\log x)^m} \cdot \frac{1}{x}$

We use the method of substitution.

Let $u = \log x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\log x)$

$\frac{du}{dx} = \frac{1}{x}$

Rearranging the terms, we get:

$du = \frac{1}{x} dx$

Now, substitute $u = \log x$ and $\frac{1}{x} dx = du$ into the integral $I$:

$I = \int \frac{1}{u^m} du$

$I = \int u^{-m} du$

Integrate $u^{-m}$ with respect to $u$ using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C'$, where $n = -m$. Since $m \neq 1$, $-m \neq -1$, so the power rule is applicable.

$I = \frac{u^{-m + 1}}{-m + 1} + C'$

$I = \frac{u^{1 - m}}{1 - m} + C'$

Substitute back $u = \log x$:

$I = \frac{(\log x)^{1 - m}}{1 - m} + C$, where $C$ is the constant of integration.

Thus, the integral of the given function is $\mathbf{\frac{(\log x)^{1 - m}}{1 - m} + C}$.

Let the given integral be $I$.

$I = \int \frac{x}{9 - 4x^2} dx$

We use the method of substitution.

Let $u = 9 - 4x^2$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(9 - 4x^2)$

$\frac{du}{dx} = 0 - 4(2x)$

$\frac{du}{dx} = -8x$

Rearranging the terms, we get:

$du = -8x \;dx$

We have $x \;dx$ in the numerator of the integrand, so we can write:

$x \;dx = -\frac{du}{8}$

Now, substitute $u = 9 - 4x^2$ and $x \;dx = -\frac{du}{8}$ into the integral $I$:

$I = \int \frac{1}{u} \cdot \left(-\frac{du}{8}\right)$

$I = -\frac{1}{8} \int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log_e |u| + C'$, where $C'$ is the constant of integration.

$I = -\frac{1}{8} (\log_e |u| + C')$

$I = -\frac{1}{8} \log_e |u| + C$, where $C = -\frac{1}{8} C'$ is the constant of integration.

Substitute back $u = 9 - 4x^2$:

$I = -\frac{1}{8} \log_e |9 - 4x^2| + C$

Thus, the integral of the given function is $\mathbf{-\frac{1}{8} \log_e |9 - 4x^2| + C}$.

Let the given integral be $I$.

$I = \int e^{2x+3} \;dx$

We use the method of substitution.

Let $u = 2x+3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(2x+3)$

$\frac{du}{dx} = 2 + 0$

$\frac{du}{dx} = 2$

Rearranging the terms, we get:

$du = 2 \;dx$

So, $dx = \frac{du}{2}$.

Now, substitute $u = 2x+3$ and $dx = \frac{du}{2}$ into the integral $I$:

$I = \int e^{u} \cdot \frac{du}{2}$

$I = \frac{1}{2} \int e^{u} \;du$

The integral of $e^u$ with respect to $u$ is $e^u + C'$, where $C'$ is the constant of integration.

$I = \frac{1}{2} (e^{u} + C')$

$I = \frac{1}{2} e^{u} + C$, where $C = \frac{C'}{2}$ is the constant of integration.

Substitute back $u = 2x+3$:

$I = \frac{1}{2} e^{2x+3} + C$

Thus, the integral of the given function is $\mathbf{\frac{1}{2} e^{2x+3} + C}$.

Let the given integral be $I$.

$I = \int \frac{x}{e^{x^2}} dx$

We can rewrite the integrand as:

$I = \int x e^{-x^2} dx$

We use the method of substitution.

Let $u = -x^2$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(-x^2)$

$\frac{du}{dx} = -2x$

Rearranging the terms, we get:

$du = -2x \;dx$

We have $x \;dx$ in the integrand, so we can write:

$x \;dx = -\frac{du}{2}$

Now, substitute $u = -x^2$ and $x \;dx = -\frac{du}{2}$ into the integral $I$:

$I = \int e^{u} \cdot \left(-\frac{du}{2}\right)$

$I = -\frac{1}{2} \int e^{u} \;du$

The integral of $e^u$ with respect to $u$ is $e^u + C'$, where $C'$ is the constant of integration.

$I = -\frac{1}{2} (e^{u} + C')$

$I = -\frac{1}{2} e^{u} + C$, where $C = -\frac{1}{2} C'$ is the constant of integration.

Substitute back $u = -x^2$:

$I = -\frac{1}{2} e^{-x^2} + C$

Thus, the integral of the given function is $\mathbf{-\frac{1}{2} e^{-x^2} + C}$ or $\mathbf{-\frac{1}{2e^{x^2}} + C}$.

Let the given integral be $I$.

$I = \int \frac{e^{\tan^{-1} x}}{1 + x^2} dx$

We use the method of substitution.

Let $u = \tan^{-1} x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\tan^{-1} x)$

$\frac{du}{dx} = \frac{1}{1 + x^2}$

Rearranging the terms, we get:

$du = \frac{1}{1 + x^2} dx$

Now, substitute $u = \tan^{-1} x$ and $\frac{1}{1 + x^2} dx = du$ into the integral $I$:

$I = \int e^{u} du$

The integral of $e^u$ with respect to $u$ is $e^u + C$, where $C$ is the constant of integration.

$I = e^{u} + C$

Substitute back $u = \tan^{-1} x$:

$I = e^{\tan^{-1} x} + C$

Thus, the integral of the given function is $\mathbf{e^{\tan^{-1} x} + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{e^{2x} - 1}{e^{2x} + 1} dx$

We can manipulate the integrand by dividing the numerator and the denominator by $e^x$:

$\frac{e^{2x} - 1}{e^{2x} + 1} = \frac{(e^{2x} - 1)/e^x}{(e^{2x} + 1)/e^x} = \frac{e^{2x}/e^x - 1/e^x}{e^{2x}/e^x + 1/e^x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

So the integral becomes:

$I = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$

We use the method of substitution.

Let $u = e^x + e^{-x}$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(e^x + e^{-x})$

$\frac{du}{dx} = e^x + (-e^{-x})$

$\frac{du}{dx} = e^x - e^{-x}$

Rearranging the terms, we get:

$du = (e^x - e^{-x}) dx$

Now, substitute $u = e^x + e^{-x}$ and $(e^x - e^{-x}) dx = du$ into the integral $I$:

$I = \int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log_e |u| + C'$, where $C'$ is the constant of integration.

$I = \log_e |u| + C'$

Substitute back $u = e^x + e^{-x}$:

$I = \log_e |e^x + e^{-x}| + C'$

Since $e^x > 0$ and $e^{-x} > 0$ for all real $x$, their sum $e^x + e^{-x}$ is always positive. Thus, the absolute value is not necessary.

$I = \log_e (e^x + e^{-x}) + C$, where $C = C'$ is the constant of integration.

Thus, the integral of the given function is $\mathbf{\log_e (e^x + e^{-x}) + C}$.

Alternate Solution:

Let the given integral be $I$.

$I = \int \frac{e^{2x} - 1}{e^{2x} + 1} dx$

We can rewrite the numerator as $(e^{2x} + 1) - 2$:

$\frac{e^{2x} - 1}{e^{2x} + 1} = \frac{(e^{2x} + 1) - 2}{e^{2x} + 1} = \frac{e^{2x} + 1}{e^{2x} + 1} - \frac{2}{e^{2x} + 1} = 1 - \frac{2}{e^{2x} + 1}$

So the integral becomes:

$I = \int \left(1 - \frac{2}{e^{2x} + 1}\right) dx$

$I = \int 1 \, dx - 2 \int \frac{1}{e^{2x} + 1} dx$

The first integral is $\int 1 \, dx = x + C_1$.

For the second integral, $\int \frac{1}{e^{2x} + 1} dx$, we can multiply the numerator and denominator by $e^{-2x}$:

$\int \frac{e^{-2x}}{e^{-2x}(e^{2x} + 1)} dx = \int \frac{e^{-2x}}{1 + e^{-2x}} dx$

Now, use substitution for this integral.

Let $w = 1 + e^{-2x}$.

Differentiating both sides with respect to $x$, we get:

$\frac{dw}{dx} = \frac{d}{dx}(1 + e^{-2x})$

$\frac{dw}{dx} = 0 + e^{-2x} \cdot \frac{d}{dx}(-2x) = e^{-2x} \cdot (-2)$

$\frac{dw}{dx} = -2e^{-2x}$

Rearranging the terms, we get:

$dw = -2e^{-2x} dx$

So, $e^{-2x} dx = -\frac{dw}{2}$.

Substitute $w = 1 + e^{-2x}$ and $e^{-2x} dx = -\frac{dw}{2}$ into the integral:

$\int \frac{e^{-2x}}{1 + e^{-2x}} dx = \int \frac{1}{w} \left(-\frac{dw}{2}\right) = -\frac{1}{2} \int \frac{1}{w} dw$

The integral of $\frac{1}{w}$ with respect to $w$ is $\log_e |w| + C_2$.

$-\frac{1}{2} \int \frac{1}{w} dw = -\frac{1}{2} (\log_e |w| + C_2) = -\frac{1}{2} \log_e |w| - \frac{C_2}{2}$

Substitute back $w = 1 + e^{-2x}$:

$-\frac{1}{2} \log_e |1 + e^{-2x}| + C_3$, where $C_3 = -\frac{C_2}{2}$.

Since $1 + e^{-2x} > 0$, the absolute value is not needed:

$-\frac{1}{2} \log_e (1 + e^{-2x}) + C_3$

Now substitute this back into the expression for $I$:

$I = (x + C_1) - 2 \left( -\frac{1}{2} \log_e (1 + e^{-2x}) + C_3 \right) + C_4$

$I = x + C_1 + \log_e (1 + e^{-2x}) - 2C_3 + C_4$

$I = x + \log_e (1 + e^{-2x}) + C$, where $C = C_1 - 2C_3 + C_4$ is the constant of integration.

Thus, the integral of the given function is $\mathbf{x + \log_e (1 + e^{-2x}) + C}$.

Note that $\log_e (e^x + e^{-x}) = \log_e (e^x(1 + e^{-2x})) = \log_e (e^x) + \log_e (1 + e^{-2x}) = x + \log_e (1 + e^{-2x})$. The two results are equivalent.

Let the given integral be $I$.

$I = \int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} dx$

We use the method of substitution.

Let $u = e^{2x} + e^{-2x}$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(e^{2x} + e^{-2x})$

Using the chain rule, $\frac{du}{dx} = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(e^{-2x})$

$\frac{du}{dx} = e^{2x} \cdot \frac{d}{dx}(2x) + e^{-2x} \cdot \frac{d}{dx}(-2x)$

$\frac{du}{dx} = e^{2x} \cdot 2 + e^{-2x} \cdot (-2)$

$\frac{du}{dx} = 2e^{2x} - 2e^{-2x}$

$\frac{du}{dx} = 2(e^{2x} - e^{-2x})$

Rearranging the terms, we get:

$du = 2(e^{2x} - e^{-2x}) dx$

We have $(e^{2x} - e^{-2x}) dx$ in the numerator of the integrand, so we can write:

$(e^{2x} - e^{-2x}) dx = \frac{du}{2}$

Now, substitute $u = e^{2x} + e^{-2x}$ and $(e^{2x} - e^{-2x}) dx = \frac{du}{2}$ into the integral $I$:

$I = \int \frac{1}{u} \cdot \frac{du}{2}$

$I = \frac{1}{2} \int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log_e |u| + C'$, where $C'$ is the constant of integration.

$I = \frac{1}{2} (\log_e |u| + C')$

$I = \frac{1}{2} \log_e |u| + C$, where $C = \frac{C'}{2}$ is the constant of integration.

Substitute back $u = e^{2x} + e^{-2x}$:

$I = \frac{1}{2} \log_e |e^{2x} + e^{-2x}| + C$

Since $e^{2x} > 0$ and $e^{-2x} > 0$ for all real values of $x$, their sum $e^{2x} + e^{-2x}$ is always positive. Thus, the absolute value is not required.

$I = \frac{1}{2} \log_e (e^{2x} + e^{-2x}) + C$

Thus, the integral of the given function is $\mathbf{\frac{1}{2} \log_e (e^{2x} + e^{-2x}) + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \tan^2(2x - 3) \;dx$

We use the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$.

Here, $\theta = 2x - 3$.

So, $\tan^2(2x - 3) = \sec^2(2x - 3) - 1$.

The integral becomes:

$I = \int (\sec^2(2x - 3) - 1) \;dx$

$I = \int \sec^2(2x - 3) \;dx - \int 1 \;dx$

Let's evaluate the first integral $\int \sec^2(2x - 3) \;dx$ using substitution.

Let $u = 2x - 3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(2x - 3)$

$\frac{du}{dx} = 2 - 0$

$\frac{du}{dx} = 2$

Rearranging the terms, we get:

$du = 2 \;dx$

So, $dx = \frac{du}{2}$.

Substitute $u = 2x - 3$ and $dx = \frac{du}{2}$ into the first integral:

$\int \sec^2(2x - 3) \;dx = \int \sec^2(u) \frac{du}{2}$

$= \frac{1}{2} \int \sec^2(u) \;du$

The integral of $\sec^2(u)$ with respect to $u$ is $\tan(u)$.

$= \frac{1}{2} \tan(u) + C_1$

Substitute back $u = 2x - 3$:

$= \frac{1}{2} \tan(2x - 3) + C_1$

Now, evaluate the second integral:

$\int 1 \;dx = x + C_2$

Combine the results of both integrals:

$I = \left(\frac{1}{2} \tan(2x - 3) + C_1\right) - (x + C_2)$

$I = \frac{1}{2} \tan(2x - 3) - x + C_1 - C_2$

Let $C = C_1 - C_2$ be the constant of integration.

$I = \frac{1}{2} \tan(2x - 3) - x + C$

Thus, the integral of the given function is $\mathbf{\frac{1}{2} \tan(2x - 3) - x + C}$.

Let the given integral be $I$.

$I = \int \sec^2(7 - 4x) \;dx$

We use the method of substitution.

Let $u = 7 - 4x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(7 - 4x)$

$\frac{du}{dx} = 0 - 4$

$\frac{du}{dx} = -4$

Rearranging the terms, we get:

$du = -4 \;dx$

So, $dx = -\frac{du}{4}$.

Now, substitute $u = 7 - 4x$ and $dx = -\frac{du}{4}$ into the integral $I$:

$I = \int \sec^2(u) \cdot \left(-\frac{du}{4}\right)$

$I = -\frac{1}{4} \int \sec^2(u) \;du$

The integral of $\sec^2(u)$ with respect to $u$ is $\tan(u) + C'$, where $C'$ is the constant of integration.

$I = -\frac{1}{4} (\tan(u) + C')$

$I = -\frac{1}{4} \tan(u) + C$, where $C = -\frac{C'}{4}$ is the constant of integration.

Substitute back $u = 7 - 4x$:

$I = -\frac{1}{4} \tan(7 - 4x) + C$

Thus, the integral of the given function is $\mathbf{-\frac{1}{4} \tan(7 - 4x) + C}$.

Let the given integral be $I$.

$I = \int \frac{\sin^{-1} x}{\sqrt{1 - x^2}} dx$

We can rewrite the integrand as:

$I = \int \sin^{-1} x \cdot \frac{1}{\sqrt{1 - x^2}} dx$

We use the method of substitution.

Let $u = \sin^{-1} x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\sin^{-1} x)$

$\frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}}$

Rearranging the terms, we get:

$du = \frac{1}{\sqrt{1 - x^2}} dx$

Now, substitute $u = \sin^{-1} x$ and $\frac{1}{\sqrt{1 - x^2}} dx = du$ into the integral $I$:

$I = \int u \, du$

Integrate $u$ with respect to $u$ using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C$, for $n \neq -1$):

$I = \frac{u^{1+1}}{1+1} + C$

$I = \frac{u^2}{2} + C$

Substitute back $u = \sin^{-1} x$:

$I = \frac{(\sin^{-1} x)^2}{2} + C$

Thus, the integral of the given function is $\mathbf{\frac{(\sin^{-1} x)^2}{2} + C}$, where $C$ is the constant of integration.

To evaluate the integral $\int \frac{2\cos x − 3 \sin x}{6\cos x + 4\sin x} dx$, we can use the method of substitution.

Let the denominator be $u$.

$u = 6\cos x + 4\sin x$

Now, we find the derivative of $u$ with respect to $x$.

$\frac{du}{dx} = \frac{d}{dx}(6\cos x + 4\sin x)$

$\frac{du}{dx} = 6 \frac{d}{dx}(\cos x) + 4 \frac{d}{dx}(\sin x)$

$\frac{du}{dx} = 6 (-\sin x) + 4 (\cos x)$

$\frac{du}{dx} = -6\sin x + 4\cos x$

$\frac{du}{dx} = 4\cos x - 6\sin x$

From this, we have $du = (4\cos x - 6\sin x) dx$.

Notice that the numerator of the integrand is $2\cos x - 3\sin x$. This is exactly half of the derivative we found ($4\cos x - 6\sin x$).

So, $2(2\cos x - 3\sin x) dx = (4\cos x - 6\sin x) dx = du$.

This implies $(2\cos x - 3\sin x) dx = \frac{1}{2} du$.

Now, substitute $u = 6\cos x + 4\sin x$ and $(2\cos x - 3\sin x) dx = \frac{1}{2} du$ into the integral:

$\int \frac{2\cos x − 3 \sin x}{6\cos x + 4\sin x} dx = \int \frac{1}{u} \left(\frac{1}{2} du\right)$

= $\frac{1}{2} \int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log |u|$.

= $\frac{1}{2} \log |u| + C$

where $C$ is the constant of integration.

Finally, substitute back $u = 6\cos x + 4\sin x$.

= $\frac{1}{2} \log |6\cos x + 4\sin x| + C$

We can factor out 2 from the argument of the logarithm: $6\cos x + 4\sin x = 2(3\cos x + 2\sin x) = 2(2\sin x + 3\cos x)$.

So the result can be written as:

= $\frac{1}{2} \log |2(2\sin x + 3\cos x)| + C$

Using the property $\log |ab| = \log |a| + \log |b|$, we get:

= $\frac{1}{2} (\log |2| + \log |2\sin x + 3\cos x|) + C$

= $\frac{1}{2}\log |2| + \frac{1}{2}\log |2\sin x + 3\cos x| + C$

Since $\frac{1}{2}\log |2|$ is a constant, we can combine it with the constant of integration $C$ to get a new constant $C'$.

The integral is equal to:

$\frac{1}{2}\log |2\sin x + 3\cos x| + C'$

Thus, the final answer is $\frac{1}{2} \log |2 \sin x + 3 \cos x| + C$, where $C$ represents the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{\cos^2 x (1 - \tan x)^2} dx$

We can rewrite $\frac{1}{\cos^2 x}$ as $\sec^2 x$.

$I = \int \frac{\sec^2 x}{(1 - \tan x)^2} dx$

We use the method of substitution.

Let $u = 1 - \tan x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(1 - \tan x)$

$\frac{du}{dx} = 0 - \sec^2 x$

$\frac{du}{dx} = -\sec^2 x$

Rearranging the terms, we get:

$du = -\sec^2 x \;dx$

So, $\sec^2 x \;dx = -du$.

Now, substitute $u = 1 - \tan x$ and $\sec^2 x \;dx = -du$ into the integral $I$:

$I = \int \frac{1}{u^2} (-du)$

$I = - \int u^{-2} du$

Integrate $u^{-2}$ with respect to $u$ using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C'$, where $n \neq -1$:

$\int u^{-2} du = \frac{u^{-2 + 1}}{-2 + 1} + C'$

$\int u^{-2} du = \frac{u^{-1}}{-1} + C'$

$\int u^{-2} du = -\frac{1}{u} + C'$

Now, substitute this back into the expression for $I$:

$I = - \left(-\frac{1}{u} + C'\right)$

$I = \frac{1}{u} - C'$

Let $C = -C'$ be the constant of integration.

$I = \frac{1}{u} + C$

Substitute back $u = 1 - \tan x$:

$I = \frac{1}{1 - \tan x} + C$

Thus, the integral of the given function is $\mathbf{\frac{1}{1 - \tan x} + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} dx$

We can rewrite the integrand as:

$I = \int \cos \sqrt{x} \cdot \frac{1}{\sqrt{x}} dx$

We use the method of substitution.

Let $u = \sqrt{x}$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\sqrt{x})$

$\frac{du}{dx} = \frac{d}{dx}(x^{1/2})$

$\frac{du}{dx} = \frac{1}{2} x^{1/2 - 1}$

$\frac{du}{dx} = \frac{1}{2} x^{-1/2}$

$\frac{du}{dx} = \frac{1}{2\sqrt{x}}$

Rearranging the terms, we get:

$du = \frac{1}{2\sqrt{x}} dx$

So, $\frac{1}{\sqrt{x}} dx = 2 \, du$.

Now, substitute $u = \sqrt{x}$ and $\frac{1}{\sqrt{x}} dx = 2 \, du$ into the integral $I$:

$I = \int \cos(u) \cdot 2 \, du$

$I = 2 \int \cos(u) \, du$

The integral of $\cos(u)$ with respect to $u$ is $\sin(u) + C'$, where $C'$ is the constant of integration.

$I = 2 (\sin(u) + C')$

$I = 2 \sin(u) + C$, where $C = 2C'$ is the constant of integration.

Substitute back $u = \sqrt{x}$:

$I = 2 \sin(\sqrt{x}) + C$

Thus, the integral of the given function is $\mathbf{2 \sin(\sqrt{x}) + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \sqrt{\sin 2x} \cos 2x \;dx$

$I = \int (\sin 2x)^{1/2} \cos 2x \;dx$

We use the method of substitution.

Let $u = \sin 2x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\sin 2x)$

Using the chain rule, $\frac{du}{dx} = \cos 2x \cdot \frac{d}{dx}(2x) = \cos 2x \cdot 2$

$\frac{du}{dx} = 2 \cos 2x$

Rearranging the terms, we get:

$du = 2 \cos 2x \;dx$

We have $\cos 2x \;dx$ in the integrand, so we can write:

$\cos 2x \;dx = \frac{du}{2}$

Now, substitute $u = \sin 2x$ and $\cos 2x \;dx = \frac{du}{2}$ into the integral $I$:

$I = \int u^{1/2} \cdot \frac{du}{2}$

$I = \frac{1}{2} \int u^{1/2} \;du$

Integrate $u^{1/2}$ with respect to $u$ using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C'$):

$\int u^{1/2} \;du = \frac{u^{1/2 + 1}}{1/2 + 1} + C'$

$\int u^{1/2} \;du = \frac{u^{3/2}}{3/2} + C'$

$\int u^{1/2} \;du = \frac{2}{3} u^{3/2} + C'$

Now, substitute this back into the expression for $I$:

$I = \frac{1}{2} \left( \frac{2}{3} u^{3/2} \right) + C$

$I = \frac{1}{3} u^{3/2} + C$, where $C = \frac{C'}{2}$ is the constant of integration.

Substitute back $u = \sin 2x$:

$I = \frac{1}{3} (\sin 2x)^{3/2} + C$

$I = \frac{1}{3} (\sin 2x)\sqrt{\sin 2x} + C$

Thus, the integral of the given function is $\mathbf{\frac{1}{3} (\sin 2x)^{3/2} + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{\cos x}{\sqrt{1 + \sin x}} dx$

We can rewrite the integrand as:

$I = \int (1 + \sin x)^{-1/2} \cos x \;dx$

We use the method of substitution.

Let $u = 1 + \sin x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(1 + \sin x)$

$\frac{du}{dx} = 0 + \cos x$

$\frac{du}{dx} = \cos x$

Rearranging the terms, we get:

$du = \cos x \;dx$

Now, substitute $u = 1 + \sin x$ and $\cos x \;dx = du$ into the integral $I$:

$I = \int u^{-1/2} du$

Integrate $u^{-1/2}$ with respect to $u$ using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C'$, where $n \neq -1$:

$\int u^{-1/2} du = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C'$

$\int u^{-1/2} du = \frac{u^{1/2}}{1/2} + C'$

$\int u^{-1/2} du = 2u^{1/2} + C'$

$\int u^{-1/2} du = 2\sqrt{u} + C'$

Now, substitute this back into the expression for $I$:

$I = 2\sqrt{u} + C$, where $C = C'$ is the constant of integration.

Substitute back $u = 1 + \sin x$:

$I = 2\sqrt{1 + \sin x} + C$

Thus, the integral of the given function is $\mathbf{2\sqrt{1 + \sin x} + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \cot x \log (\sin x) \;dx$

We use the method of substitution.

Let $u = \log (\sin x)$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\log (\sin x))$

Using the chain rule, $\frac{du}{dx} = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)$

$\frac{du}{dx} = \frac{1}{\sin x} \cdot \cos x$

$\frac{du}{dx} = \cot x$

Rearranging the terms, we get:

$du = \cot x \;dx$

Now, substitute $u = \log (\sin x)$ and $\cot x \;dx = du$ into the integral $I$:

$I = \int u \, du$

Integrate $u$ with respect to $u$ using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C'$):

$I = \frac{u^{1+1}}{1+1} + C'$

$I = \frac{u^2}{2} + C'$

Substitute back $u = \log (\sin x)$:

$I = \frac{(\log (\sin x))^2}{2} + C$

Thus, the integral of the given function is $\mathbf{\frac{(\log (\sin x))^2}{2} + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{\sin x}{1 + \cos x} dx$

We use the method of substitution.

Let $u = 1 + \cos x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(1 + \cos x)$

$\frac{du}{dx} = 0 - \sin x$

$\frac{du}{dx} = -\sin x$

Rearranging the terms, we get:

$du = -\sin x \;dx$

So, $\sin x \;dx = -du$.

Now, substitute $u = 1 + \cos x$ and $\sin x \;dx = -du$ into the integral $I$:

$I = \int \frac{1}{u} (-du)$

$I = - \int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log_e |u| + C'$, where $C'$ is the constant of integration.

$I = - (\log_e |u| + C')$

$I = -\log_e |u| - C'$

Let $C = -C'$ be the constant of integration.

$I = -\log_e |u| + C$

Substitute back $u = 1 + \cos x$:

$I = -\log_e |1 + \cos x| + C$

Thus, the integral of the given function is $\mathbf{-\log_e |1 + \cos x| + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{\sin x}{(1 + \cos x)^2} dx$

We can rewrite the integrand as:

$I = \int (1 + \cos x)^{-2} \sin x \;dx$

We use the method of substitution.

Let $u = 1 + \cos x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(1 + \cos x)$

$\frac{du}{dx} = 0 - \sin x$

$\frac{du}{dx} = -\sin x$

Rearranging the terms, we get:

$du = -\sin x \;dx$

So, $\sin x \;dx = -du$.

Now, substitute $u = 1 + \cos x$ and $\sin x \;dx = -du$ into the integral $I$:

$I = \int u^{-2} (-du)$

$I = - \int u^{-2} du$

Integrate $u^{-2}$ with respect to $u$ using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C'$, where $n \neq -1$:

$\int u^{-2} du = \frac{u^{-2 + 1}}{-2 + 1} + C'$

$\int u^{-2} du = \frac{u^{-1}}{-1} + C'$

$\int u^{-2} du = -\frac{1}{u} + C'$

Now, substitute this back into the expression for $I$:

$I = - \left(-\frac{1}{u} + C'\right)$

$I = \frac{1}{u} - C'$

Let $C = -C'$ be the constant of integration.

$I = \frac{1}{u} + C$

Substitute back $u = 1 + \cos x$:

$I = \frac{1}{1 + \cos x} + C$

Thus, the integral of the given function is $\mathbf{\frac{1}{1 + \cos x} + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{1 + \cot x} dx$

We can rewrite the integrand by expressing $\cot x$ in terms of $\sin x$ and $\cos x$:

$\cot x = \frac{\cos x}{\sin x}$

So, the integrand is $\frac{1}{1 + \frac{\cos x}{\sin x}} = \frac{1}{\frac{\sin x + \cos x}{\sin x}} = \frac{\sin x}{\sin x + \cos x}$.

The integral becomes:

$I = \int \frac{\sin x}{\sin x + \cos x} dx$

We can write the numerator as a linear combination of the denominator and its derivative.

Let $D = \sin x + \cos x$. Then $D' = \cos x - \sin x$.

We want to find constants $A$ and $B$ such that $\sin x = A(\sin x + \cos x) + B(\cos x - \sin x)$.

$\sin x = A\sin x + A\cos x + B\cos x - B\sin x$

$\sin x = (A - B)\sin x + (A + B)\cos x$

Comparing the coefficients of $\sin x$ and $\cos x$ on both sides:

$1 = A - B$

$0 = A + B$

Solving these two equations, we get $A = \frac{1}{2}$ and $B = -\frac{1}{2}$.

Now, rewrite the integrand using these values of $A$ and $B$:

$\frac{\sin x}{\sin x + \cos x} = \frac{\frac{1}{2}(\sin x + \cos x) - \frac{1}{2}(\cos x - \sin x)}{\sin x + \cos x}$

$= \frac{1}{2} \frac{\sin x + \cos x}{\sin x + \cos x} - \frac{1}{2} \frac{\cos x - \sin x}{\sin x + \cos x}$

$= \frac{1}{2} - \frac{1}{2} \frac{\cos x - \sin x}{\sin x + \cos x}$

So the integral becomes:

$I = \int \left(\frac{1}{2} - \frac{1}{2} \frac{\cos x - \sin x}{\sin x + \cos x}\right) dx$

$I = \int \frac{1}{2} dx - \frac{1}{2} \int \frac{\cos x - \sin x}{\sin x + \cos x} dx$

The first integral is $\int \frac{1}{2} dx = \frac{1}{2} x + C_1$.

For the second integral, $\int \frac{\cos x - \sin x}{\sin x + \cos x} dx$, let $u = \sin x + \cos x$.

Then, $du = (\cos x - (-\sin x)) dx = (\cos x - \sin x) dx$.

The integral becomes $\int \frac{1}{u} du = \log_e |u| + C_2 = \log_e |\sin x + \cos x| + C_2$.

Substitute these results back into the expression for $I$:

$I = \frac{1}{2} x - \frac{1}{2} (\log_e |\sin x + \cos x|) + C$

Where $C$ is the constant of integration.

Thus, the integral of the given function is $\mathbf{\frac{1}{2} x - \frac{1}{2} \log_e |\sin x + \cos x| + C}$.

Let the given integral be $I$.

$I = \int \frac{1}{1 - \tan x} dx$

We can rewrite the integrand by expressing $\tan x$ in terms of $\sin x$ and $\cos x$:

$\tan x = \frac{\sin x}{\cos x}$

So, the integrand is $\frac{1}{1 - \frac{\sin x}{\cos x}} = \frac{1}{\frac{\cos x - \sin x}{\cos x}} = \frac{\cos x}{\cos x - \sin x}$.

The integral becomes:

$I = \int \frac{\cos x}{\cos x - \sin x} dx$

We can rewrite the numerator as a combination of the denominator and its derivative.

Let the numerator be $N = \cos x$ and the denominator be $D = \cos x - \sin x$.

The derivative of the denominator is $D' = \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x = -(\sin x + \cos x)$.

We can write $N$ as $N = A \cdot D + B \cdot D'$ for some constants $A$ and $B$.

$\cos x = A(\cos x - \sin x) + B(-(\sin x + \cos x))$

$\cos x = A\cos x - A\sin x - B\sin x - B\cos x$

$\cos x = (A - B)\cos x + (-A - B)\sin x$

Comparing the coefficients of $\cos x$ and $\sin x$ on both sides:

$1 = A - B$

$0 = -A - B$

Adding these two equations:

$(A - B) + (-A - B) = 1 + 0$

$-2B = 1$

$B = -\frac{1}{2}$

Substitute $B = -\frac{1}{2}$ into the first equation $A - B = 1$:

$A - \left(-\frac{1}{2}\right) = 1$

$A + \frac{1}{2} = 1$

$A = 1 - \frac{1}{2} = \frac{1}{2}$

So, $A = \frac{1}{2}$ and $B = -\frac{1}{2}$.

Now, rewrite the integrand using these values of $A$ and $B$:

$\frac{\cos x}{\cos x - \sin x} = \frac{\frac{1}{2}(\cos x - \sin x) - \frac{1}{2}(-\sin x - \cos x)}{\cos x - \sin x}$

$= \frac{1}{2} \frac{\cos x - \sin x}{\cos x - \sin x} - \frac{1}{2} \frac{-(\sin x + \cos x)}{\cos x - \sin x}$

$= \frac{1}{2} + \frac{1}{2} \frac{\sin x + \cos x}{\cos x - \sin x}$

So the integral becomes:

$I = \int \left(\frac{1}{2} + \frac{1}{2} \frac{\sin x + \cos x}{\cos x - \sin x}\right) dx$

$I = \int \frac{1}{2} dx + \frac{1}{2} \int \frac{\sin x + \cos x}{\cos x - \sin x} dx$

Evaluate the first integral:

$\int \frac{1}{2} dx = \frac{1}{2} x + C_1$

Evaluate the second integral $\int \frac{\sin x + \cos x}{\cos x - \sin x} dx$ using substitution.

Let $u = \cos x - \sin x$.

Differentiating both sides with respect to $x$:

$du = (-\sin x - \cos x) dx$

$du = -(\sin x + \cos x) dx$

So, $(\sin x + \cos x) dx = -du$.

Substitute into the second integral:

$\int \frac{1}{u} (-du) = - \int \frac{1}{u} du$

$= -\log_e |u| + C_2$

Substitute back $u = \cos x - \sin x$:

$= -\log_e |\cos x - \sin x| + C_2$

Combine the results of both integrals:

$I = \left(\frac{1}{2} x + C_1\right) + \frac{1}{2} \left(-\log_e |\cos x - \sin x| + C_2\right)$

$I = \frac{1}{2} x - \frac{1}{2} \log_e |\cos x - \sin x| + C_1 + \frac{C_2}{2}$

Let $C = C_1 + \frac{C_2}{2}$ be the constant of integration.

$I = \frac{1}{2} x - \frac{1}{2} \log_e |\cos x - \sin x| + C$

Thus, the integral of the given function is $\mathbf{\frac{1}{2} x - \frac{1}{2} \log_e |\cos x - \sin x| + C}$.

Let the given integral be $I$.

$I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$

We can manipulate the integrand to prepare for substitution.

Divide the numerator and the denominator by $\cos^2 x$:

$\frac{\sqrt{\tan x}}{\sin x \cos x} = \frac{\frac{\sqrt{\tan x}}{\cos^2 x}}{\frac{\sin x \cos x}{\cos^2 x}}$

$= \frac{\sqrt{\tan x} \cdot \frac{1}{\cos^2 x}}{\frac{\sin x}{\cos x}}$

$= \frac{\sqrt{\tan x} \sec^2 x}{\tan x}$

$= \frac{\sqrt{\tan x}}{\tan x} \sec^2 x$

Using the property $\frac{\sqrt{a}}{a} = \frac{1}{\sqrt{a}}$ (for $a>0$), we get:

$= \frac{1}{\sqrt{\tan x}} \sec^2 x$

So the integral becomes:

$I = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx$

We use the method of substitution.

Let $u = \tan x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\tan x)$

$\frac{du}{dx} = \sec^2 x$

Rearranging the terms, we get:

$du = \sec^2 x \;dx$

Now, substitute $u = \tan x$ and $\sec^2 x \;dx = du$ into the integral $I$:

$I = \int \frac{1}{\sqrt{u}} du$

$I = \int u^{-1/2} du$

Integrate $u^{-1/2}$ with respect to $u$ using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C'$:

$\int u^{-1/2} du = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C'$

$\int u^{-1/2} du = \frac{u^{1/2}}{1/2} + C'$

$\int u^{-1/2} du = 2u^{1/2} + C'$

$\int u^{-1/2} du = 2\sqrt{u} + C'$

Now, substitute this back into the expression for $I$:

$I = 2\sqrt{u} + C$, where $C = C'$ is the constant of integration.

Substitute back $u = \tan x$:

$I = 2\sqrt{\tan x} + C$

Thus, the integral of the given function is $\mathbf{2\sqrt{\tan x} + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{(1 + \log x)^2}{x} dx$

We can rewrite the integrand as:

$I = \int (1 + \log x)^2 \cdot \frac{1}{x} dx$

We use the method of substitution.

Let $u = 1 + \log x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(1 + \log x)$

$\frac{du}{dx} = 0 + \frac{1}{x}$

$\frac{du}{dx} = \frac{1}{x}$

Rearranging the terms, we get:

$du = \frac{1}{x} dx$

Now, substitute $u = 1 + \log x$ and $\frac{1}{x} dx = du$ into the integral $I$:

$I = \int u^2 du$

Integrate $u^2$ with respect to $u$ using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C'$):

$I = \frac{u^{2+1}}{2+1} + C'$

$I = \frac{u^3}{3} + C'$

Substitute back $u = 1 + \log x$:

$I = \frac{(1 + \log x)^3}{3} + C$, where $C = C'$ is the constant of integration.

Thus, the integral of the given function is $\mathbf{\frac{(1 + \log x)^3}{3} + C}$, where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{(x + 1) (x + \log x)^2}{x} dx$

Rewrite the integrand by dividing the term $(x+1)$ by $x$:

$\frac{(x + 1)}{x} (x + \log x)^2 = \left(\frac{x}{x} + \frac{1}{x}\right) (x + \log x)^2 = \left(1 + \frac{1}{x}\right) (x + \log x)^2$

So the integral becomes:

$I = \int \left(1 + \frac{1}{x}\right) (x + \log x)^2 dx$

We use the method of substitution.

Let $u = x + \log x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x + \log x)$

$\frac{du}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\log x)$

$\frac{du}{dx} = 1 + \frac{1}{x}$

Rearranging the terms, we get:

$du = \left(1 + \frac{1}{x}\right) dx$

Now, substitute $u = x + \log x$ and $\left(1 + \frac{1}{x}\right) dx = du$ into the integral $I$:

$I = \int u^2 du$

Integrate $u^2$ with respect to $u$ using the power rule for integration ($\int u^n du = \frac{u^{n+1}}{n+1} + C'$, for $n \neq -1$):

$I = \frac{u^{2+1}}{2+1} + C'$

$I = \frac{u^3}{3} + C'$

Substitute back $u = x + \log x$:

$I = \frac{(x + \log x)^3}{3} + C$, where $C = C'$ is the constant of integration.

Thus, the integral of the given function is $\frac{(x + \log x)^3}{3} + C$.

Let the given integral be $I$.

$I = \int \frac{x^3 \sin (\tan^{−1} x^4)}{1 + x^8} dx$

We use the method of substitution.

Let $u = \tan^{−1} x^4$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(\tan^{−1} x^4)$

Using the chain rule, $\frac{du}{dx} = \frac{1}{1 + (x^4)^2} \cdot \frac{d}{dx}(x^4)$

$\frac{du}{dx} = \frac{1}{1 + x^8} \cdot 4x^3$

$\frac{du}{dx} = \frac{4x^3}{1 + x^8}$

Rearranging the terms, we get:

$du = \frac{4x^3}{1 + x^8} dx$

We have $\frac{x^3}{1 + x^8} dx$ in the integrand, so we can write:

$\frac{x^3}{1 + x^8} dx = \frac{du}{4}$

Now, substitute $u = \tan^{−1} x^4$ and $\frac{x^3}{1 + x^8} dx = \frac{du}{4}$ into the integral $I$:

$I = \int \sin(u) \cdot \frac{du}{4}$

$I = \frac{1}{4} \int \sin(u) du$

The integral of $\sin(u)$ with respect to $u$ is $-\cos(u) + C'$, where $C'$ is the constant of integration.

$I = \frac{1}{4} (-\cos(u)) + C'$

$I = -\frac{1}{4} \cos(u) + C$, where $C = \frac{C'}{4}$ is the constant of integration.

Substitute back $u = \tan^{−1} x^4$:

$I = -\frac{1}{4} \cos(\tan^{−1} x^4) + C$

Thus, the integral of the given function is $-\frac{1}{4} \cos(\tan^{−1} x^4) + C$.

Let the given integral be $I$.

$I = \int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x} dx$

We use the method of substitution.

Let $u = x^{10} + 10^x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \frac{d}{dx}(x^{10} + 10^x)$

Using the power rule and the derivative of exponential function $\frac{d}{dx}(a^x) = a^x \log_e a$, we have:

$\frac{du}{dx} = 10x^{10-1} + 10^x \log_e 10$

$\frac{du}{dx} = 10x^9 + 10^x \log_e 10$

Rearranging the terms, we get:

$du = (10x^9 + 10^x \log_e 10) dx$

Observe that the expression $(10x^9 + 10^x \log_e 10) dx$ is exactly the numerator of the integrand multiplied by $dx$.

Now, substitute $u = x^{10} + 10^x$ and $(10x^9 + 10^x \log_e 10) dx = du$ into the integral $I$:

$I = \int \frac{1}{u} du$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log_e |u| + C'$, where $C'$ is the constant of integration.

$I = \log_e |u| + C'$

Substitute back $u = x^{10} + 10^x$:

$I = \log_e |x^{10} + 10^x| + C'$, where $C'$ is the constant of integration.

Since $x^{10} \geq 0$ (for real $x$) and $10^x > 0$ for all real $x$, the term $x^{10} + 10^x$ is always positive for real $x$. Therefore, the absolute value is not necessary.

$I = \log_e (x^{10} + 10^x) + C$, where $C = C'$ is the constant of integration.

Comparing this result with the given options, we see that it matches option (D).

The correct answer is (D) log (10^x + x¹⁰) + C.

Let the given integral be $I$.

$I = \int \frac{1}{\sin^2 x \cos^2 x} dx$

We can rewrite the integrand using the identity $1 = \sin^2 x + \cos^2 x$ in the numerator:

$I = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} dx$

Now, split the fraction into two parts:

$I = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) dx$

Simplify each term:

$I = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) dx$

Using the identities $\frac{1}{\cos^2 x} = \sec^2 x$ and $\frac{1}{\sin^2 x} = \text{cosec}^2 x$, the integral becomes:

$I = \int (\sec^2 x + \text{cosec}^2 x) dx$

Integrate term by term:

$I = \int \sec^2 x \;dx + \int \text{cosec}^2 x \;dx$

The integral of $\sec^2 x$ is $\tan x$, and the integral of $\text{cosec}^2 x$ is $-\cot x$.

$I = \tan x + (-\cot x) + C$

$I = \tan x - \cot x + C$

Thus, the integral of the given function is $\mathbf{\tan x - \cot x + C}$, where $C$ is the constant of integration.

Comparing this result with the given options, we see that it matches option (B).

The correct answer is (B) tan x – cot x + C.

(i) $\int \cos^2 x \;dx$

Let the given integral be $I$.

$I = \int \cos^2 x \;dx$

We use the trigonometric identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$.

Substitute the identity into the integral:

$I = \int \frac{1 + \cos(2x)}{2} dx$

$I = \frac{1}{2} \int (1 + \cos(2x)) dx$

$I = \frac{1}{2} \left( \int 1 \;dx + \int \cos(2x) \;dx \right)$

Integrating term by term, we get:

$\int 1 \;dx = x$

$\int \cos(2x) \;dx = \frac{1}{2} \sin(2x)$ (using substitution $u=2x$)

Combine the results:

$I = \frac{1}{2} \left( x + \frac{1}{2} \sin(2x) \right) + C$

$I = \frac{x}{2} + \frac{1}{4} \sin(2x) + C$, where $C$ is the constant of integration.

Thus, $\int \cos^2 x \;dx = \mathbf{\frac{x}{2} + \frac{1}{4} \sin(2x) + C}$.

(ii) $\int \sin 2x \cos 3x \;dx$

Let the given integral be $I$.

$I = \int \sin 2x \cos 3x \;dx$

We use the product-to-sum trigonometric identity: $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$.

Here $A = 2x$ and $B = 3x$.

$\sin 2x \cos 3x = \frac{1}{2}[\sin(2x+3x) + \sin(2x-3x)]$

$= \frac{1}{2}[\sin(5x) + \sin(-x)]$

Using the identity $\sin(-x) = -\sin x$, we get:

$= \frac{1}{2}[\sin(5x) - \sin x]$

Substitute the identity into the integral:

$I = \int \frac{1}{2}[\sin(5x) - \sin x] dx$

$I = \frac{1}{2} \int (\sin(5x) - \sin x) dx$

$I = \frac{1}{2} \left( \int \sin(5x) \;dx - \int \sin x \;dx \right)$

Integrate term by term.

$\int \sin x \;dx = -\cos x$

$\int \sin(5x) \;dx = -\frac{1}{5} \cos(5x)$ (using substitution $u=5x$)

Combine the results:

$I = \frac{1}{2} \left( -\frac{1}{5} \cos(5x) - (-\cos x) \right) + C$

$I = \frac{1}{2} \left( -\frac{1}{5} \cos(5x) + \cos x \right) + C$

$I = \frac{1}{2} \cos x - \frac{1}{10} \cos(5x) + C$, where $C$ is the constant of integration.

Thus, $\int \sin 2x \cos 3x \;dx = \mathbf{\frac{1}{2} \cos x - \frac{1}{10} \cos(5x) + C}$.

(iii) $\int \sin^3 x \;dx$

Let the given integral be $I$.

$I = \int \sin^3 x \;dx$

We can rewrite $\sin^3 x$ as $\sin^2 x \cdot \sin x$.

$I = \int \sin^2 x \sin x \;dx$

We use the identity $\sin^2 x = 1 - \cos^2 x$.

$I = \int (1 - \cos^2 x) \sin x \;dx$

We use the method of substitution.

Let $u = \cos x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = -\sin x$

So, $\sin x \;dx = -du$.

Now, substitute $u = \cos x$ and $\sin x \;dx = -du$ into the integral $I$:

$I = \int (1 - u^2) (-du)$

$I = - \int (1 - u^2) du$

$I = - \int 1 \;du + \int u^2 \;du$

Integrate term by term using the power rule for integration:

$I = -u + \frac{u^3}{3} + C$, where $C$ is the constant of integration.

Substitute back $u = \cos x$:

$I = -\cos x + \frac{(\cos x)^3}{3} + C$

$I = -\cos x + \frac{\cos^3 x}{3} + C$

Thus, $\int \sin^3 x \;dx = \mathbf{-\cos x + \frac{\cos^3 x}{3} + C}$.

Solution:

We need to find the integral of $\sin^2(2x+5)$ with respect to $x$.

The integral is given by:

$\int \sin^2(2x+5) \, dx$

We use the trigonometric identity:

$\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$

Here, $\theta = 2x+5$. So, $2\theta = 2(2x+5) = 4x+10$.

Substituting this into the identity, we get:

$\sin^2(2x+5) = \frac{1 - \cos(4x+10)}{2}$

Now, we can rewrite the integral:

$\int \sin^2(2x+5) \, dx = \int \frac{1 - \cos(4x+10)}{2} \, dx$

= $\frac{1}{2} \int (1 - \cos(4x+10)) \, dx$

We can split this into two separate integrals:

= $\frac{1}{2} \left( \int 1 \, dx - \int \cos(4x+10) \, dx \right)$

First integral is straightforward:

$\int 1 \, dx = x$

For the second integral, $\int \cos(4x+10) \, dx$, we use a substitution method.

Let $u = 4x+10$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(4x+10) = 4$

So, $du = 4 \, dx$, which means $dx = \frac{1}{4} \, du$.

Substituting $u$ and $dx$ into the second integral:

$\int \cos(4x+10) \, dx = \int \cos(u) \frac{1}{4} \, du$

= $\frac{1}{4} \int \cos(u) \, du$

The integral of $\cos(u)$ is $\sin(u)$.

= $\frac{1}{4} \sin(u) + C'$

Substitute back $u = 4x+10$:

= $\frac{1}{4} \sin(4x+10) + C'$

Now, combine the results of both integrals:

$\int \sin^2(2x+5) \, dx = \frac{1}{2} \left( x - \left( \frac{1}{4} \sin(4x+10) \right) \right) + C$

= $\frac{1}{2} \left( x - \frac{1}{4} \sin(4x+10) \right) + C$

Distribute the $\frac{1}{2}$:

= $\frac{x}{2} - \frac{1}{8} \sin(4x+10) + C$

The final answer is:

$\boxed{\frac{x}{2} - \frac{1}{8} \sin(4x+10) + C}$

Solution:

We need to find the integral of $\sin(3x) \cos(4x)$ with respect to $x$.

The integral is given by:

$\int \sin(3x) \cos(4x) \, dx$

We use the product-to-sum trigonometric identity:

$2 \sin A \cos B = \sin(A+B) + \sin(A-B)$

Dividing by 2, we get:

$\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$

Let $A = 3x$ and $B = 4x$. Then $A+B = 3x+4x = 7x$ and $A-B = 3x-4x = -x$.

So, $\sin(3x) \cos(4x) = \frac{1}{2} [\sin(7x) + \sin(-x)]$.

Since $\sin(-x) = -\sin(x)$, the expression becomes:

$\sin(3x) \cos(4x) = \frac{1}{2} [\sin(7x) - \sin(x)]$

Now, we can rewrite the integral:

$\int \sin(3x) \cos(4x) \, dx = \int \frac{1}{2} [\sin(7x) - \sin(x)] \, dx$

= $\frac{1}{2} \int [\sin(7x) - \sin(x)] \, dx$

We can integrate each term separately:

= $\frac{1}{2} \left( \int \sin(7x) \, dx - \int \sin(x) \, dx \right)$

For $\int \sin(7x) \, dx$, let $u = 7x$, so $du = 7 \, dx$, or $dx = \frac{1}{7} \, du$.

$\int \sin(7x) \, dx = \int \sin(u) \frac{1}{7} \, du = \frac{1}{7} \int \sin(u) \, du = \frac{1}{7} (-\cos(u)) + C_1 = -\frac{1}{7} \cos(7x) + C_1$

For $\int \sin(x) \, dx$, the integral is $-\cos(x) + C_2$.

Substituting these back into the main integral:

= $\frac{1}{2} \left( -\frac{1}{7} \cos(7x) - (-\cos(x)) \right) + C$

= $\frac{1}{2} \left( -\frac{1}{7} \cos(7x) + \cos(x) \right) + C$

= $\frac{1}{2} \cos(x) - \frac{1}{14} \cos(7x) + C$

The final answer is:

$\boxed{\frac{1}{2} \cos(x) - \frac{1}{14} \cos(7x) + C}$

Solution:

We need to find the integral of $\cos(2x) \cos(4x) \cos(6x)$ with respect to $x$.

The integral is given by:

$\int \cos(2x) \cos(4x) \cos(6x) \, dx$

We use the product-to-sum trigonometric identity: $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.

This gives $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$.

First, consider the product of two terms, say $\cos(4x) \cos(6x)$:

$\cos(4x) \cos(6x) = \frac{1}{2} [\cos(4x+6x) + \cos(4x-6x)]$

= $\frac{1}{2} [\cos(10x) + \cos(-2x)]$

Since $\cos(-\theta) = \cos(\theta)$, we have:

= $\frac{1}{2} [\cos(10x) + \cos(2x)]$

Now, multiply this result by the remaining term $\cos(2x)$:

$\cos(2x) \cos(4x) \cos(6x) = \cos(2x) \cdot \frac{1}{2} [\cos(10x) + \cos(2x)]$

= $\frac{1}{2} [\cos(2x)\cos(10x) + \cos(2x)\cos(2x)]$

= $\frac{1}{2} [\cos(2x)\cos(10x) + \cos^2(2x)]$

Apply the product-to-sum identity again to $\cos(2x)\cos(10x)$ (with $A=2x, B=10x$):

$\cos(2x)\cos(10x) = \frac{1}{2} [\cos(2x+10x) + \cos(2x-10x)]$

= $\frac{1}{2} [\cos(12x) + \cos(-8x)]$

= $\frac{1}{2} [\cos(12x) + \cos(8x)]$

Also, use the identity $\cos^2 \theta = \frac{1+\cos(2\theta)}{2}$ for $\cos^2(2x)$:

$\cos^2(2x) = \frac{1+\cos(2 \cdot 2x)}{2} = \frac{1+\cos(4x)}{2}$

Substitute these back into the expression for $\cos(2x) \cos(4x) \cos(6x)$:

= $\frac{1}{2} \left[ \frac{1}{2}(\cos(12x) + \cos(8x)) + \frac{1+\cos(4x)}{2} \right]$

= $\frac{1}{2} \cdot \frac{1}{2} [(\cos(12x) + \cos(8x)) + (1+\cos(4x))]$

= $\frac{1}{4} [1 + \cos(4x) + \cos(8x) + \cos(12x)]$

Now, integrate this simplified expression:

$\int \cos(2x) \cos(4x) \cos(6x) \, dx = \int \frac{1}{4} [1 + \cos(4x) + \cos(8x) + \cos(12x)] \, dx$

= $\frac{1}{4} \int [1 + \cos(4x) + \cos(8x) + \cos(12x)] \, dx$

Integrate term by term, recalling that $\int \cos(ax) \, dx = \frac{1}{a}\sin(ax) + C'$:

= $\frac{1}{4} \left[ \int 1 \, dx + \int \cos(4x) \, dx + \int \cos(8x) \, dx + \int \cos(12x) \, dx \right]$

= $\frac{1}{4} \left[ x + \frac{1}{4}\sin(4x) + \frac{1}{8}\sin(8x) + \frac{1}{12}\sin(12x) \right] + C$

= $\frac{x}{4} + \frac{1}{16}\sin(4x) + \frac{1}{32}\sin(8x) + \frac{1}{48}\sin(12x) + C$

The final answer is:

$\boxed{\frac{x}{4} + \frac{1}{16}\sin(4x) + \frac{1}{32}\sin(8x) + \frac{1}{48}\sin(12x) + C}$

Solution:

We need to find the integral of $\sin^3(2x+1)$ with respect to $x$.

The integral is given by:

$\int \sin^3(2x+1) \, dx$

We can rewrite $\sin^3(2x+1)$ as $\sin^2(2x+1) \cdot \sin(2x+1)$.

Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$, we get:

$\sin^3(2x+1) = (1 - \cos^2(2x+1)) \sin(2x+1)$

Now, the integral becomes:

$\int (1 - \cos^2(2x+1)) \sin(2x+1) \, dx$

We use the substitution method.

Let $u = \cos(2x+1)$.

Differentiating $u$ with respect to $x$ using the chain rule:

$\frac{du}{dx} = \frac{d}{dx}(\cos(2x+1)) = -\sin(2x+1) \cdot \frac{d}{dx}(2x+1)$

= $-\sin(2x+1) \cdot 2$

= $-2 \sin(2x+1)$

Rearranging to find $\sin(2x+1) \, dx$:

$du = -2 \sin(2x+1) \, dx$

$\sin(2x+1) \, dx = -\frac{1}{2} \, du$

Substitute $u = \cos(2x+1)$ and $\sin(2x+1) \, dx = -\frac{1}{2} \, du$ into the integral:

$\int (1 - \cos^2(2x+1)) \sin(2x+1) \, dx = \int (1 - u^2) \left(-\frac{1}{2}\right) \, du$

= $-\frac{1}{2} \int (1 - u^2) \, du$

Now, integrate with respect to $u$:

= $-\frac{1}{2} \left( \int 1 \, du - \int u^2 \, du \right)$

= $-\frac{1}{2} \left( u - \frac{u^3}{3} \right) + C'$

Distribute the $-\frac{1}{2}$:

= $-\frac{1}{2} u + \frac{1}{6} u^3 + C'$

Substitute back $u = \cos(2x+1)$:

= $-\frac{1}{2} \cos(2x+1) + \frac{1}{6} \cos^3(2x+1) + C$ (where $C$ is the constant of integration)

We can rearrange the terms:

= $\frac{1}{6} \cos^3(2x+1) - \frac{1}{2} \cos(2x+1) + C$

The final answer is:

$\boxed{\frac{1}{6} \cos^3(2x+1) - \frac{1}{2} \cos(2x+1) + C}$

To evaluate the integral $\int \sin^3 x \cos^3 x \;dx$.

We can rewrite the integrand $\sin^3 x \cos^3 x$ as follows:

$\sin^3 x \cos^3 x = \sin^2 x \cos^3 x \sin x$

Using the trigonometric identity $\sin^2 x = 1 - \cos^2 x$, we substitute it into the expression:

$= (1 - \cos^2 x) \cos^3 x \sin x$

Now, we use the method of substitution to evaluate the integral.

Let $u = \cos x$.

Differentiating $u$ with respect to $x$ gives:

$\frac{du}{dx} = \frac{d}{dx}(\cos x) = -\sin x$

Rearranging this differential, we get $du = -\sin x \;dx$, which means $\sin x \;dx = -du$.

Substitute $u = \cos x$ and $\sin x \;dx = -du$ into the integral:

$\int (1 - \cos^2 x) \cos^3 x \sin x \;dx = \int (1 - u^2) u^3 (-du)$

$= - \int (u^3 \times 1 - u^3 \times u^2) du$

$= - \int (u^3 - u^{3+2}) du$

$= - \int (u^3 - u^5) du$

We can distribute the negative sign:

$= \int -(u^3 - u^5) du$

$= \int (-u^3 + u^5) du$

$= \int (u^5 - u^3) du$

Now, we integrate term by term using the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C'$ (where $n \neq -1$):

$= \frac{u^{5+1}}{5+1} - \frac{u^{3+1}}{3+1} + C$

$= \frac{u^6}{6} - \frac{u^4}{4} + C$

Substitute back $u = \cos x$ to get the result in terms of the original variable $x$:

$= \frac{(\cos x)^6}{6} - \frac{(\cos x)^4}{4} + C$

$= \frac{1}{6} \cos^6 x - \frac{1}{4} \cos^4 x + C$

Where $C$ is the constant of integration.

Thus, the integral of $\sin^3 x \cos^3 x$ is $\mathbf{\frac{1}{6} \cos^6 x - \frac{1}{4} \cos^4 x + C}$.

Alternate Solution:

We can also solve this integral by using a different substitution.

Start with the integral $\int \sin^3 x \cos^3 x \;dx$.

We can rewrite the integrand as:

$\sin^3 x \cos^3 x = \sin^3 x \cos^2 x \cos x$

Using the trigonometric identity $\cos^2 x = 1 - \sin^2 x$, we substitute it into the expression:

$= \sin^3 x (1 - \sin^2 x) \cos x$

Now, we use the method of substitution.

Let $v = \sin x$.

Differentiating $v$ with respect to $x$ gives:

$\frac{dv}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Rearranging, we get $dv = \cos x \;dx$.

Substitute $v = \sin x$ and $\cos x \;dx = dv$ into the integral:

$\int \sin^3 x (1 - \sin^2 x) \cos x \;dx = \int v^3 (1 - v^2) dv$

$= \int (v^3 \times 1 - v^3 \times v^2) dv$

$= \int (v^3 - v^5) dv$

Integrate term by term using the power rule:

$= \frac{v^{3+1}}{3+1} - \frac{v^{5+1}}{5+1} + C'$

$= \frac{v^4}{4} - \frac{v^6}{6} + C'$

Substitute back $v = \sin x$ to get the result in terms of $x$:

$= \frac{(\sin x)^4}{4} - \frac{(\sin x)^6}{6} + C'$

$= \frac{1}{4} \sin^4 x - \frac{1}{6} \sin^6 x + C'$

Where $C'$ is the constant of integration.

This form of the solution, $\frac{1}{4} \sin^4 x - \frac{1}{6} \sin^6 x + C'$, is equivalent to the first form $\frac{1}{6} \cos^6 x - \frac{1}{4} \cos^4 x + C$, as they differ only by a constant value.

Solution:

We need to find the integral of $\sin x \sin 2x \sin 3x$ with respect to $x$.

The integral is given by:

$\int \sin x \sin 2x \sin 3x \, dx$

We use the product-to-sum trigonometric identity:

$2 \sin A \sin B = \cos(A-B) - \cos(A+B)$

Let's group $\sin x \sin 2x$ first. Using the identity with $A=2x$ and $B=x$:

$2 \sin 2x \sin x = \cos(2x-x) - \cos(2x+x) = \cos x - \cos 3x$

So, $\sin x \sin 2x = \frac{1}{2}(\cos x - \cos 3x)$.

Now multiply by $\sin 3x$:

$\sin x \sin 2x \sin 3x = \frac{1}{2}(\cos x - \cos 3x) \sin 3x$

= $\frac{1}{2}(\cos x \sin 3x - \cos 3x \sin 3x)$

Now, we use another product-to-sum identity for the first term in the bracket:

$2 \cos A \sin B = \sin(A+B) - \sin(A-B)$

For $\cos x \sin 3x$, let $A=x$ and $B=3x$:

$2 \cos x \sin 3x = \sin(x+3x) - \sin(x-3x) = \sin 4x - \sin(-2x)$

Since $\sin(-\theta) = -\sin \theta$, this becomes:

= $\sin 4x - (-\sin 2x) = \sin 4x + \sin 2x$

So, $\cos x \sin 3x = \frac{1}{2}(\sin 4x + \sin 2x)$.

For the second term in the bracket, $\cos 3x \sin 3x$, we use the double angle identity:

$\sin(2\theta) = 2 \sin \theta \cos \theta$

Let $\theta = 3x$:

$\sin(2 \cdot 3x) = 2 \sin 3x \cos 3x$

So, $\cos 3x \sin 3x = \frac{1}{2}\sin 6x$.

Substitute these results back into the expression for $\sin x \sin 2x \sin 3x$:

= $\frac{1}{2} \left[ \frac{1}{2}(\sin 4x + \sin 2x) - \frac{1}{2}\sin 6x \right]$

= $\frac{1}{2} \cdot \frac{1}{2} [\sin 4x + \sin 2x - \sin 6x]$

= $\frac{1}{4} [\sin 4x + \sin 2x - \sin 6x]$

Now, we integrate this expression:

$\int \sin x \sin 2x \sin 3x \, dx = \int \frac{1}{4} [\sin 4x + \sin 2x - \sin 6x] \, dx$

= $\frac{1}{4} \int (\sin 4x + \sin 2x - \sin 6x) \, dx$

We integrate each term using the standard integral $\int \sin(ax) \, dx = -\frac{1}{a}\cos(ax) + C'$:

= $\frac{1}{4} \left[ \int \sin 4x \, dx + \int \sin 2x \, dx - \int \sin 6x \, dx \right]$

= $\frac{1}{4} \left[ -\frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x - \left(-\frac{1}{6}\cos 6x\right) \right] + C$

= $\frac{1}{4} \left[ -\frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x + \frac{1}{6}\cos 6x \right] + C$

Distribute the $\frac{1}{4}$:

= $-\frac{1}{16}\cos 4x - \frac{1}{8}\cos 2x + \frac{1}{24}\cos 6x + C$

Rearranging the terms:

= $\frac{1}{24}\cos 6x - \frac{1}{8}\cos 2x - \frac{1}{16}\cos 4x + C$

The final answer is:

$\boxed{\frac{1}{24}\cos 6x - \frac{1}{8}\cos 2x - \frac{1}{16}\cos 4x + C}$

Solution:

We need to find the integral of $\sin(4x) \sin(8x)$ with respect to $x$.

The integral is given by:

$\int \sin(4x) \sin(8x) \, dx$

We use the product-to-sum trigonometric identity:

$2 \sin A \sin B = \cos(A-B) - \cos(A+B)$

Let $A = 8x$ and $B = 4x$. Then $A-B = 8x-4x = 4x$ and $A+B = 8x+4x = 12x$.

So, $2 \sin(8x) \sin(4x) = \cos(4x) - \cos(12x)$.

Dividing by 2, we get:

$\sin(4x) \sin(8x) = \frac{1}{2} [\cos(4x) - \cos(12x)]$

Now, we can rewrite the integral:

$\int \sin(4x) \sin(8x) \, dx = \int \frac{1}{2} [\cos(4x) - \cos(12x)] \, dx$

= $\frac{1}{2} \int [\cos(4x) - \cos(12x)] \, dx$

We can integrate each term separately:

= $\frac{1}{2} \left( \int \cos(4x) \, dx - \int \cos(12x) \, dx \right)$

Using the standard integral $\int \cos(ax) \, dx = \frac{1}{a}\sin(ax) + C'$:

= $\frac{1}{2} \left( \frac{1}{4}\sin(4x) - \frac{1}{12}\sin(12x) \right) + C$

Distribute the $\frac{1}{2}$:

= $\frac{1}{8}\sin(4x) - \frac{1}{24}\sin(12x) + C$

The final answer is:

$\boxed{\frac{1}{8}\sin(4x) - \frac{1}{24}\sin(12x) + C}$

Solution:

We need to find the integral of $\frac{1 - \cos x}{1 + \cos x}$ with respect to $x$.

The integral is given by:

$\int \frac{1 - \cos x}{1 + \cos x} \, dx$

We use the half-angle identities:

$1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$

$1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)$

Substitute these into the integrand:

$\frac{1 - \cos x}{1 + \cos x} = \frac{2 \sin^2 \left(\frac{x}{2}\right)}{2 \cos^2 \left(\frac{x}{2}\right)}$

Cancel out the factor of 2:

= $\frac{\sin^2 \left(\frac{x}{2}\right)}{\cos^2 \left(\frac{x}{2}\right)}$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we get:

= $\tan^2 \left(\frac{x}{2}\right)$

Now, we use the identity $\tan^2 \theta = \sec^2 \theta - 1$. Let $\theta = \frac{x}{2}$:

= $\sec^2 \left(\frac{x}{2}\right) - 1$

Now, the integral becomes:

$\int \left(\sec^2 \left(\frac{x}{2}\right) - 1\right) \, dx$

We can integrate term by term:

= $\int \sec^2 \left(\frac{x}{2}\right) \, dx - \int 1 \, dx$

For the first integral, $\int \sec^2 \left(\frac{x}{2}\right) \, dx$, we use substitution.

Let $u = \frac{x}{2}$.

Differentiating with respect to $x$, $\frac{du}{dx} = \frac{1}{2}$. So, $dx = 2 \, du$.

Substitute $u$ and $dx$ into the first integral:

$\int \sec^2(u) \, (2 \, du) = 2 \int \sec^2(u) \, du$

The integral of $\sec^2(u)$ is $\tan(u)$.

= $2 \tan(u) + C_1$

Substitute back $u = \frac{x}{2}$:

= $2 \tan \left(\frac{x}{2}\right) + C_1$

The second integral is straightforward:

$\int 1 \, dx = x + C_2$

Combining the results and adding the constant of integration $C$ ($C = C_1 + C_2$):

= $2 \tan \left(\frac{x}{2}\right) - x + C$

The final answer is:

$\boxed{2 \tan \left(\frac{x}{2}\right) - x + C}$

Solution:

We need to find the integral of $\frac{\cos x}{1 + \cos x}$ with respect to $x$.

The integral is given by:

$\int \frac{\cos x}{1 + \cos x} \, dx$

We can simplify the integrand by adding and subtracting 1 in the numerator:

$\frac{\cos x}{1 + \cos x} = \frac{1 + \cos x - 1}{1 + \cos x}$

= $\frac{1 + \cos x}{1 + \cos x} - \frac{1}{1 + \cos x}$

= $1 - \frac{1}{1 + \cos x}$

Now, we integrate this simplified expression:

$\int \left(1 - \frac{1}{1 + \cos x}\right) \, dx$

= $\int 1 \, dx - \int \frac{1}{1 + \cos x} \, dx$

The first integral is:

$\int 1 \, dx = x$

For the second integral, we use the half-angle identity $1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)$:

$\int \frac{1}{1 + \cos x} \, dx = \int \frac{1}{2 \cos^2 \left(\frac{x}{2}\right)} \, dx$

= $\int \frac{1}{2} \sec^2 \left(\frac{x}{2}\right) \, dx$

Let $u = \frac{x}{2}$. Then differentiating with respect to $x$, $\frac{du}{dx} = \frac{1}{2}$. This gives $dx = 2 \, du$.

Substituting into the integral:

$\int \frac{1}{2} \sec^2(u) \, (2 \, du) = \int \sec^2(u) \, du$

The integral of $\sec^2(u)$ is $\tan(u)$.

= $\tan u + C'$

Substitute back $u = \frac{x}{2}$:

= $\tan \left(\frac{x}{2}\right) + C'$

Combining the results of the two integrals:

$\int \frac{\cos x}{1 + \cos x} \, dx = x - \tan \left(\frac{x}{2}\right) + C$ (where $C$ is the constant of integration)

The final answer is:

$\boxed{x - \tan \left(\frac{x}{2}\right) + C}$

Solution:

We need to find the integral of $\sin^4 x$ with respect to $x$.

The integral is given by:

$\int \sin^4 x \, dx$

We can rewrite $\sin^4 x$ using trigonometric identities:

$\sin^4 x = (\sin^2 x)^2$

Using the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$:

$\sin^4 x = \left(\frac{1 - \cos(2x)}{2}\right)^2$

= $\frac{(1 - \cos(2x))^2}{4}$

= $\frac{1 - 2\cos(2x) + \cos^2(2x)}{4}$

Now, use the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$ for $\cos^2(2x)$, where $\theta = 2x$:

$\cos^2(2x) = \frac{1 + \cos(2 \cdot 2x)}{2} = \frac{1 + \cos(4x)}{2}$

Substitute this back into the expression for $\sin^4 x$:

= $\frac{1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}}{4}$

Combine terms in the numerator by finding a common denominator:

= $\frac{\frac{2}{2} - \frac{4\cos(2x)}{2} + \frac{1 + \cos(4x)}{2}}{4}$

= $\frac{\frac{2 - 4\cos(2x) + 1 + \cos(4x)}{2}}{4}$

= $\frac{3 - 4\cos(2x) + \cos(4x)}{8}$

= $\frac{1}{8} (3 - 4\cos(2x) + \cos(4x))$

Now, integrate this simplified expression:

$\int \sin^4 x \, dx = \int \frac{1}{8} (3 - 4\cos(2x) + \cos(4x)) \, dx$

= $\frac{1}{8} \int (3 - 4\cos(2x) + \cos(4x)) \, dx$

Integrate term by term, using the standard integral $\int \cos(ax) \, dx = \frac{1}{a}\sin(ax) + C'$:

= $\frac{1}{8} \left( \int 3 \, dx - \int 4\cos(2x) \, dx + \int \cos(4x) \, dx \right)$

= $\frac{1}{8} \left( 3x - 4 \left(\frac{\sin(2x)}{2}\right) + \frac{\sin(4x)}{4} \right) + C$

= $\frac{1}{8} \left( 3x - 2\sin(2x) + \frac{1}{4}\sin(4x) \right) + C$

Distribute the $\frac{1}{8}$:

= $\frac{3}{8}x - \frac{2}{8}\sin(2x) + \frac{1}{32}\sin(4x) + C$

= $\frac{3}{8}x - \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) + C$

The final answer is:

$\boxed{\frac{3}{8}x - \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) + C}$

Solution:

We need to find the integral of $\cos^4(2x)$ with respect to $x$.

The integral is given by:

$\int \cos^4(2x) \, dx$

We can rewrite $\cos^4(2x)$ using trigonometric identities:

$\cos^4(2x) = (\cos^2(2x))^2$

Using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$, with $\theta = 2x$, we get:

$\cos^2(2x) = \frac{1 + \cos(2 \cdot 2x)}{2} = \frac{1 + \cos(4x)}{2}$

Substitute this back into the expression for $\cos^4(2x)$:

$\cos^4(2x) = \left(\frac{1 + \cos(4x)}{2}\right)^2$

= $\frac{(1 + \cos(4x))^2}{4}$

= $\frac{1 + 2\cos(4x) + \cos^2(4x)}{4}$

Now, use the identity $\cos^2 \phi = \frac{1 + \cos(2\phi)}{2}$ for $\cos^2(4x)$, where $\phi = 4x$:

$\cos^2(4x) = \frac{1 + \cos(2 \cdot 4x)}{2} = \frac{1 + \cos(8x)}{2}$

Substitute this back into the expression for $\cos^4(2x)$:

= $\frac{1 + 2\cos(4x) + \frac{1 + \cos(8x)}{2}}{4}$

Combine terms in the numerator by finding a common denominator:

= $\frac{\frac{2}{2} + \frac{4\cos(4x)}{2} + \frac{1 + \cos(8x)}{2}}{4}$

= $\frac{\frac{2 + 4\cos(4x) + 1 + \cos(8x)}{2}}{4}$

= $\frac{3 + 4\cos(4x) + \cos(8x)}{8}$

= $\frac{1}{8} (3 + 4\cos(4x) + \cos(8x))$

Now, integrate this simplified expression:

$\int \cos^4(2x) \, dx = \int \frac{1}{8} (3 + 4\cos(4x) + \cos(8x)) \, dx$

= $\frac{1}{8} \int (3 + 4\cos(4x) + \cos(8x)) \, dx$

Integrate term by term, using the standard integral $\int \cos(ax) \, dx = \frac{1}{a}\sin(ax) + C'$:

= $\frac{1}{8} \left( \int 3 \, dx + \int 4\cos(4x) \, dx + \int \cos(8x) \, dx \right)$

= $\frac{1}{8} \left( 3x + 4 \left(\frac{\sin(4x)}{4}\right) + \left(\frac{\sin(8x)}{8}\right) \right) + C$

= $\frac{1}{8} \left( 3x + \sin(4x) + \frac{1}{8}\sin(8x) \right) + C$

Distribute the $\frac{1}{8}$:

= $\frac{3}{8}x + \frac{1}{8}\sin(4x) + \frac{1}{64}\sin(8x) + C$

The final answer is:

$\boxed{\frac{3}{8}x + \frac{1}{8}\sin(4x) + \frac{1}{64}\sin(8x) + C}$

Solution:

We need to find the integral of $\frac{\sin^2 x}{1 + \cos x}$ with respect to $x$.

The integral is given by:

$\int \frac{\sin^2 x}{1 + \cos x} \, dx$

We use the identity $\sin^2 x = 1 - \cos^2 x$ in the numerator:

$\int \frac{1 - \cos^2 x}{1 + \cos x} \, dx$

The numerator is a difference of squares ($a^2 - b^2 = (a-b)(a+b)$), where $a=1$ and $b=\cos x$:

$\int \frac{(1 - \cos x)(1 + \cos x)}{1 + \cos x} \, dx$

Assuming $1 + \cos x \neq 0$, we can cancel out the $(1 + \cos x)$ term:

$\int (1 - \cos x) \, dx$

Now, integrate term by term:

= $\int 1 \, dx - \int \cos x \, dx$

The integral of 1 with respect to $x$ is $x$.

The integral of $\cos x$ with respect to $x$ is $\sin x$.

= $x - \sin x + C$

where $C$ is the constant of integration.

The final answer is:

$\boxed{x - \sin x + C}$

Solution:

We need to find the integral of $\frac{\cos 2x − \cos 2α}{\cos x − \cos α}$ with respect to $x$. Note that $\alpha$ is a constant.

The integral is given by:

$\int \frac{\cos 2x − \cos 2α}{\cos x − \cos α} \, dx$

We use the double angle identity $\cos 2\theta = 2\cos^2 \theta - 1$ in the numerator:

Numerator = $\cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1)$

= $2\cos^2 x - 1 - 2\cos^2 \alpha + 1$

= $2\cos^2 x - 2\cos^2 \alpha$

= $2(\cos^2 x - \cos^2 \alpha)$

Now substitute this back into the integrand:

$\frac{\cos 2x − \cos 2α}{\cos x − \cos α} = \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x - \cos α}$

We use the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$ on the numerator, with $a = \cos x$ and $b = \cos \alpha$:

= $\frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha}$

Assuming $\cos x - \cos \alpha \neq 0$, we can cancel the term $(\cos x - \cos \alpha)$:

= $2(\cos x + \cos \alpha)$

Now, we integrate the simplified expression:

$\int 2(\cos x + \cos \alpha) \, dx = \int (2\cos x + 2\cos \alpha) \, dx$

= $\int 2\cos x \, dx + \int 2\cos \alpha \, dx$

Since $\alpha$ is a constant, $2\cos \alpha$ is also a constant. We can take constants out of the integral:

= $2 \int \cos x \, dx + 2\cos \alpha \int 1 \, dx$

The integral of $\cos x$ is $\sin x$, and the integral of 1 is $x$.

= $2(\sin x) + 2\cos \alpha (x) + C$

= $2\sin x + 2x\cos \alpha + C$

where $C$ is the constant of integration.

The final answer is:

$\boxed{2\sin x + 2x\cos \alpha + C}$

Solution:

We need to find the integral of $\frac{\cos x − \sin x}{1 + \sin 2x}$ with respect to $x$.

The integral is given by:

$\int \frac{\cos x − \sin x}{1 + \sin 2x} \, dx$

We use the identity $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$ in the denominator:

$1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x$

= $(\sin x + \cos x)^2$

Substitute this into the integrand:

$\int \frac{\cos x − \sin x}{(\sin x + \cos x)^2} \, dx$

Now, we use the substitution method.

Let $u = \sin x + \cos x$.

Differentiating $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$

So, $du = (\cos x - \sin x) \, dx$.

Substitute $u = \sin x + \cos x$ and $du = (\cos x - \sin x) \, dx$ into the integral:

$\int \frac{1}{u^2} \, du$

Rewrite $\frac{1}{u^2}$ as $u^{-2}$:

= $\int u^{-2} \, du$

Now, integrate with respect to $u$ using the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1} + C'$:

= $\frac{u^{-2+1}}{-2+1} + C'$

= $\frac{u^{-1}}{-1} + C'$

= $-\frac{1}{u} + C'$

Substitute back $u = \sin x + \cos x$:

= $-\frac{1}{\sin x + \cos x} + C$ (where $C$ is the constant of integration)

The final answer is:

$\boxed{-\frac{1}{\sin x + \cos x} + C}$

Solution:

We need to find the integral of $\tan^3(2x) \sec(2x)$ with respect to $x$.

The integral is given by:

$\int \tan^3(2x) \sec(2x) \, dx$

We can rewrite the integrand as follows:

$\int \tan^2(2x) \cdot (\tan(2x) \sec(2x)) \, dx$

Using the identity $\tan^2 \theta = \sec^2 \theta - 1$, with $\theta = 2x$, we get:

$\tan^2(2x) = \sec^2(2x) - 1$

Substitute this into the integral:

$\int (\sec^2(2x) - 1) (\tan(2x) \sec(2x)) \, dx$

Now, we use the substitution method.

Let $u = \sec(2x)$.

Differentiating $u$ with respect to $x$ using the chain rule:

$\frac{du}{dx} = \frac{d}{dx}(\sec(2x)) = \sec(2x) \tan(2x) \cdot \frac{d}{dx}(2x)$

= $\sec(2x) \tan(2x) \cdot 2$

= $2 \sec(2x) \tan(2x)$

Rearranging to find $\sec(2x) \tan(2x) \, dx$:

$du = 2 \sec(2x) \tan(2x) \, dx$

$\sec(2x) \tan(2x) \, dx = \frac{1}{2} \, du$

Substitute $u = \sec(2x)$ and $\sec(2x) \tan(2x) \, dx = \frac{1}{2} \, du$ into the integral:

$\int (u^2 - 1) \left(\frac{1}{2}\right) \, du$

= $\frac{1}{2} \int (u^2 - 1) \, du$

Now, integrate with respect to $u$:

= $\frac{1}{2} \left( \int u^2 \, du - \int 1 \, du \right)$

= $\frac{1}{2} \left( \frac{u^3}{3} - u \right) + C'$

Distribute the $\frac{1}{2}$:

= $\frac{u^3}{6} - \frac{u}{2} + C'$

Substitute back $u = \sec(2x)$:

= $\frac{\sec^3(2x)}{6} - \frac{\sec(2x)}{2} + C$ (where $C$ is the constant of integration)

The final answer is:

$\boxed{\frac{\sec^3(2x)}{6} - \frac{\sec(2x)}{2} + C}$

Solution:

We need to find the integral of $\tan^4 x$ with respect to $x$.

The integral is given by:

$\int \tan^4 x \, dx$

We can rewrite $\tan^4 x$ using trigonometric identities:

$\tan^4 x = \tan^2 x \cdot \tan^2 x$

Using the identity $\tan^2 x = \sec^2 x - 1$:

$\tan^4 x = \tan^2 x (\sec^2 x - 1)$

Distribute $\tan^2 x$:

= $\tan^2 x \sec^2 x - \tan^2 x$

Use the identity $\tan^2 x = \sec^2 x - 1$ again for the second term:

= $\tan^2 x \sec^2 x - (\sec^2 x - 1)$

= $\tan^2 x \sec^2 x - \sec^2 x + 1$

Now, the integral becomes:

$\int (\tan^2 x \sec^2 x - \sec^2 x + 1) \, dx$

We can integrate term by term:

= $\int \tan^2 x \sec^2 x \, dx - \int \sec^2 x \, dx + \int 1 \, dx$

For the first integral, $\int \tan^2 x \sec^2 x \, dx$, we use substitution.

Let $u = \tan x$.

Differentiating with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

So, $du = \sec^2 x \, dx$.

Substitute $u = \tan x$ and $du = \sec^2 x \, dx$ into the first integral:

$\int u^2 \, du$

Integrate with respect to $u$ using the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1} + C_1$:

= $\frac{u^3}{3} + C_1$

Substitute back $u = \tan x$:

= $\frac{\tan^3 x}{3} + C_1$

The second integral is the standard integral of $\sec^2 x$:

$\int \sec^2 x \, dx = \tan x + C_2$

The third integral is the standard integral of 1:

$\int 1 \, dx = x + C_3$

Combine the results of the three integrals and add the constant of integration $C$ ($C = C_1 + C_2 + C_3$):

= $\frac{\tan^3 x}{3} - (\tan x) + x + C$

= $\frac{1}{3}\tan^3 x - \tan x + x + C$

The final answer is:

$\boxed{\frac{1}{3}\tan^3 x - \tan x + x + C}$

Solution:

We need to find the integral of $\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x}$ with respect to $x$.

The integral is given by:

$\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \, dx$

We can split the fraction into two terms by dividing each term in the numerator by the denominator:

$\frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} = \frac{\sin^3 x}{\sin^2 x \cos^2 x} + \frac{\cos^3 x}{\sin^2 x \cos^2 x}$

Simplify each term by cancelling common factors:

= $\frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x}$

We can rewrite these terms using reciprocal and quotient identities:

$\frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x$

$\frac{\cos x}{\sin^2 x} = \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \cot x \text{ cosec } x$

So, the integrand becomes:

= $\tan x \sec x + \cot x \text{ cosec } x$

Now, integrate this simplified expression:

$\int (\tan x \sec x + \cot x \text{ cosec } x) \, dx$

We can integrate term by term:

= $\int \tan x \sec x \, dx + \int \cot x \text{ cosec } x \, dx$

These are standard integrals:

$\int \tan x \sec x \, dx = \sec x + C_1$

$\int \cot x \text{ cosec } x \, dx = -\text{ cosec } x + C_2$

Combining the results and adding the constant of integration $C$ ($C = C_1 + C_2$):

= $\sec x - \text{ cosec } x + C$

The final answer is:

$\boxed{\sec x - \text{ cosec } x + C}$

Solution:

We need to find the integral of $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$ with respect to $x$.

The integral is given by:

$\int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} \, dx$

We can simplify the numerator using the double angle identity $\cos 2x = 1 - 2\sin^2 x$:

Numerator = $\cos 2x + 2\sin^2 x = (1 - 2\sin^2 x) + 2\sin^2 x$

= $1 - 2\sin^2 x + 2\sin^2 x$

= $1$

Substitute this back into the integrand:

$\frac{\cos 2x + 2\sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}$

Using the reciprocal identity $\sec x = \frac{1}{\cos x}$, we get:

= $\sec^2 x$

Now, we integrate this simplified expression:

$\int \sec^2 x \, dx$

This is a standard integral:

= $\tan x + C$

where $C$ is the constant of integration.

The final answer is:

$\boxed{\tan x + C}$

Solution:

We need to find the integral of $\frac{1}{\sin x \cos^3 x}$ with respect to $x$.

The integral is given by:

$\int \frac{1}{\sin x \cos^3 x} \, dx$

We can rewrite the integrand by multiplying the numerator and the denominator by $\sec^4 x$:

$\frac{1}{\sin x \cos^3 x} = \frac{1}{\sin x \cos^3 x} \cdot \frac{\sec^4 x}{\sec^4 x}$

= $\frac{\sec^4 x}{\sin x \cos^3 x \cdot \frac{1}{\cos^4 x}}$

= $\frac{\sec^4 x}{\frac{\sin x}{\cos x}}$

= $\frac{\sec^4 x}{\tan x}$

We can rewrite $\sec^4 x$ as $\sec^2 x \cdot \sec^2 x$. Using the identity $\sec^2 x = 1 + \tan^2 x$, we get:

= $\frac{\sec^2 x \cdot (1 + \tan^2 x)}{\tan x}$

= $\frac{\sec^2 x + \tan^2 x \sec^2 x}{\tan x}$

Split the fraction into two terms:

= $\frac{\sec^2 x}{\tan x} + \frac{\tan^2 x \sec^2 x}{\tan x}$

= $\frac{\sec^2 x}{\tan x} + \tan x \sec^2 x$

Now, the integral becomes:

$\int \left( \frac{\sec^2 x}{\tan x} + \tan x \sec^2 x \right) \, dx$

We can integrate each term separately.

For both integrals, we use the substitution method. Let $u = \tan x$.

Differentiating $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

So, $du = \sec^2 x \, dx$.

Substitute $u = \tan x$ and $du = \sec^2 x \, dx$ into the integral:

= $\int \left( \frac{1}{u} + u \right) \, du$

Now, integrate with respect to $u$:

= $\int \frac{1}{u} \, du + \int u \, du$

= $\log_e |u| + \frac{u^{1+1}}{1+1} + C'$

= $\log_e |u| + \frac{u^2}{2} + C'$

Substitute back $u = \tan x$:

= $\log_e |\tan x| + \frac{\tan^2 x}{2} + C$ (where $C$ is the constant of integration)

The final answer is:

$\boxed{\log_e |\tan x| + \frac{\tan^2 x}{2} + C}$

To evaluate the integral $\int \frac{\cos 2x}{(\cos x + \sin x)^2} dx$, we can simplify the integrand first.

We use the trigonometric identity for $\cos 2x$: $\cos 2x = \cos^2 x - \sin^2 x$.

The denominator is $(\cos x + \sin x)^2$.

So, the integrand becomes:

$\frac{\cos^2 x - \sin^2 x}{(\cos x + \sin x)^2}$

The numerator $\cos^2 x - \sin^2 x$ is a difference of squares, which can be factored as $(\cos x - \sin x)(\cos x + \sin x)$.

The integrand is now:

$\frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2}$

Assuming $\cos x + \sin x \neq 0$, we can cancel one term of $(\cos x + \sin x)$ from the numerator and the denominator:

$= \frac{\cos x - \sin x}{\cos x + \sin x}$

Now we need to evaluate the integral $\int \frac{\cos x - \sin x}{\cos x + \sin x} dx$.

We can use the method of substitution.

Let $u = \cos x + \sin x$.

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\cos x + \sin x)$

$\frac{du}{dx} = -\sin x + \cos x$

$\frac{du}{dx} = \cos x - \sin x$

From this, we get $du = (\cos x - \sin x) dx$.

Now substitute $u = \cos x + \sin x$ and $du = (\cos x - \sin x) dx$ into the integral $\int \frac{\cos x - \sin x}{\cos x + \sin x} dx$:

$\int \frac{du}{u}$

The integral of $\frac{1}{u}$ with respect to $u$ is $\log |u|$.

$\int \frac{du}{u} = \log |u| + C$

where $C$ is the constant of integration.

Finally, substitute back $u = \cos x + \sin x$ to get the result in terms of $x$:

The integral is equal to:

$\log |\cos x + \sin x| + C$

Thus, $\int \frac{\cos 2x}{(\cos x + \sin x)^2} dx = \mathbf{\log |\cos x + \sin x| + C}$.

Solution:

We need to find the integral of $\sin^{-1}(\cos x)$ with respect to $x$.

The integral is given by:

$\int \sin^{-1}(\cos x) \, dx$

We use the trigonometric identity $\cos x = \sin\left(\frac{\pi}{2} - x\right)$.

Substituting this into the integrand:

$\sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - x\right)\right)$

For $\sin^{-1}(\sin \theta) = \theta$, provided $\theta$ is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$. The range of $\frac{\pi}{2} - x$ depends on the domain of $x$. However, for the purpose of finding a general integral, we can use this simplification, assuming $x$ is such that $\frac{\pi}{2} - x$ is within this range. For instance, if $x \in [0, \pi]$, then $\frac{\pi}{2} - x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

So, $\sin^{-1}(\cos x) = \frac{\pi}{2} - x$.

Now, we integrate this simplified expression:

$\int \left(\frac{\pi}{2} - x\right) \, dx$

We can integrate term by term:

= $\int \frac{\pi}{2} \, dx - \int x \, dx$

The integral of a constant $\frac{\pi}{2}$ is $\frac{\pi}{2}x$.

The integral of $x$ is $\frac{x^{1+1}}{1+1} = \frac{x^2}{2}$.

= $\frac{\pi}{2}x - \frac{x^2}{2} + C$

where $C$ is the constant of integration.

Note: This solution assumes the principal value of $\sin^{-1}(\cos x)$, which is $\frac{\pi}{2} - x$ for $x \in [0, \pi]$. If the domain of $x$ is different, the expression for $\sin^{-1}(\cos x)$ might be different (e.g., piecewise). However, for a general indefinite integral, $\frac{\pi}{2} - x$ is typically used.

The final answer is:

$\boxed{\frac{\pi}{2}x - \frac{x^2}{2} + C}$

Solution:

We need to find the integral of $\frac{1}{\cos (x − a) \cos (x − b)}$ with respect to $x$.

The integral is given by:

$\int \frac{1}{\cos (x − a) \cos (x − b)} \, dx$

We can multiply the numerator and the denominator by a constant factor, say $\sin(b-a)$, assuming $a \neq b$. This will allow us to use trigonometric identities to simplify the expression.

$\int \frac{1}{\sin(b-a)} \frac{\sin(b-a)}{\cos (x − a) \cos (x − b)} \, dx$

We can rewrite the term $\sin(b-a)$ in the numerator using the difference of the arguments in the denominator:

$b - a = (x - a) - (x - b)$

So, $\sin(b-a) = \sin((x-a) - (x-b))$.

Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, where $A = x-a$ and $B = x-b$:

$\sin((x-a) - (x-b)) = \sin(x-a)\cos(x-b) - \cos(x-a)\sin(x-b)$

Substitute this back into the integral:

= $\frac{1}{\sin(b-a)} \int \frac{\sin(x-a)\cos(x-b) - \cos(x-a)\sin(x-b)}{\cos (x − a) \cos (x − b)} \, dx$

Now, split the fraction into two terms:

= $\frac{1}{\sin(b-a)} \int \left[ \frac{\sin(x-a)\cos(x-b)}{\cos (x − a) \cos (x − b)} - \frac{\cos(x-a)\sin(x-b)}{\cos (x − a) \cos (x − b)} \right] \, dx$

Simplify each term:

= $\frac{1}{\sin(b-a)} \int \left[ \frac{\sin(x-a)}{\cos (x − a)} - \frac{\sin(x-b)}{\cos (x − b)} \right] \, dx$

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$:

= $\frac{1}{\sin(b-a)} \int [\tan(x-a) - \tan(x-b)] \, dx$

Now, integrate term by term:

= $\frac{1}{\sin(b-a)} \left[ \int \tan(x-a) \, dx - \int \tan(x-b) \, dx \right]$

Recall the standard integral $\int \tan \theta \, d\theta = -\log_e |\cos \theta| + C'$.

$\int \tan(x-a) \, dx = -\log_e |\cos(x-a)|$ (using substitution $u = x-a$, $du = dx$)

$\int \tan(x-b) \, dx = -\log_e |\cos(x-b)|$ (using substitution $v = x-b$, $dv = dx$)

Substitute these results back into the expression:

= $\frac{1}{\sin(b-a)} \left[ (-\log_e |\cos(x-a)|) - (-\log_e |\cos(x-b)|) \right] + C$

= $\frac{1}{\sin(b-a)} \left[ -\log_e |\cos(x-a)| + \log_e |\cos(x-b)| \right] + C$

Using the logarithm property $\log_e P - \log_e Q = \log_e \frac{P}{Q}$:

= $\frac{1}{\sin(b-a)} \log_e \left| \frac{\cos(x-b)}{\cos(x-a)} \right| + C$

where $C$ is the constant of integration. This result is valid for $a \neq b$. If $a=b$, the integral is $\int \sec^2(x-a) dx = \tan(x-a) + C$.

The final answer (for $a \neq b$) is:

$\boxed{\frac{1}{\sin(b-a)} \log_e \left| \frac{\cos(x-b)}{\cos(x-a)} \right| + C}$

Solution:

We need to find the integral of $\frac{\sin^2 x − \cos^2 x}{\sin^2 x \;\cos^2 x}$ with respect to $x$.

The integral is given by:

$\int \frac{\sin^2 x − \cos^2 x}{\sin^2 x \;\cos^2 x} \, dx$

We can split the fraction by dividing each term in the numerator by the denominator:

$\frac{\sin^2 x − \cos^2 x}{\sin^2 x \;\cos^2 x} = \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x}$

Simplify each term by cancelling common factors:

= $\frac{1}{\cos^2 x} - \frac{1}{\sin^2 x}$

Using reciprocal identities $\sec x = \frac{1}{\cos x}$ and $\text{ cosec } x = \frac{1}{\sin x}$, we get:

= $\sec^2 x - \text{ cosec}^2 x$

Now, integrate this simplified expression:

$\int (\sec^2 x - \text{ cosec}^2 x) \, dx$

We can integrate term by term:

= $\int \sec^2 x \, dx - \int \text{ cosec}^2 x \, dx$

Using the standard integrals $\int \sec^2 x \, dx = \tan x + C_1$ and $\int \text{ cosec}^2 x \, dx = -\cot x + C_2$:

= $\tan x - (-\cot x) + C$

= $\tan x + \cot x + C$

where $C$ is the constant of integration ($C = C_1 + C_2$).

Comparing the result with the given options:

(A) tan x + cot x + C

(B) tan x + cosec x + C

(C) – tan x + cot x + C

(D) tan x + sec x + C

The calculated integral matches option (A).

The final answer is:

$\boxed{\text{tan x + cot x + C}}$

Solution:

We need to find the integral of $\frac{e^x (1 + x)}{\cos^2 (e^x x)}$ with respect to $x$.

The integral is given by:

$\int \frac{e^x (1 + x)}{\cos^2 (e^x x)} \, dx$

We use the substitution method.

Let $u = e^x x$.

Differentiating $u$ with respect to $x$ using the product rule:

$\frac{du}{dx} = \frac{d}{dx}(e^x x) = \frac{d}{dx}(e^x) \cdot x + e^x \cdot \frac{d}{dx}(x)$

= $e^x \cdot x + e^x \cdot 1$

= $e^x(x + 1)$

So, $du = e^x(1+x) \, dx$.

Substitute $u = e^x x$ and $e^x(1+x) \, dx = du$ into the integral:

$\int \frac{1}{\cos^2 u} \, du$

Using the reciprocal identity $\frac{1}{\cos^2 u} = \sec^2 u$, the integral becomes:

= $\int \sec^2 u \, du$

The integral of $\sec^2 u$ is $\tan u$.

= $\tan u + C'$

Substitute back $u = e^x x$:

= $\tan(e^x x) + C$ (where $C$ is the constant of integration)

Comparing the result with the given options:

(A) – cot (ex^x) + C

(B) tan (xe^x) + C

(C) tan (e^x) + C

(D) cot (e^x) + C

Our result $\tan(e^x x) + C$ matches option (B) (since $e^x x$ is the same as $xe^x$).

The final answer is:

$\boxed{\text{tan } (xe^x) + C}$

Solution:

Let's solve each integral separately.

(i) $\int \frac{dx}{x^2 − 16}$

We can write the denominator as $x^2 - 4^2$.

This integral is in the standard form $\int \frac{dx}{x^2 - a^2}$, where $a=4$.

The formula for this integral is $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log_e \left| \frac{x-a}{x+a} \right| + C$.

Substituting $a=4$ into the formula, we get:

$\int \frac{dx}{x^2 − 16} = \frac{1}{2(4)} \log_e \left| \frac{x-4}{x+4} \right| + C$

= $\frac{1}{8} \log_e \left| \frac{x-4}{x+4} \right| + C$

where $C$ is the constant of integration.

(ii) $\int \frac{dx}{\sqrt{2x − x^2}}$

We need to complete the square in the denominator $2x - x^2$.

$2x - x^2 = -(x^2 - 2x)$

To complete the square for $x^2 - 2x$, we add and subtract $\left(\frac{-2}{2}\right)^2 = (-1)^2 = 1$.

$x^2 - 2x = (x^2 - 2x + 1) - 1 = (x-1)^2 - 1$

So, $2x - x^2 = -((x-1)^2 - 1) = 1 - (x-1)^2$.

The integral becomes:

$\int \frac{dx}{\sqrt{1 - (x-1)^2}}$

This integral is in the standard form $\int \frac{du}{\sqrt{a^2 - u^2}}$, where $a=1$ and $u=x-1$.

Let $u = x-1$. Then $du = dx$.

The integral is $\int \frac{du}{\sqrt{1^2 - u^2}}$.

The formula for this integral is $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C$.

Substituting $a=1$ and $u=x-1$, we get:

$\int \frac{dx}{\sqrt{2x − x^2}} = \sin^{-1}\left(\frac{x-1}{1}\right) + C$

= $\sin^{-1}(x-1) + C$

where $C$ is the constant of integration.

Solution:

Let's solve each integral separately.

(i) $\int \frac{dx}{x^2 − 6x + 13}$

We complete the square in the denominator $x^2 - 6x + 13$.

$x^2 - 6x + 13 = (x^2 - 6x + 9) + 13 - 9$

= $(x - 3)^2 + 4$

= $(x - 3)^2 + 2^2$

The integral becomes:

$\int \frac{dx}{(x - 3)^2 + 2^2}$

This integral is in the standard form $\int \frac{du}{u^2 + a^2}$, where $u = x-3$ and $a=2$.

Let $u = x-3$. Then $du = dx$.

The integral is $\int \frac{du}{u^2 + 2^2}$.

The formula for this integral is $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$.

Substituting $u = x-3$ and $a=2$, we get:

$\int \frac{dx}{x^2 − 6x + 13} = \frac{1}{2} \tan^{-1}\left(\frac{x-3}{2}\right) + C$

where $C$ is the constant of integration.

(ii) $\int \frac{dx}{3x^2 + 13x − 10}$

First, factor out 3 from the quadratic in the denominator:

$3x^2 + 13x - 10 = 3\left(x^2 + \frac{13}{3}x - \frac{10}{3}\right)$

The integral becomes:

$\int \frac{dx}{3\left(x^2 + \frac{13}{3}x - \frac{10}{3}\right)} = \frac{1}{3} \int \frac{dx}{x^2 + \frac{13}{3}x - \frac{10}{3}}$

Now, complete the square for $x^2 + \frac{13}{3}x - \frac{10}{3}$.

Add and subtract $\left(\frac{13/3}{2}\right)^2 = \left(\frac{13}{6}\right)^2 = \frac{169}{36}$.

$x^2 + \frac{13}{3}x - \frac{10}{3} = \left(x^2 + \frac{13}{3}x + \frac{169}{36}\right) - \frac{10}{3} - \frac{169}{36}$

= $\left(x + \frac{13}{6}\right)^2 - \left(\frac{10 \cdot 12}{3 \cdot 12} + \frac{169}{36}\right)$

= $\left(x + \frac{13}{6}\right)^2 - \left(\frac{120}{36} + \frac{169}{36}\right)$

= $\left(x + \frac{13}{6}\right)^2 - \frac{289}{36}$

= $\left(x + \frac{13}{6}\right)^2 - \left(\frac{17}{6}\right)^2$

The integral becomes:

$\frac{1}{3} \int \frac{dx}{\left(x + \frac{13}{6}\right)^2 - \left(\frac{17}{6}\right)^2}$

This is in the standard form $\int \frac{du}{u^2 - a^2}$, where $u = x + \frac{13}{6}$ and $a = \frac{17}{6}$.

Let $u = x + \frac{13}{6}$. Then $du = dx$.

The integral is $\frac{1}{3} \int \frac{du}{u^2 - \left(\frac{17}{6}\right)^2}$.

The formula for this integral is $\int \frac{du}{u^2 - a^2} = \frac{1}{2a} \log_e \left| \frac{u-a}{u+a} \right| + C'$.

Substituting $u = x + \frac{13}{6}$ and $a = \frac{17}{6}$, we get:

= $\frac{1}{3} \cdot \frac{1}{2\left(\frac{17}{6}\right)} \log_e \left| \frac{x + \frac{13}{6} - \frac{17}{6}}{x + \frac{13}{6} + \frac{17}{6}} \right| + C$

= $\frac{1}{3} \cdot \frac{1}{\frac{17}{3}} \log_e \left| \frac{x - \frac{4}{6}}{x + \frac{30}{6}} \right| + C$

= $\frac{1}{3} \cdot \frac{3}{17} \log_e \left| \frac{x - \frac{2}{3}}{x + 5} \right| + C$

= $\frac{1}{17} \log_e \left| \frac{\frac{3x-2}{3}}{x + 5} \right| + C$

= $\frac{1}{17} \log_e \left| \frac{3x-2}{3(x + 5)} \right| + C$

Using logarithm properties, $\log_e \left| \frac{3x-2}{3(x + 5)} \right| = \log_e |3x-2| - \log_e |3(x+5)| = \log_e |3x-2| - (\log_e |3| + \log_e |x+5|)$.

So, $\frac{1}{17} (\log_e |3x-2| - \log_e |x+5|) - \frac{1}{17}\log_e |3| + C$. The term $-\frac{1}{17}\log_e |3|$ can be absorbed into the constant $C$.

= $\frac{1}{17} \log_e \left| \frac{3x-2}{x+5} \right| + C$

(iii) $\int \frac{dx}{\sqrt{5x^2 − 2x}}$

First, factor out 5 from the term under the square root:

$5x^2 - 2x = 5\left(x^2 - \frac{2}{5}x\right)$

The integral becomes:

$\int \frac{dx}{\sqrt{5\left(x^2 - \frac{2}{5}x\right)}} = \int \frac{dx}{\sqrt{5} \sqrt{x^2 - \frac{2}{5}x}}$

= $\frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{x^2 - \frac{2}{5}x}}$

Complete the square for $x^2 - \frac{2}{5}x$.

Add and subtract $\left(\frac{-2/5}{2}\right)^2 = \left(-\frac{1}{5}\right)^2 = \frac{1}{25}$.

$x^2 - \frac{2}{5}x = \left(x^2 - \frac{2}{5}x + \frac{1}{25}\right) - \frac{1}{25} = \left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2$

The integral becomes:

= $\frac{1}{\sqrt{5}} \int \frac{dx}{\sqrt{\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2}}$

This is in the standard form $\int \frac{du}{\sqrt{u^2 - a^2}}$, where $u = x - \frac{1}{5}$ and $a = \frac{1}{5}$.

Let $u = x - \frac{1}{5}$. Then $du = dx$.

The integral is $\frac{1}{\sqrt{5}} \int \frac{du}{\sqrt{u^2 - \left(\frac{1}{5}\right)^2}}$.

The formula for this integral is $\int \frac{du}{\sqrt{u^2 - a^2}} = \log_e \left| u + \sqrt{u^2 - a^2} \right| + C''$.

Substituting $u = x - \frac{1}{5}$ and $a = \frac{1}{5}$, we get:

= $\frac{1}{\sqrt{5}} \log_e \left| \left(x - \frac{1}{5}\right) + \sqrt{\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2} \right| + C$

We know that $\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2 = x^2 - \frac{2}{5}x$. So, $\sqrt{\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2} = \sqrt{x^2 - \frac{2}{5}x}$.

Also, $\sqrt{x^2 - \frac{2}{5}x} = \sqrt{\frac{5x^2 - 2x}{5}} = \frac{\sqrt{5x^2 - 2x}}{\sqrt{5}}$.

So the result is:

= $\frac{1}{\sqrt{5}} \log_e \left| x - \frac{1}{5} + \sqrt{x^2 - \frac{2}{5}x} \right| + C$

This can also be written as:

= $\frac{1}{\sqrt{5}} \log_e \left| \frac{5x - 1}{5} + \frac{\sqrt{5x^2 - 2x}}{\sqrt{5}} \right| + C$

= $\frac{1}{\sqrt{5}} \log_e \left| \frac{\sqrt{5}(5x - 1) + 5\sqrt{5x^2 - 2x}}{5\sqrt{5}} \right| + C$

= $\frac{1}{\sqrt{5}} \left( \log_e \left| \sqrt{5}(5x - 1) + 5\sqrt{5x^2 - 2x} \right| - \log_e |5\sqrt{5}| \right) + C$

The term $-\frac{1}{\sqrt{5}}\log_e |5\sqrt{5}|$ can be absorbed into the constant of integration.

Let's simplify the term inside the absolute value:

$\sqrt{5}(5x - 1) + 5\sqrt{5x^2 - 2x} = \sqrt{5}(5x - 1) + 5\sqrt{\frac{5x^2 - 2x}{5} \cdot 5} = \sqrt{5}(5x - 1) + 5\sqrt{5} \sqrt{x^2 - \frac{2}{5}x}$

= $\sqrt{5}(5x - 1) + 5\sqrt{5} \sqrt{\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2}$

= $\sqrt{5} \left[ (5x - 1) + 5 \sqrt{\left(x - \frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^2} \right]$

= $\sqrt{5} \left[ 5x - 1 + 5 \sqrt{x^2 - \frac{2}{5}x} \right]$

= $\sqrt{5} \left[ 5x - 1 + \sqrt{25x^2 - 10x} \right]$

So the integral is:

= $\frac{1}{\sqrt{5}} \log_e \left| \sqrt{5} (5x - 1 + \sqrt{25x^2 - 10x}) \right| + C$

= $\frac{1}{\sqrt{5}} \left( \log_e |\sqrt{5}| + \log_e |5x - 1 + \sqrt{25x^2 - 10x}| \right) + C$

The term $\frac{1}{\sqrt{5}}\log_e |\sqrt{5}|$ can be absorbed into the constant.

Alternatively, we can use the formula $\int \frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}} \log_e \left| x + \frac{b}{2a} + \sqrt{x^2 + \frac{b}{a}x + \frac{c}{a}} \right| + C$ for $a > 0$.

Here, $a=5$, $b=-2$, $c=0$.

= $\frac{1}{\sqrt{5}} \log_e \left| x + \frac{-2}{2(5)} + \sqrt{x^2 + \frac{-2}{5}x + \frac{0}{5}} \right| + C$

= $\frac{1}{\sqrt{5}} \log_e \left| x - \frac{1}{5} + \sqrt{x^2 - \frac{2}{5}x} \right| + C$

This matches the previous result.

Solution:

We solve each integral separately.

(i) $\int \frac{x + 2}{2x^2 + 6x + 5} \;dx$

Let the integral be $I = \int \frac{x + 2}{2x^2 + 6x + 5} \, dx$.

We use the form $\int \frac{px+q}{ax^2+bx+c} \, dx$. The derivative of the denominator is $\frac{d}{dx}(2x^2 + 6x + 5) = 4x + 6$.

We want to write the numerator $x+2$ in the form $A(4x+6) + B$.

$x+2 = 4Ax + 6A + B$

Comparing coefficients of $x$: $4A = 1 \implies A = \frac{1}{4}$.

Comparing constant terms: $6A + B = 2$. Substitute $A = \frac{1}{4}$: $6\left(\frac{1}{4}\right) + B = 2 \implies \frac{3}{2} + B = 2 \implies B = 2 - \frac{3}{2} = \frac{4-3}{2} = \frac{1}{2}$.

So, $x+2 = \frac{1}{4}(4x+6) + \frac{1}{2}$.

Substitute this into the integral:

$I = \int \frac{\frac{1}{4}(4x+6) + \frac{1}{2}}{2x^2 + 6x + 5} \, dx$

= $\int \left( \frac{\frac{1}{4}(4x+6)}{2x^2 + 6x + 5} + \frac{\frac{1}{2}}{2x^2 + 6x + 5} \right) \, dx$

= $\frac{1}{4} \int \frac{4x+6}{2x^2 + 6x + 5} \, dx + \frac{1}{2} \int \frac{dx}{2x^2 + 6x + 5}$

Let the first integral be $I_1$ and the second integral be $I_2$.

$I_1 = \int \frac{4x+6}{2x^2 + 6x + 5} \, dx$. Let $u = 2x^2 + 6x + 5$. Then $du = (4x+6) \, dx$.

$I_1 = \int \frac{du}{u} = \log_e |u| + C_1 = \log_e |2x^2 + 6x + 5| + C_1$.

For $I_2 = \int \frac{dx}{2x^2 + 6x + 5}$, complete the square in the denominator.

$2x^2 + 6x + 5 = 2\left(x^2 + 3x + \frac{5}{2}\right)$

$x^2 + 3x + \frac{5}{2} = \left(x^2 + 3x + \left(\frac{3}{2}\right)^2\right) + \frac{5}{2} - \left(\frac{3}{2}\right)^2$

= $\left(x + \frac{3}{2}\right)^2 + \frac{5}{2} - \frac{9}{4} = \left(x + \frac{3}{2}\right)^2 + \frac{10-9}{4} = \left(x + \frac{3}{2}\right)^2 + \frac{1}{4}$

= $\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2$

So, $2x^2 + 6x + 5 = 2\left(\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2\right)$.

$I_2 = \int \frac{dx}{2\left(\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2\right)} = \frac{1}{2} \int \frac{dx}{\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2}$

This is in the standard form $\int \frac{du}{u^2 + a^2}$, with $u = x + \frac{3}{2}$ ($du=dx$) and $a = \frac{1}{2}$.

$I_2 = \frac{1}{2} \cdot \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C_2 = \frac{1}{2} \cdot \frac{1}{\frac{1}{2}} \tan^{-1}\left(\frac{x + \frac{3}{2}}{\frac{1}{2}}\right) + C_2$

= $\frac{1}{2} \cdot 2 \tan^{-1}\left(\frac{\frac{2x+3}{2}}{\frac{1}{2}}\right) + C_2 = \tan^{-1}(2x+3) + C_2$.

Combining $I_1$ and $I_2$:

$I = \frac{1}{4} I_1 + \frac{1}{2} I_2$

= $\frac{1}{4} \log_e |2x^2 + 6x + 5| + \frac{1}{2} \tan^{-1}(2x+3) + C$ (where $C = \frac{1}{4}C_1 + \frac{1}{2}C_2$)

The quadratic $2x^2 + 6x + 5$ has discriminant $\Delta = b^2 - 4ac = 6^2 - 4(2)(5) = 36 - 40 = -4 < 0$. Since the leading coefficient is positive, the quadratic is always positive, so we can remove the absolute value.

$I = \frac{1}{4} \log_e (2x^2 + 6x + 5) + \frac{1}{2} \tan^{-1}(2x+3) + C$

(ii) $\int \frac{x + 3}{\sqrt{5 − 4x − x^2}} \;dx$

Let the integral be $I = \int \frac{x + 3}{\sqrt{5 − 4x − x^2}} \, dx$.

We use the form $\int \frac{px+q}{\sqrt{ax^2+bx+c}} \, dx$. The derivative of the quadratic inside the square root is $\frac{d}{dx}(5 - 4x - x^2) = -4 - 2x$.

We want to write the numerator $x+3$ in the form $A(-4-2x) + B$.

$x+3 = -4A - 2Ax + B$

Comparing coefficients of $x$: $-2A = 1 \implies A = -\frac{1}{2}$.

Comparing constant terms: $-4A + B = 3$. Substitute $A = -\frac{1}{2}$: $-4\left(-\frac{1}{2}\right) + B = 3 \implies 2 + B = 3 \implies B = 1$.

So, $x+3 = -\frac{1}{2}(-4-2x) + 1$.

Substitute this into the integral:

$I = \int \frac{-\frac{1}{2}(-4-2x) + 1}{\sqrt{5 − 4x − x^2}} \, dx$

= $\int \left( \frac{-\frac{1}{2}(-4-2x)}{\sqrt{5 − 4x − x^2}} + \frac{1}{\sqrt{5 − 4x − x^2}} \right) \, dx$

= $-\frac{1}{2} \int \frac{-4-2x}{\sqrt{5 − 4x − x^2}} \, dx + \int \frac{dx}{\sqrt{5 − 4x − x^2}}$

Let the first integral be $I_1$ and the second integral be $I_2$.

$I_1 = \int \frac{-4-2x}{\sqrt{5 − 4x − x^2}} \, dx$. Let $u = 5 - 4x - x^2$. Then $du = (-4-2x) \, dx$.

$I_1 = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} \, du = \frac{u^{-1/2+1}}{-1/2+1} + C_1 = \frac{u^{1/2}}{1/2} + C_1 = 2\sqrt{u} + C_1 = 2\sqrt{5 - 4x - x^2} + C_1$.

For $I_2 = \int \frac{dx}{\sqrt{5 − 4x − x^2}}$, complete the square in the term under the square root.

$5 - 4x - x^2 = 5 - (x^2 + 4x)$

Complete the square for $x^2 + 4x$: add and subtract $\left(\frac{4}{2}\right)^2 = 2^2 = 4$.

$x^2 + 4x = (x^2 + 4x + 4) - 4 = (x+2)^2 - 4$

So, $5 - (x^2 + 4x) = 5 - ((x+2)^2 - 4) = 5 - (x+2)^2 + 4 = 9 - (x+2)^2 = 3^2 - (x+2)^2$.

$I_2 = \int \frac{dx}{\sqrt{3^2 - (x+2)^2}}$

This is in the standard form $\int \frac{du}{\sqrt{a^2 - u^2}}$, with $u = x+2$ ($du=dx$) and $a=3$.

$I_2 = \sin^{-1}\left(\frac{u}{a}\right) + C_2 = \sin^{-1}\left(\frac{x+2}{3}\right) + C_2$.

Combining $I_1$ and $I_2$:

$I = -\frac{1}{2} I_1 + I_2$

= $-\frac{1}{2} (2\sqrt{5 - 4x - x^2}) + \sin^{-1}\left(\frac{x+2}{3}\right) + C$ (where $C = -\frac{1}{2}C_1 + C_2$)

= $-\sqrt{5 - 4x - x^2} + \sin^{-1}\left(\frac{x+2}{3}\right) + C$

Let the given integral be $I$.

$I = \int \frac{3x^2}{x^6 + 1} dx$

We can rewrite the denominator as $(x^3)^2 + 1$. The integral becomes:

$I = \int \frac{3x^2}{(x^3)^2 + 1} dx$

Let us use the substitution method.

Let $u = x^3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 3x^2$

So, $du = 3x^2 dx$.

Now, substitute $u$ and $du$ into the integral:

$I = \int \frac{du}{u^2 + 1^2}$

This is a standard integral of the form $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.

Here, $x$ is $u$ and $a$ is 1.

So, $I = \frac{1}{1} \tan^{-1}\left(\frac{u}{1}\right) + C$

$I = \tan^{-1}(u) + C$

Substitute back $u = x^3$:

$I = \tan^{-1}(x^3) + C$

Therefore, the integral of the given function is:

$\int \frac{3x^2}{x^6 + 1} dx = \tan^{-1}(x^3) + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{\sqrt{1 + 4x^2}} dx$

We can rewrite the term under the square root as $1^2 + (2x)^2$.

$I = \int \frac{1}{\sqrt{1^2 + (2x)^2}} dx$

Let us use the substitution method.

Let $u = 2x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 2$

So, $du = 2 dx$, which means $dx = \frac{1}{2} du$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{1}{\sqrt{1^2 + u^2}} \left(\frac{1}{2} du\right)$

$I = \frac{1}{2} \int \frac{1}{\sqrt{1^2 + u^2}} du$

This is a standard integral of the form $\int \frac{1}{\sqrt{a^2 + x^2}} dx = \log_e |x + \sqrt{a^2 + x^2}| + C$.

Here, $x$ is $u$ and $a$ is 1.

So, $I = \frac{1}{2} \log_e |u + \sqrt{1^2 + u^2}| + C$

$I = \frac{1}{2} \log_e |u + \sqrt{1 + u^2}| + C$

Substitute back $u = 2x$:

$I = \frac{1}{2} \log_e |2x + \sqrt{1 + (2x)^2}| + C$

$I = \frac{1}{2} \log_e |2x + \sqrt{1 + 4x^2}| + C$

Therefore, the integral of the given function is:

$\int \frac{1}{\sqrt{1 + 4x^2}} dx = \frac{1}{2} \log_e |2x + \sqrt{1 + 4x^2}| + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{\sqrt{(2 − x)^2 + 1}} dx$

We can rewrite the term under the square root as $(2-x)^2 + 1^2$.

$I = \int \frac{1}{\sqrt{(2-x)^2 + 1^2}} dx$

Let us use the substitution method.

Let $u = 2 - x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = -1$

So, $du = -dx$, which means $dx = -du$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{1}{\sqrt{u^2 + 1^2}} (-du)$

$I = - \int \frac{1}{\sqrt{u^2 + 1^2}} du$

This is a standard integral of the form $\int \frac{1}{\sqrt{x^2 + a^2}} dx = \log_e |x + \sqrt{x^2 + a^2}| + C$.

Here, $x$ is $u$ and $a$ is 1.

So, $I = - \left( \log_e |u + \sqrt{u^2 + 1^2}| \right) + C$

$I = - \log_e |u + \sqrt{u^2 + 1}| + C$

Substitute back $u = 2 - x$:

$I = - \log_e |(2 - x) + \sqrt{(2 - x)^2 + 1}| + C$

Alternatively, using properties of logarithms, $- \log_e A = \log_e (A^{-1}) = \log_e \left(\frac{1}{A}\right)$.

$I = \log_e \left| \frac{1}{(2 - x) + \sqrt{(2 - x)^2 + 1}} \right| + C$

Therefore, the integral of the given function is:

$\int \frac{1}{\sqrt{(2 − x)^2 + 1}} dx = - \log_e |(2 - x) + \sqrt{(2 - x)^2 + 1}| + C$

or

$\int \frac{1}{\sqrt{(2 − x)^2 + 1}} dx = \log_e \left| \frac{1}{(2 - x) + \sqrt{(2 - x)^2 + 1}} \right| + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{\sqrt{9 − 25x^2}} dx$

We can rewrite the term under the square root as $3^2 - (5x)^2$.

$I = \int \frac{1}{\sqrt{3^2 - (5x)^2}} dx$

Let us use the substitution method.

Let $u = 5x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 5$

So, $du = 5 dx$, which means $dx = \frac{1}{5} du$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{1}{\sqrt{3^2 - u^2}} \left(\frac{1}{5} du\right)$

$I = \frac{1}{5} \int \frac{1}{\sqrt{3^2 - u^2}} du$

This is a standard integral of the form $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C$.

Here, $x$ is $u$ and $a$ is 3.

So, $I = \frac{1}{5} \sin^{-1}\left(\frac{u}{3}\right) + C$

Substitute back $u = 5x$:

$I = \frac{1}{5} \sin^{-1}\left(\frac{5x}{3}\right) + C$

Therefore, the integral of the given function is:

$\int \frac{1}{\sqrt{9 − 25x^2}} dx = \frac{1}{5} \sin^{-1}\left(\frac{5x}{3}\right) + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{3x}{1 + 2x^4} dx$

We can rewrite the denominator as $1^2 + (\sqrt{2}x^2)^2$. The integral becomes:

$I = \int \frac{3x}{1^2 + (\sqrt{2}x^2)^2} dx$

Let us use the substitution method.

Let $u = \sqrt{2}x^2$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \sqrt{2} (2x) = 2\sqrt{2}x$

So, $du = 2\sqrt{2}x dx$. This means $x dx = \frac{1}{2\sqrt{2}} du$.

The numerator has $3x dx$, so $3x dx = 3 \left(\frac{1}{2\sqrt{2}} du\right) = \frac{3}{2\sqrt{2}} du$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{1}{1^2 + u^2} \left(\frac{3}{2\sqrt{2}} du\right)$

$I = \frac{3}{2\sqrt{2}} \int \frac{1}{1^2 + u^2} du$

This is a standard integral of the form $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.

Here, $x$ is $u$ and $a$ is 1.

So, $I = \frac{3}{2\sqrt{2}} \left( \frac{1}{1} \tan^{-1}\left(\frac{u}{1}\right) \right) + C$

$I = \frac{3}{2\sqrt{2}} \tan^{-1}(u) + C$

Substitute back $u = \sqrt{2}x^2$:

$I = \frac{3}{2\sqrt{2}} \tan^{-1}(\sqrt{2}x^2) + C$

We can rationalize the denominator $\frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{2\sqrt{2}\sqrt{2}} = \frac{3\sqrt{2}}{4}$.

$I = \frac{3\sqrt{2}}{4} \tan^{-1}(\sqrt{2}x^2) + C$

Therefore, the integral of the given function is:

$\int \frac{3x}{1 + 2x^4} dx = \frac{3\sqrt{2}}{4} \tan^{-1}(\sqrt{2}x^2) + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{x^2}{1 − x^6} dx$

We can rewrite the denominator as $1^2 - (x^3)^2$. The integral becomes:

$I = \int \frac{x^2}{1^2 - (x^3)^2} dx$

Let us use the substitution method.

Let $u = x^3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 3x^2$

So, $du = 3x^2 dx$. This means $x^2 dx = \frac{1}{3} du$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{\frac{1}{3} du}{1^2 - u^2}$

$I = \frac{1}{3} \int \frac{1}{1^2 - u^2} du$

This is a standard integral of the form $\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \log_e \left| \frac{a+x}{a-x} \right| + C$.

Here, $x$ is $u$ and $a$ is 1.

So, $I = \frac{1}{3} \left( \frac{1}{2 \times 1} \log_e \left| \frac{1+u}{1-u} \right| \right) + C$

$I = \frac{1}{6} \log_e \left| \frac{1+u}{1-u} \right| + C$

Substitute back $u = x^3$:

$I = \frac{1}{6} \log_e \left| \frac{1+x^3}{1-x^3} \right| + C$

Therefore, the integral of the given function is:

$\int \frac{x^2}{1 − x^6} dx = \frac{1}{6} \log_e \left| \frac{1+x^3}{1-x^3} \right| + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{x − 1}{\sqrt{x^2 − 1}} dx$

We can split the integral into two parts:

$I = \int \frac{x}{\sqrt{x^2 − 1}} dx - \int \frac{1}{\sqrt{x^2 − 1}} dx$

Let's evaluate the first integral, $I_1 = \int \frac{x}{\sqrt{x^2 − 1}} dx$.

Let $u = x^2 - 1$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 2x$

So, $du = 2x dx$, which means $x dx = \frac{1}{2} du$.

Substituting $u$ and $dx$ into $I_1$:

$I_1 = \int \frac{1}{\sqrt{u}} \left(\frac{1}{2} du\right)$

$I_1 = \frac{1}{2} \int u^{-1/2} du$

$I_1 = \frac{1}{2} \left( \frac{u^{-1/2 + 1}}{-1/2 + 1} \right) + C_1$

$I_1 = \frac{1}{2} \left( \frac{u^{1/2}}{1/2} \right) + C_1$

$I_1 = \frac{1}{2} (2 \sqrt{u}) + C_1$

$I_1 = \sqrt{u} + C_1$

Substitute back $u = x^2 - 1$:

$I_1 = \sqrt{x^2 - 1} + C_1$

Now let's evaluate the second integral, $I_2 = - \int \frac{1}{\sqrt{x^2 − 1}} dx$.

This is a standard integral of the form $\int \frac{1}{\sqrt{x^2 - a^2}} dx = \log_e |x + \sqrt{x^2 - a^2}| + C$.

Here, $a=1$.

So, $I_2 = - \left( \log_e |x + \sqrt{x^2 - 1^2}| \right) + C_2$

$I_2 = - \log_e |x + \sqrt{x^2 - 1}| + C_2$

Combining $I_1$ and $I_2$ to get the total integral $I$:

$I = I_1 + I_2$

$I = (\sqrt{x^2 - 1} + C_1) + (- \log_e |x + \sqrt{x^2 - 1}| + C_2)$

$I = \sqrt{x^2 - 1} - \log_e |x + \sqrt{x^2 - 1}| + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

Therefore, the integral of the given function is:

$\int \frac{x − 1}{\sqrt{x^2 − 1}} dx = \sqrt{x^2 - 1} - \log_e |x + \sqrt{x^2 - 1}| + C$

Let the given integral be $I$.

$I = \int \frac{x^2}{\sqrt{x^6 + a^6}} dx$

We can rewrite the terms under the square root as $(x^3)^2 + (a^3)^2$. The integral becomes:

$I = \int \frac{x^2}{\sqrt{(x^3)^2 + (a^3)^2}} dx$

Let us use the substitution method.

Let $u = x^3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 3x^2$

So, $du = 3x^2 dx$. This means $x^2 dx = \frac{1}{3} du$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{\frac{1}{3} du}{\sqrt{u^2 + (a^3)^2}}$

$I = \frac{1}{3} \int \frac{1}{\sqrt{u^2 + (a^3)^2}} du$

This is a standard integral of the form $\int \frac{1}{\sqrt{x^2 + b^2}} dx = \log_e |x + \sqrt{x^2 + b^2}| + C$.

Here, $x$ is $u$ and $b$ is $a^3$.

So, $I = \frac{1}{3} \left( \log_e |u + \sqrt{u^2 + (a^3)^2}| \right) + C$

$I = \frac{1}{3} \log_e |u + \sqrt{u^2 + a^6}| + C$

Substitute back $u = x^3$:

$I = \frac{1}{3} \log_e |x^3 + \sqrt{(x^3)^2 + a^6}| + C$

$I = \frac{1}{3} \log_e |x^3 + \sqrt{x^6 + a^6}| + C$

Therefore, the integral of the given function is:

$\int \frac{x^2}{\sqrt{x^6 + a^6}} dx = \frac{1}{3} \log_e |x^3 + \sqrt{x^6 + a^6}| + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} dx$

Let us use the substitution method.

Let $u = \tan x$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = \sec^2 x$

So, $du = \sec^2 x dx$.

Now, substitute $u$ and $du$ into the integral:

$I = \int \frac{du}{\sqrt{u^2 + 4}}$

We can rewrite the term under the square root as $u^2 + 2^2$.

$I = \int \frac{du}{\sqrt{u^2 + 2^2}}$

This is a standard integral of the form $\int \frac{1}{\sqrt{x^2 + a^2}} dx = \log_e |x + \sqrt{x^2 + a^2}| + C$.

Here, $x$ is $u$ and $a$ is 2.

So, $I = \log_e |u + \sqrt{u^2 + 2^2}| + C$

$I = \log_e |u + \sqrt{u^2 + 4}| + C$

Substitute back $u = \tan x$:

$I = \log_e |\tan x + \sqrt{\tan^2 x + 4}| + C$

Therefore, the integral of the given function is:

$\int \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}} dx = \log_e |\tan x + \sqrt{\tan^2 x + 4}| + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{\sqrt{x^2 + 2x + 2}} dx$

First, we complete the square for the quadratic expression in the denominator:

$x^2 + 2x + 2 = (x^2 + 2x + 1) + 1$

$= (x+1)^2 + 1^2$

Substitute this back into the integral:

$I = \int \frac{1}{\sqrt{(x+1)^2 + 1^2}} dx$

Let us use the substitution method.

Let $u = x+1$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 1$

So, $du = dx$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{1}{\sqrt{u^2 + 1^2}} du$

This is a standard integral of the form $\int \frac{1}{\sqrt{x^2 + a^2}} dx = \log_e |x + \sqrt{x^2 + a^2}| + C$.

Here, $x$ is $u$ and $a$ is 1.

So, $I = \log_e |u + \sqrt{u^2 + 1^2}| + C$

$I = \log_e |u + \sqrt{u^2 + 1}| + C$

Substitute back $u = x+1$:

$I = \log_e |(x+1) + \sqrt{(x+1)^2 + 1}| + C$

Simplify the term under the square root:

$(x+1)^2 + 1 = x^2 + 2x + 1 + 1 = x^2 + 2x + 2$

So, $I = \log_e |x+1 + \sqrt{x^2 + 2x + 2}| + C$

Therefore, the integral of the given function is:

$\int \frac{1}{\sqrt{x^2 + 2x + 2}} dx = \log_e |x+1 + \sqrt{x^2 + 2x + 2}| + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{9x^2 + 6x + 5} dx$

First, we complete the square for the quadratic expression in the denominator.

$9x^2 + 6x + 5 = 9\left(x^2 + \frac{6}{9}x + \frac{5}{9}\right)$

$= 9\left(x^2 + \frac{2}{3}x + \frac{5}{9}\right)$

To complete the square for $x^2 + \frac{2}{3}x$, we add and subtract $\left(\frac{1}{2} \times \frac{2}{3}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$.

$= 9\left(x^2 + \frac{2}{3}x + \frac{1}{9} - \frac{1}{9} + \frac{5}{9}\right)$

$= 9\left(\left(x + \frac{1}{3}\right)^2 + \frac{4}{9}\right)$

$= 9\left(x + \frac{1}{3}\right)^2 + 9 \times \frac{4}{9}$

$= 9\left(x + \frac{1}{3}\right)^2 + 4$

We can rewrite $9\left(x + \frac{1}{3}\right)^2$ as $\left(3\left(x + \frac{1}{3}\right)\right)^2 = (3x + 1)^2$.

So, the denominator is $(3x + 1)^2 + 4 = (3x+1)^2 + 2^2$.

Substitute this back into the integral:

$I = \int \frac{1}{(3x + 1)^2 + 2^2} dx$

Let us use the substitution method.

Let $u = 3x + 1$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 3$

So, $du = 3 dx$, which means $dx = \frac{1}{3} du$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{1}{u^2 + 2^2} \left(\frac{1}{3} du\right)$

$I = \frac{1}{3} \int \frac{1}{u^2 + 2^2} du$

This is a standard integral of the form $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.

Here, $x$ is $u$ and $a$ is 2.

So, $I = \frac{1}{3} \left( \frac{1}{2} \tan^{-1}\left(\frac{u}{2}\right) \right) + C$

$I = \frac{1}{6} \tan^{-1}\left(\frac{u}{2}\right) + C$

Substitute back $u = 3x + 1$:

$I = \frac{1}{6} \tan^{-1}\left(\frac{3x + 1}{2}\right) + C$

Therefore, the integral of the given function is:

$\int \frac{1}{9x^2 + 6x + 5} dx = \frac{1}{6} \tan^{-1}\left(\frac{3x + 1}{2}\right) + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{\sqrt{7 − 6x − x^2}} dx$

First, we complete the square for the quadratic expression in the denominator:

$7 - 6x - x^2 = - (x^2 + 6x - 7)$

Complete the square for $x^2 + 6x$ by adding and subtracting $\left(\frac{6}{2}\right)^2 = 3^2 = 9$:

$x^2 + 6x - 7 = (x^2 + 6x + 9) - 9 - 7$

$= (x+3)^2 - 16$

So, $7 - 6x - x^2 = - ((x+3)^2 - 16) = 16 - (x+3)^2$.

We can write $16$ as $4^2$.

The denominator is $\sqrt{4^2 - (x+3)^2}$.

Substitute this back into the integral:

$I = \int \frac{1}{\sqrt{4^2 - (x+3)^2}} dx$

Let us use the substitution method.

Let $u = x+3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 1$

So, $du = dx$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{1}{\sqrt{4^2 - u^2}} du$

This is a standard integral of the form $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C$.

Here, $x$ is $u$ and $a$ is 4.

So, $I = \sin^{-1}\left(\frac{u}{4}\right) + C$

Substitute back $u = x+3$:

$I = \sin^{-1}\left(\frac{x+3}{4}\right) + C$

Therefore, the integral of the given function is:

$\int \frac{1}{\sqrt{7 − 6x − x^2}} dx = \sin^{-1}\left(\frac{x+3}{4}\right) + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{\sqrt{(x − 1) (x − 2)}} dx$

First, expand the expression inside the square root:

$(x − 1) (x − 2) = x^2 - 2x - x + 2 = x^2 - 3x + 2$

The integral becomes:

$I = \int \frac{1}{\sqrt{x^2 - 3x + 2}} dx$

Next, we complete the square for the quadratic expression $x^2 - 3x + 2$.

The coefficient of $x$ is -3. Half of this coefficient is $-\frac{3}{2}$. Squaring it gives $\left(-\frac{3}{2}\right)^2 = \frac{9}{4}$.

$x^2 - 3x + 2 = \left(x^2 - 3x + \frac{9}{4}\right) - \frac{9}{4} + 2$

$= \left(x - \frac{3}{2}\right)^2 + \frac{-9 + 8}{4}$

$= \left(x - \frac{3}{2}\right)^2 - \frac{1}{4}$

We can write $\frac{1}{4}$ as $\left(\frac{1}{2}\right)^2$.

So, $x^2 - 3x + 2 = \left(x - \frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2$.

Substitute this back into the integral:

$I = \int \frac{1}{\sqrt{\left(x - \frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} dx$

This is a standard integral of the form $\int \frac{1}{\sqrt{u^2 - a^2}} du = \log_e |u + \sqrt{u^2 - a^2}| + C$.

Here, $u = x - \frac{3}{2}$ and $a = \frac{1}{2}$.

Note that if we substitute $u = x - \frac{3}{2}$, then $du = dx$.

So, the integral is:

$I = \log_e \left| \left(x - \frac{3}{2}\right) + \sqrt{\left(x - \frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2} \right| + C$

Substitute back the original quadratic expression inside the square root:

$\left(x - \frac{3}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = x^2 - 3x + 2 = (x-1)(x-2)$

So, $I = \log_e \left| x - \frac{3}{2} + \sqrt{x^2 - 3x + 2} \right| + C$

or

$I = \log_e \left| x - \frac{3}{2} + \sqrt{(x-1)(x-2)} \right| + C$

Therefore, the integral of the given function is:

$\int \frac{1}{\sqrt{(x − 1) (x − 2)}} dx = \log_e \left| x - \frac{3}{2} + \sqrt{(x-1)(x-2)} \right| + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{\sqrt{8 + 3x − x^2}} dx$

First, we complete the square for the quadratic expression under the square root:

$8 + 3x - x^2 = - (x^2 - 3x - 8)$

To complete the square for $x^2 - 3x$, we add and subtract $\left(\frac{-3}{2}\right)^2 = \frac{9}{4}$.

$x^2 - 3x - 8 = \left(x^2 - 3x + \frac{9}{4}\right) - \frac{9}{4} - 8$

$= \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{32}{4}$

$= \left(x - \frac{3}{2}\right)^2 - \frac{41}{4}$

Now, substitute back the negative sign:

$8 + 3x - x^2 = - \left( \left(x - \frac{3}{2}\right)^2 - \frac{41}{4} \right) = \frac{41}{4} - \left(x - \frac{3}{2}\right)^2$

We can write $\frac{41}{4}$ as $\left(\frac{\sqrt{41}}{2}\right)^2$.

So, the expression under the square root is $\left(\frac{\sqrt{41}}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2$.

Substitute this back into the integral:

$I = \int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2}} dx$

This integral is in the form $\int \frac{1}{\sqrt{a^2 - u^2}} du$, where $a = \frac{\sqrt{41}}{2}$ and $u = x - \frac{3}{2}$.

Let $u = x - \frac{3}{2}$. Then $du = dx$.

Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}\left(\frac{u}{a}\right) + C$, we get:

$I = \sin^{-1}\left(\frac{x - \frac{3}{2}}{\frac{\sqrt{41}}{2}}\right) + C$

$I = \sin^{-1}\left(\frac{\frac{2x - 3}{2}}{\frac{\sqrt{41}}{2}}\right) + C$

$I = \sin^{-1}\left(\frac{2x - 3}{\sqrt{41}}\right) + C$

Therefore, the integral of the given function is:

$\int \frac{1}{\sqrt{8 + 3x − x^2}} dx = \sin^{-1}\left(\frac{2x - 3}{\sqrt{41}}\right) + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{1}{\sqrt{(x − a) (x − b)}} dx$

First, we expand the expression inside the square root:

$(x − a) (x − b) = x^2 - bx - ax + ab = x^2 - (a+b)x + ab$

The integral becomes:

$I = \int \frac{1}{\sqrt{x^2 - (a+b)x + ab}} dx$

Next, we complete the square for the quadratic expression $x^2 - (a+b)x + ab$.

The coefficient of $x$ is $-(a+b)$. Half of this coefficient is $-\frac{a+b}{2}$. Squaring it gives $\left(-\frac{a+b}{2}\right)^2 = \left(\frac{a+b}{2}\right)^2$.

$x^2 - (a+b)x + ab = \left(x^2 - (a+b)x + \left(\frac{a+b}{2}\right)^2\right) - \left(\frac{a+b}{2}\right)^2 + ab$

$= \left(x - \frac{a+b}{2}\right)^2 - \frac{(a+b)^2 - 4ab}{4}$

$= \left(x - \frac{a+b}{2}\right)^2 - \frac{a^2 + 2ab + b^2 - 4ab}{4}$

$= \left(x - \frac{a+b}{2}\right)^2 - \frac{a^2 - 2ab + b^2}{4}$

$= \left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2$

Substitute this back into the integral:

$I = \int \frac{1}{\sqrt{\left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2}} dx$

This is a standard integral of the form $\int \frac{1}{\sqrt{u^2 - c^2}} du = \log_e |u + \sqrt{u^2 - c^2}| + C$.

Here, $u = x - \frac{a+b}{2}$ and $c = \frac{a-b}{2}$ (assuming $a>b$, or we can use $\left|\frac{a-b}{2}\right|$). The substitution $u = x - \frac{a+b}{2}$ gives $du = dx$.

Using the standard integral formula:

$I = \log_e \left| \left(x - \frac{a+b}{2}\right) + \sqrt{\left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2} \right| + C$

Substitute back the original quadratic expression inside the square root:

$\left(x - \frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2 = x^2 - (a+b)x + ab = (x-a)(x-b)$

So, $I = \log_e \left| x - \frac{a+b}{2} + \sqrt{(x-a)(x-b)} \right| + C$

Therefore, the integral of the given function is:

$\int \frac{1}{\sqrt{(x − a) (x − b)}} dx = \log_e \left| x - \frac{a+b}{2} + \sqrt{(x-a)(x-b)} \right| + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{4x + 1}{\sqrt{2x^2 + x - 3}} dx$

Observe the numerator and the expression under the square root.

Let the expression under the square root be $f(x) = 2x^2 + x - 3$.

The derivative of $f(x)$ with respect to $x$ is $f'(x) = \frac{d}{dx}(2x^2 + x - 3) = 4x + 1$.

The numerator is exactly the derivative of the expression under the square root.

This integral is of the form $\int \frac{f'(x)}{\sqrt{f(x)}} dx$.

Let us use the substitution method.

Let $u = 2x^2 + x - 3$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 4x + 1$

So, $du = (4x + 1) dx$.

Now, substitute $u$ and $du$ into the integral:

$I = \int \frac{du}{\sqrt{u}}$

$I = \int u^{-1/2} du$

Using the power rule for integration, $\int u^n du = \frac{u^{n+1}}{n+1} + C$ (for $n \neq -1$).

Here, $n = -1/2$.

$I = \frac{u^{-1/2 + 1}}{-1/2 + 1} + C$

$I = \frac{u^{1/2}}{1/2} + C$

$I = 2u^{1/2} + C$

$I = 2\sqrt{u} + C$

Substitute back $u = 2x^2 + x - 3$:

$I = 2\sqrt{2x^2 + x - 3} + C$

Therefore, the integral of the given function is:

$\int \frac{4x + 1}{\sqrt{2x^2 + x - 3}} dx = 2\sqrt{2x^2 + x - 3} + C$

where $C$ is the constant of integration.

Let the given integral be $I$.

$I = \int \frac{x + 2}{\sqrt{x^2 − 1}} dx$

We can split the integral into two parts:

$I = \int \frac{x}{\sqrt{x^2 − 1}} dx + \int \frac{2}{\sqrt{x^2 − 1}} dx$

Let's evaluate the first integral, $I_1 = \int \frac{x}{\sqrt{x^2 − 1}} dx$.

Let $u = x^2 - 1$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 2x$

So, $du = 2x dx$, which means $x dx = \frac{1}{2} du$.

Substituting $u$ and $dx$ into $I_1$:

$I_1 = \int \frac{1}{\sqrt{u}} \left(\frac{1}{2} du\right)$

$I_1 = \frac{1}{2} \int u^{-1/2} du$

$I_1 = \frac{1}{2} \left( \frac{u^{-1/2 + 1}}{-1/2 + 1} \right) + C_1$

$I_1 = \frac{1}{2} \left( \frac{u^{1/2}}{1/2} \right) + C_1$

$I_1 = \frac{1}{2} (2 \sqrt{u}) + C_1$

$I_1 = \sqrt{u} + C_1$

Substitute back $u = x^2 - 1$:

$I_1 = \sqrt{x^2 - 1} + C_1$

Now let's evaluate the second integral, $I_2 = \int \frac{2}{\sqrt{x^2 − 1}} dx$.

$I_2 = 2 \int \frac{1}{\sqrt{x^2 − 1^2}} dx$

This is a standard integral of the form $\int \frac{1}{\sqrt{x^2 - a^2}} dx = \log_e |x + \sqrt{x^2 - a^2}| + C$.

Here, $a=1$.

So, $I_2 = 2 \left( \log_e |x + \sqrt{x^2 - 1^2}| \right) + C_2$

$I_2 = 2 \log_e |x + \sqrt{x^2 - 1}| + C_2$

Combining $I_1$ and $I_2$ to get the total integral $I$: a

$I = I_1 + I_2$

$I = (\sqrt{x^2 - 1} + C_1) + (2 \log_e |x + \sqrt{x^2 - 1}| + C_2)$

$I = \sqrt{x^2 - 1} + 2 \log_e |x + \sqrt{x^2 - 1}| + (C_1 + C_2)$

Let $C = C_1 + C_2$ be the constant of integration.

Therefore, the integral of the given function is:

$\int \frac{x + 2}{\sqrt{x^2 − 1}} dx = \sqrt{x^2 - 1} + 2 \log_e |x + \sqrt{x^2 - 1}| + C$

Let the given integral be $I$.

$I = \int \frac{5x − 2}{3x^2 + 2x + 1} dx$

We write the numerator $5x - 2$ in terms of the derivative of the denominator $3x^2 + 2x + 1$.

The derivative of $3x^2 + 2x + 1$ is $\frac{d}{dx}(3x^2 + 2x + 1) = 6x + 2$.

We need to find constants $A$ and $B$ such that:

$5x - 2 = A(6x + 2) + B$

$5x - 2 = 6Ax + 2A + B$

Comparing the coefficients of $x$ on both sides:

$5 = 6A \implies A = \frac{5}{6}$

Comparing the constant terms on both sides:

$-2 = 2A + B$

Substitute $A = \frac{5}{6}$:

$-2 = 2\left(\frac{5}{6}\right) + B$

$-2 = \frac{10}{6} + B$

$-2 = \frac{5}{3} + B$

$B = -2 - \frac{5}{3} = -\frac{6}{3} - \frac{5}{3} = -\frac{11}{3}$

So, we can write the numerator as $5x - 2 = \frac{5}{6}(6x + 2) - \frac{11}{3}$.

The integral becomes:

$I = \int \frac{\frac{5}{6}(6x + 2) - \frac{11}{3}}{3x^2 + 2x + 1} dx$

$I = \frac{5}{6} \int \frac{6x + 2}{3x^2 + 2x + 1} dx - \frac{11}{3} \int \frac{1}{3x^2 + 2x + 1} dx$

Let's evaluate the first integral: $I_1 = \int \frac{6x + 2}{3x^2 + 2x + 1} dx$.

Let $u = 3x^2 + 2x + 1$. Then $du = (6x + 2) dx$.

$I_1 = \int \frac{du}{u} = \log_e |u| + C_1$

Since the discriminant of $3x^2 + 2x + 1$ is $2^2 - 4(3)(1) = 4 - 12 = -8 < 0$ and the coefficient of $x^2$ is positive, the quadratic is always positive. So, $|3x^2 + 2x + 1| = 3x^2 + 2x + 1$.

$I_1 = \log_e (3x^2 + 2x + 1) + C_1$

Now, let's evaluate the second integral: $I_2 = \int \frac{1}{3x^2 + 2x + 1} dx$.

We complete the square for the denominator $3x^2 + 2x + 1$:

$3x^2 + 2x + 1 = 3\left(x^2 + \frac{2}{3}x + \frac{1}{3}\right)$

$= 3\left(\left(x + \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2 + \frac{1}{3}\right)$

$= 3\left(\left(x + \frac{1}{3}\right)^2 - \frac{1}{9} + \frac{3}{9}\right)$

$= 3\left(\left(x + \frac{1}{3}\right)^2 + \frac{2}{9}\right)$

$= 3\left(x + \frac{1}{3}\right)^2 + \frac{2}{3}$

$= (\sqrt{3}(x + \frac{1}{3}))^2 + \left(\sqrt{\frac{2}{3}}\right)^2$

Substitute this back into $I_2$:

$I_2 = \int \frac{1}{3\left(x + \frac{1}{3}\right)^2 + \frac{2}{3}} dx = \frac{1}{3} \int \frac{1}{\left(x + \frac{1}{3}\right)^2 + \frac{2}{9}} dx$

$I_2 = \frac{1}{3} \int \frac{1}{\left(x + \frac{1}{3}\right)^2 + \left(\frac{\sqrt{2}}{3}\right)^2} dx$

This integral is of the form $\int \frac{1}{u^2 + a^2} du$, where $u = x + \frac{1}{3}$ and $a = \frac{\sqrt{2}}{3}$. Note that $du = dx$.

Using the standard integral $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C_2$, we get:

$I_2 = \frac{1}{3} \left( \frac{1}{\frac{\sqrt{2}}{3}} \tan^{-1}\left(\frac{x + \frac{1}{3}}{\frac{\sqrt{2}}{3}}\right) \right) + C_2$

$I_2 = \frac{1}{3} \left( \frac{3}{\sqrt{2}} \tan^{-1}\left(\frac{\frac{3x + 1}{3}}{\frac{\sqrt{2}}{3}}\right) \right) + C_2$

$I_2 = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{3x + 1}{\sqrt{2}}\right) + C_2$

Combining the results for $I_1$ and $I_2$ for the total integral $I = \frac{5}{6} I_1 - \frac{11}{3} I_2$:

$I = \frac{5}{6} (\log_e (3x^2 + 2x + 1) + C_1) - \frac{11}{3} \left( \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{3x + 1}{\sqrt{2}}\right) + C_2 \right)$

$I = \frac{5}{6} \log_e (3x^2 + 2x + 1) - \frac{11}{3\sqrt{2}} \tan^{-1}\left(\frac{3x + 1}{\sqrt{2}}\right) + \frac{5}{6}C_1 - \frac{11}{3}C_2$

Let $C = \frac{5}{6}C_1 - \frac{11}{3}C_2$ be the constant of integration.

We can rationalize $\frac{11}{3\sqrt{2}} = \frac{11\sqrt{2}}{3\sqrt{2}\sqrt{2}} = \frac{11\sqrt{2}}{6}$.

Therefore, the integral of the given function is:

$\int \frac{5x − 2}{1 + 2x + 3x^2} dx = \frac{5}{6} \log_e (3x^2 + 2x + 1) - \frac{11\sqrt{2}}{6} \tan^{-1}\left(\frac{3x + 1}{\sqrt{2}}\right) + C$

Let the given integral be $I$.

$I = \int \frac{6x + 7}{\sqrt{(x − 5)(x − 4)}} dx$

First, we expand the expression inside the square root:

$(x − 5) (x − 4) = x^2 - 4x - 5x + 20 = x^2 - 9x + 20$

The integral becomes:

$I = \int \frac{6x + 7}{\sqrt{x^2 - 9x + 20}} dx$

We write the numerator $6x + 7$ in terms of the derivative of the quadratic expression under the square root, $x^2 - 9x + 20$.

The derivative of $x^2 - 9x + 20$ is $\frac{d}{dx}(x^2 - 9x + 20) = 2x - 9$.

We need to find constants $A$ and $B$ such that:

$6x + 7 = A(2x - 9) + B$

$6x + 7 = 2Ax - 9A + B$

Comparing the coefficients of $x$ on both sides:

$6 = 2A \implies A = 3$

Comparing the constant terms on both sides:

$7 = -9A + B$

Substitute $A = 3$:

$7 = -9(3) + B$

$7 = -27 + B$

$B = 7 + 27 = 34$

So, we can write the numerator as $6x + 7 = 3(2x - 9) + 34$.

The integral becomes:

$I = \int \frac{3(2x - 9) + 34}{\sqrt{x^2 - 9x + 20}} dx$

We split the integral into two parts:

$I = 3 \int \frac{2x - 9}{\sqrt{x^2 - 9x + 20}} dx + 34 \int \frac{1}{\sqrt{x^2 - 9x + 20}} dx$

Let's evaluate the first integral: $I_1 = 3 \int \frac{2x - 9}{\sqrt{x^2 - 9x + 20}} dx$.

Let $u = x^2 - 9x + 20$. Then $du = (2x - 9) dx$.

$I_1 = 3 \int \frac{du}{\sqrt{u}} = 3 \int u^{-1/2} du$

$I_1 = 3 \left(\frac{u^{-1/2 + 1}}{-1/2 + 1}\right) + C_1 = 3 \left(\frac{u^{1/2}}{1/2}\right) + C_1$

$I_1 = 6\sqrt{u} + C_1$

Substitute back $u = x^2 - 9x + 20$:

$I_1 = 6\sqrt{x^2 - 9x + 20} + C_1$

Now, let's evaluate the second integral: $I_2 = 34 \int \frac{1}{\sqrt{x^2 - 9x + 20}} dx$.

We complete the square for the quadratic expression $x^2 - 9x + 20$:

$x^2 - 9x + 20 = x^2 - 9x + \left(\frac{9}{2}\right)^2 - \left(\frac{9}{2}\right)^2 + 20$

$= \left(x - \frac{9}{2}\right)^2 - \frac{81}{4} + \frac{80}{4}$

$= \left(x - \frac{9}{2}\right)^2 - \frac{1}{4} = \left(x - \frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2$

Substitute this back into $I_2$:

$I_2 = 34 \int \frac{1}{\sqrt{\left(x - \frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2}} dx$

This is a standard integral of the form $\int \frac{1}{\sqrt{u^2 - a^2}} du = \log_e |u + \sqrt{u^2 - a^2}| + C$.

Here, $u = x - \frac{9}{2}$ and $a = \frac{1}{2}$. The substitution $u = x - \frac{9}{2}$ gives $du = dx$.

Using the standard integral formula:

$I_2 = 34 \log_e \left| \left(x - \frac{9}{2}\right) + \sqrt{\left(x - \frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2} \right| + C_2$

Substitute back the original quadratic expression inside the square root:

$\left(x - \frac{9}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = x^2 - 9x + 20 = (x-5)(x-4)$

So, $I_2 = 34 \log_e \left| x - \frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + C_2$

or

$I_2 = 34 \log_e \left| x - \frac{9}{2} + \sqrt{(x-5)(x-4)} \right| + C_2$

Combining the results for $I_1$ and $I_2$ for the total integral $I = I_1 + I_2$:

$I = (6\sqrt{x^2 - 9x + 20} + C_1) + (34 \log_e \left| x - \frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + C_2)$

$I = 6\sqrt{x^2 - 9x + 20} + 34 \log_e \left| x - \frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral of the given function is:

$\int \frac{6x + 7}{\sqrt{(x − 5)(x − 4)}} dx = 6\sqrt{x^2 - 9x + 20} + 34 \log_e \left| x - \frac{9}{2} + \sqrt{x^2 - 9x + 20} \right| + C$

Let the given integral be $I$.

$I = \int \frac{x + 2}{\sqrt{4x − x^2}} dx$

We write the numerator $x + 2$ in terms of the derivative of the quadratic expression under the square root, $4x - x^2$.

The derivative of $4x - x^2$ is $\frac{d}{dx}(4x - x^2) = 4 - 2x$.

We need to find constants $A$ and $B$ such that:

$x + 2 = A(4 - 2x) + B$

$x + 2 = 4A - 2Ax + B$

$x + 2 = -2Ax + (4A + B)$

Comparing the coefficients of $x$ on both sides:

$1 = -2A \implies A = -\frac{1}{2}$

Comparing the constant terms on both sides:

$2 = 4A + B$

Substitute $A = -\frac{1}{2}$:

$2 = 4\left(-\frac{1}{2}\right) + B$

$2 = -2 + B$

$B = 2 + 2 = 4$

So, we can write the numerator as $x + 2 = -\frac{1}{2}(4 - 2x) + 4$.

The integral becomes:

$I = \int \frac{-\frac{1}{2}(4 - 2x) + 4}{\sqrt{4x - x^2}} dx$

We split the integral into two parts:

$I = -\frac{1}{2} \int \frac{4 - 2x}{\sqrt{4x - x^2}} dx + 4 \int \frac{1}{\sqrt{4x - x^2}} dx$

Let's evaluate the first integral: $I_1 = -\frac{1}{2} \int \frac{4 - 2x}{\sqrt{4x - x^2}} dx$.

Let $u = 4x - x^2$. Differentiating both sides with respect to $x$, we get $du = (4 - 2x) dx$.

$I_1 = -\frac{1}{2} \int \frac{du}{\sqrt{u}} = -\frac{1}{2} \int u^{-1/2} du$

$I_1 = -\frac{1}{2} \left(\frac{u^{-1/2 + 1}}{-1/2 + 1}\right) + C_1 = -\frac{1}{2} \left(\frac{u^{1/2}}{1/2}\right) + C_1$

$I_1 = -\sqrt{u} + C_1$

Substitute back $u = 4x - x^2$:

$I_1 = -\sqrt{4x - x^2} + C_1$

Now, let's evaluate the second integral: $I_2 = 4 \int \frac{1}{\sqrt{4x - x^2}} dx$.

We complete the square for the quadratic expression $4x - x^2$:

$4x - x^2 = - (x^2 - 4x)$

$= - (x^2 - 4x + 4 - 4)$

$= - ((x - 2)^2 - 4)$

$= 4 - (x - 2)^2$

So, the expression under the square root is $2^2 - (x-2)^2$.

Substitute this back into $I_2$:

$I_2 = 4 \int \frac{1}{\sqrt{2^2 - (x - 2)^2}} dx$

This is a standard integral of the form $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}\left(\frac{u}{a}\right) + C$.

Here, $u = x - 2$ and $a = 2$. The substitution $u = x - 2$ gives $du = dx$.

Using the standard integral formula:

$I_2 = 4 \sin^{-1}\left(\frac{x - 2}{2}\right) + C_2$

Combining the results for $I_1$ and $I_2$ for the total integral $I = I_1 + I_2$:

$I = (-\sqrt{4x - x^2} + C_1) + (4 \sin^{-1}\left(\frac{x - 2}{2}\right) + C_2)$

$I = -\sqrt{4x - x^2} + 4 \sin^{-1}\left(\frac{x - 2}{2}\right) + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral of the given function is:

$\int \frac{x + 2}{\sqrt{4x − x^2}} dx = -\sqrt{4x - x^2} + 4 \sin^{-1}\left(\frac{x - 2}{2}\right) + C$

Let the given integral be $I$.

$I = \int \frac{x + 2}{\sqrt{x^2 + 2x + 3}} dx$

We write the numerator $x + 2$ in terms of the derivative of the quadratic expression under the square root, $x^2 + 2x + 3$.

The derivative of $x^2 + 2x + 3$ is $\frac{d}{dx}(x^2 + 2x + 3) = 2x + 2$.

We need to find constants $A$ and $B$ such that:

$x + 2 = A(2x + 2) + B$

$x + 2 = 2Ax + 2A + B$

Comparing the coefficients of $x$ on both sides:

$1 = 2A \implies A = \frac{1}{2}$

Comparing the constant terms on both sides:

$2 = 2A + B$

Substitute $A = \frac{1}{2}$:

$2 = 2\left(\frac{1}{2}\right) + B$

$2 = 1 + B$

$B = 2 - 1 = 1$

So, we can write the numerator as $x + 2 = \frac{1}{2}(2x + 2) + 1$.

The integral becomes:

$I = \int \frac{\frac{1}{2}(2x + 2) + 1}{\sqrt{x^2 + 2x + 3}} dx$

We split the integral into two parts:

$I = \frac{1}{2} \int \frac{2x + 2}{\sqrt{x^2 + 2x + 3}} dx + \int \frac{1}{\sqrt{x^2 + 2x + 3}} dx$

Let's evaluate the first integral: $I_1 = \frac{1}{2} \int \frac{2x + 2}{\sqrt{x^2 + 2x + 3}} dx$.

Let $u = x^2 + 2x + 3$. Differentiating both sides with respect to $x$, we get $du = (2x + 2) dx$.

$I_1 = \frac{1}{2} \int \frac{du}{\sqrt{u}} = \frac{1}{2} \int u^{-1/2} du$

$I_1 = \frac{1}{2} \left(\frac{u^{-1/2 + 1}}{-1/2 + 1}\right) + C_1 = \frac{1}{2} \left(\frac{u^{1/2}}{1/2}\right) + C_1$

$I_1 = \sqrt{u} + C_1$

Substitute back $u = x^2 + 2x + 3$:

$I_1 = \sqrt{x^2 + 2x + 3} + C_1$

Now, let's evaluate the second integral: $I_2 = \int \frac{1}{\sqrt{x^2 + 2x + 3}} dx$.

We complete the square for the quadratic expression $x^2 + 2x + 3$:

$x^2 + 2x + 3 = (x^2 + 2x + 1) + 2$

$= (x+1)^2 + (\sqrt{2})^2$

Substitute this back into $I_2$:

$I_2 = \int \frac{1}{\sqrt{(x+1)^2 + (\sqrt{2})^2}} dx$

This is a standard integral of the form $\int \frac{1}{\sqrt{u^2 + a^2}} du = \log_e |u + \sqrt{u^2 + a^2}| + C$.

Here, $u = x+1$ and $a = \sqrt{2}$. The substitution $u = x+1$ gives $du = dx$.

Using the standard integral formula:

$I_2 = \log_e |(x+1) + \sqrt{(x+1)^2 + (\sqrt{2})^2}| + C_2$

Substitute back the original quadratic expression inside the square root:

$(x+1)^2 + (\sqrt{2})^2 = x^2 + 2x + 1 + 2 = x^2 + 2x + 3$

So, $I_2 = \log_e |x+1 + \sqrt{x^2 + 2x + 3}| + C_2$

Combining the results for $I_1$ and $I_2$ for the total integral $I = I_1 + I_2$:

$I = (\sqrt{x^2 + 2x + 3} + C_1) + (\log_e |x+1 + \sqrt{x^2 + 2x + 3}| + C_2)$

$I = \sqrt{x^2 + 2x + 3} + \log_e |x+1 + \sqrt{x^2 + 2x + 3}| + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral of the given function is:

$\int \frac{x + 2}{\sqrt{x^2 + 2x + 3}} dx = \sqrt{x^2 + 2x + 3} + \log_e |x+1 + \sqrt{x^2 + 2x + 3}| + C$

Let the given integral be $I$.

$I = \int \frac{x + 3}{x^2 − 2x − 5} dx$

We write the numerator $x + 3$ in terms of the derivative of the denominator $x^2 - 2x - 5$.

The derivative of $x^2 - 2x - 5$ is $\frac{d}{dx}(x^2 - 2x - 5) = 2x - 2$.

We need to find constants $A$ and $B$ such that:

$x + 3 = A(2x - 2) + B$

$x + 3 = 2Ax - 2A + B$

Comparing the coefficients of $x$ on both sides:

From (i), $A = \frac{1}{2}$.

Comparing the constant terms on both sides:

Substitute $A = \frac{1}{2}$ into (ii):

So, we can write the numerator as $x + 3 = \frac{1}{2}(2x - 2) + 4$.

The integral becomes:

$I = \int \frac{\frac{1}{2}(2x - 2) + 4}{x^2 - 2x - 5} dx$

We split the integral into two parts:

$I = \frac{1}{2} \int \frac{2x - 2}{x^2 - 2x - 5} dx + 4 \int \frac{1}{x^2 - 2x - 5} dx$

$I = I_1 + I_2$ (say)

Let's evaluate the first integral: $I_1 = \frac{1}{2} \int \frac{2x - 2}{x^2 - 2x - 5} dx$.

Let $u = x^2 - 2x - 5$. Differentiating with respect to $x$, $du = (2x - 2) dx$.

$I_1 = \frac{1}{2} \int \frac{du}{u}$

$I_1 = \frac{1}{2} \log_e |u| + C_1$

$I_1 = \frac{1}{2} \log_e |x^2 - 2x - 5| + C_1$

Now, let's evaluate the second integral: $I_2 = 4 \int \frac{1}{x^2 - 2x - 5} dx$.

We complete the square for the denominator $x^2 - 2x - 5$:

$x^2 - 2x - 5 = (x^2 - 2x + 1) - 1 - 5$

$= (x - 1)^2 - 6$

$= (x - 1)^2 - (\sqrt{6})^2$

Substitute this back into $I_2$:

$I_2 = 4 \int \frac{1}{(x - 1)^2 - (\sqrt{6})^2} dx$

This is a standard integral of the form $\int \frac{1}{u^2 - a^2} du = \frac{1}{2a} \log_e \left| \frac{u-a}{u+a} \right| + C$.

Here, $u = x - 1$ and $a = \sqrt{6}$. If $u = x-1$, then $du = dx$.

Using the standard integral formula:

$I_2 = 4 \left( \frac{1}{2\sqrt{6}} \log_e \left| \frac{(x - 1) - \sqrt{6}}{(x - 1) + \sqrt{6}} \right| \right) + C_2$

$I_2 = \frac{4}{2\sqrt{6}} \log_e \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right| + C_2$

$I_2 = \frac{2}{\sqrt{6}} \log_e \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right| + C_2$

Rationalizing the coefficient:

$I_2 = \frac{2\sqrt{6}}{6} \log_e \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right| + C_2$

$I_2 = \frac{\sqrt{6}}{3} \log_e \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right| + C_2$

Combining the results for $I_1$ and $I_2$ for the total integral $I = I_1 + I_2$:

$I = \left(\frac{1}{2} \log_e |x^2 - 2x - 5| + C_1\right) + \left(\frac{\sqrt{6}}{3} \log_e \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right| + C_2\right)$

$I = \frac{1}{2} \log_e |x^2 - 2x - 5| + \frac{\sqrt{6}}{3} \log_e \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right| + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral of the given function is:

$\int \frac{x + 3}{x^2 − 2x − 5} dx = \frac{1}{2} \log_e |x^2 - 2x - 5| + \frac{\sqrt{6}}{3} \log_e \left| \frac{x - 1 - \sqrt{6}}{x - 1 + \sqrt{6}} \right| + C$

Let the given integral be $I$.

$I = \int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} dx$

We write the numerator $5x + 3$ in terms of the derivative of the quadratic expression under the square root, $x^2 + 4x + 10$.

The derivative of $x^2 + 4x + 10$ is $\frac{d}{dx}(x^2 + 4x + 10) = 2x + 4$.

We need to find constants $A$ and $B$ such that:

$5x + 3 = A(2x + 4) + B$

$5x + 3 = 2Ax + 4A + B$

Comparing the coefficients of $x$ on both sides:

From (i), $A = \frac{5}{2}$.

Comparing the constant terms on both sides:

Substitute $A = \frac{5}{2}$ into (ii):

So, we can write the numerator as $5x + 3 = \frac{5}{2}(2x + 4) - 7$.

The integral becomes:

$I = \int \frac{\frac{5}{2}(2x + 4) - 7}{\sqrt{x^2 + 4x + 10}} dx$

We split the integral into two parts:

$I = \frac{5}{2} \int \frac{2x + 4}{\sqrt{x^2 + 4x + 10}} dx - 7 \int \frac{1}{\sqrt{x^2 + 4x + 10}} dx$

$I = I_1 + I_2$ (say)

Let's evaluate the first integral: $I_1 = \frac{5}{2} \int \frac{2x + 4}{\sqrt{x^2 + 4x + 10}} dx$.

Let $u = x^2 + 4x + 10$. Differentiating with respect to $x$, $du = (2x + 4) dx$.

$I_1 = \frac{5}{2} \int \frac{du}{\sqrt{u}} = \frac{5}{2} \int u^{-1/2} du$

$I_1 = \frac{5}{2} \left(\frac{u^{-1/2 + 1}}{-1/2 + 1}\right) + C_1 = \frac{5}{2} \left(\frac{u^{1/2}}{1/2}\right) + C_1$

$I_1 = \frac{5}{2} (2\sqrt{u}) + C_1 = 5\sqrt{u} + C_1$

Substitute back $u = x^2 + 4x + 10$:

$I_1 = 5\sqrt{x^2 + 4x + 10} + C_1$

Now, let's evaluate the second integral: $I_2 = - 7 \int \frac{1}{\sqrt{x^2 + 4x + 10}} dx$.

We complete the square for the denominator $x^2 + 4x + 10$:

$x^2 + 4x + 10 = (x^2 + 4x + 4) + 6$

$= (x + 2)^2 + (\sqrt{6})^2$

Substitute this back into $I_2$:

$I_2 = - 7 \int \frac{1}{\sqrt{(x + 2)^2 + (\sqrt{6})^2}} dx$

This is a standard integral of the form $\int \frac{1}{\sqrt{u^2 + a^2}} du = \log_e |u + \sqrt{u^2 + a^2}| + C$.

Here, $u = x + 2$ and $a = \sqrt{6}$. If $u = x+2$, then $du = dx$.

Using the standard integral formula:

$I_2 = - 7 \left( \log_e |(x + 2) + \sqrt{(x + 2)^2 + (\sqrt{6})^2}| \right) + C_2$

Substitute back the original quadratic expression inside the square root:

$(x + 2)^2 + (\sqrt{6})^2 = x^2 + 4x + 4 + 6 = x^2 + 4x + 10$

So, $I_2 = - 7 \log_e |x + 2 + \sqrt{x^2 + 4x + 10}| + C_2$

Combining the results for $I_1$ and $I_2$ for the total integral $I = I_1 + I_2$:

$I = (5\sqrt{x^2 + 4x + 10} + C_1) + (- 7 \log_e |x + 2 + \sqrt{x^2 + 4x + 10}| + C_2)$

$I = 5\sqrt{x^2 + 4x + 10} - 7 \log_e |x + 2 + \sqrt{x^2 + 4x + 10}| + C$

where $C = C_1 + C_2$ is the constant of integration.

Therefore, the integral of the given function is:

$\int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} dx = 5\sqrt{x^2 + 4x + 10} - 7 \log_e |x + 2 + \sqrt{x^2 + 4x + 10}| + C$

Let the given integral be $I$.

$I = \int \frac{1}{x^2 + 2x + 2} dx$

First, we complete the square for the quadratic expression in the denominator:

$x^2 + 2x + 2 = (x^2 + 2x + 1) + 1$

$= (x+1)^2 + 1^2$

Substitute this back into the integral:

$I = \int \frac{1}{(x+1)^2 + 1^2} dx$

Let us use the substitution method.

Let $u = x+1$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 1$

So, $du = dx$.

Now, substitute $u$ and $dx$ into the integral:

$I = \int \frac{1}{u^2 + 1^2} du$

This is a standard integral of the form $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.

Here, $x$ is $u$ and $a$ is 1.

So, $I = \frac{1}{1} \tan^{-1}\left(\frac{u}{1}\right) + C$

$I = \tan^{-1}(u) + C$

Substitute back $u = x+1$:

$I = \tan^{-1}(x+1) + C$

Comparing this result with the given options, we find that it matches option (B).

The correct answer is (B) $\tan^{-1} (x + 1) + C$.

Let the given integral be $I$.

$I = \int \frac{dx}{\sqrt{9x − 4x^2}}$

First, we work with the expression under the square root, $9x - 4x^2$. We complete the square.

$9x - 4x^2 = -4(x^2 - \frac{9}{4}x)$

To complete the square for $x^2 - \frac{9}{4}x$, we add and subtract $\left(\frac{1}{2} \times -\frac{9}{4}\right)^2 = \left(-\frac{9}{8}\right)^2 = \frac{81}{64}$.

$-4\left(x^2 - \frac{9}{4}x + \frac{81}{64} - \frac{81}{64}\right)$

$= -4\left(\left(x - \frac{9}{8}\right)^2 - \frac{81}{64}\right)$

$= -4\left(x - \frac{9}{8}\right)^2 + 4 \times \frac{81}{64}$

$= \frac{81}{16} - 4\left(x - \frac{9}{8}\right)^2$

We can rewrite this as $\left(\frac{9}{4}\right)^2 - \left(2\left(x - \frac{9}{8}\right)\right)^2 = \left(\frac{9}{4}\right)^2 - \left(2x - \frac{18}{8}\right)^2 = \left(\frac{9}{4}\right)^2 - \left(2x - \frac{9}{4}\right)^2$.

Substitute this completed square form back into the integral:

$I = \int \frac{dx}{\sqrt{\left(\frac{9}{4}\right)^2 - \left(2x - \frac{9}{4}\right)^2}}$

This integral is in the form $\int \frac{1}{\sqrt{a^2 - u^2}} du$.

Let $u = 2x - \frac{9}{4}$.

Differentiating both sides with respect to $x$, we get:

$\frac{du}{dx} = 2$

So, $du = 2 dx$, which means $dx = \frac{1}{2} du$.

Also, let $a = \frac{9}{4}$.

Now, substitute $u$, $a$, and $dx$ into the integral:

$I = \int \frac{\frac{1}{2} du}{\sqrt{a^2 - u^2}}$

$I = \frac{1}{2} \int \frac{1}{\sqrt{a^2 - u^2}} du$

Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}\left(\frac{u}{a}\right) + C$, we get:

$I = \frac{1}{2} \sin^{-1}\left(\frac{u}{a}\right) + C$

Substitute back $u = 2x - \frac{9}{4}$ and $a = \frac{9}{4}$:

$I = \frac{1}{2} \sin^{-1}\left(\frac{2x - \frac{9}{4}}{\frac{9}{4}}\right) + C$

$I = \frac{1}{2} \sin^{-1}\left(\frac{\frac{8x - 9}{4}}{\frac{9}{4}}\right) + C$

$I = \frac{1}{2} \sin^{-1}\left(\frac{8x - 9}{9}\right) + C$

Comparing this result with the given options, we find that it matches option (B).

The correct answer is (B) $\frac{1}{2} \sin^{-1} \left( \frac{8x − 9}{9} \right) + C$

Given:

We are asked to find the integral $\int \frac{dx}{(x + 1)(x + 2)}$.

Solution:

The integrand is a rational function $\frac{1}{(x + 1)(x + 2)}$. Since the denominator is a product of distinct linear factors, we use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x + 1)(x + 2)$, we get:

To find the values of A and B, we can substitute convenient values for $x$ in the equation $1 = A(x + 2) + B(x + 1)$.

Setting $x = -1$:

Thus, $\mathbf{A = 1}$.

Setting $x = -2$:

Thus, $\mathbf{B = -1}$.

Substituting the values of A and B back into equation (1), we get:

$\frac{1}{(x + 1)(x + 2)} = \frac{1}{x + 1} - \frac{1}{x + 2}$

Now, we can integrate the expression:

$\int \frac{dx}{(x + 1)(x + 2)} = \int \left( \frac{1}{x + 1} - \frac{1}{x + 2} \right) dx$

Using the linearity of the integral, we have:

$\int \left( \frac{1}{x + 1} - \frac{1}{x + 2} \right) dx = \int \frac{1}{x + 1} dx - \int \frac{1}{x + 2} dx$

We know that $\int \frac{1}{ax+b} dx = \frac{1}{a} \log_e |ax+b| + C$.

So, $\int \frac{1}{x + 1} dx = \log_e |x + 1|$ (with $a=1, b=1$)

and $\int \frac{1}{x + 2} dx = \log_e |x + 2|$ (with $a=1, b=2$)

(We add the constant of integration at the end).

Therefore, the integral is:

$\int \frac{dx}{(x + 1)(x + 2)} = \log_e |x + 1| - \log_e |x + 2| + C$

where $C$ is the constant of integration.

Using the logarithm property $\log_e a - \log_e b = \log_e \frac{a}{b}$, we can simplify the result:

$\int \frac{dx}{(x + 1)(x + 2)} = \log_e \left|\frac{x + 1}{x + 2}\right| + C$

Final Answer:

The integral is $\int \frac{dx}{(x + 1)(x + 2)} = \log_e \left|\frac{x + 1}{x + 2}\right| + C$.

Given:

We are asked to find the integral $\int \frac{x^2 + 1}{x^2 − 5x + 6} \;dx$.

Solution:

The integrand is a rational function where the degree of the numerator ($2$) is equal to the degree of the denominator ($2$). In such cases, we first perform polynomial long division.

Let's divide $x^2 + 1$ by $x^2 - 5x + 6$.

So, we can write the integrand as:

Now, we need to integrate the remainder term, which requires partial fraction decomposition for $\frac{5x - 5}{x^2 − 5x + 6}$.

First, factor the denominator: $x^2 - 5x + 6 = (x - 2)(x - 3)$.

So, we decompose $\frac{5x - 5}{(x - 2)(x - 3)}$:

Multiplying both sides by $(x - 2)(x - 3)$, we get:

To find A and B, substitute values for $x$:

Set $x = 2$:

Set $x = 3$:

Thus, $\mathbf{A = -5}$ and $\mathbf{B = 10}$.

Substituting these values back into the partial fraction decomposition:

Now, substitute the result of the long division (1) and partial fraction decomposition (2) back into the original integral:

$\int \frac{x^2 + 1}{x^2 − 5x + 6} \;dx = \int \left( 1 + \frac{-5}{x - 2} + \frac{10}{x - 3} \right) dx$

Using the linearity of the integral:

$\int \frac{x^2 + 1}{x^2 − 5x + 6} \;dx = \int 1 \;dx - 5 \int \frac{1}{x - 2} \;dx + 10 \int \frac{1}{x - 3} \;dx$

We know that $\int 1 \;dx = x$, $\int \frac{1}{x - a} \;dx = \log_e |x - a| + C$.

So, $\int 1 \;dx = x$

$\int \frac{1}{x - 2} \;dx = \log_e |x - 2|$

$\int \frac{1}{x - 3} \;dx = \log_e |x - 3|$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{x^2 + 1}{x^2 − 5x + 6} \;dx = x - 5 \log_e |x - 2| + 10 \log_e |x - 3| + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{x^2 + 1}{x^2 − 5x + 6} \;dx = x - 5 \log_e |x - 2| + 10 \log_e |x - 3| + C$.

Given:

We are asked to find the integral $\int \frac{3x − 2}{(x + 1)^2 (x + 3)} \;dx$.

Solution:

The integrand is a rational function with a repeated linear factor $(x+1)^2$ and a distinct linear factor $(x+3)$ in the denominator. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x + 1)^2 (x + 3)$, we get:

We can find the values of A, B, and C by substituting convenient values for $x$ or by comparing coefficients.

Setting $x = -1$ in equation (2):

Thus, $\mathbf{B = -\frac{5}{2}}$.

Setting $x = -3$ in equation (2):

Thus, $\mathbf{C = -\frac{11}{4}}$.

To find A, we can compare the coefficients of $x^2$ in equation (2). Expanding the right side:

$3x - 2 = A(x^2 + 4x + 3) + B(x + 3) + C(x^2 + 2x + 1)$

$3x - 2 = Ax^2 + 4Ax + 3A + Bx + 3B + Cx^2 + 2Cx + C$

$3x - 2 = (A + C)x^2 + (4A + B + 2C)x + (3A + 3B + C)$

Comparing the coefficients of $x^2$ on both sides:

Since $C = -\frac{11}{4}$, we have:

Thus, $\mathbf{A = \frac{11}{4}}$.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we integrate each term:

$\int \frac{3x − 2}{(x + 1)^2 (x + 3)} \;dx = \int \left( \frac{11}{4(x + 1)} - \frac{5}{2(x + 1)^2} - \frac{11}{4(x + 3)} \right) dx$

Using the linearity of the integral:

$\int \frac{3x − 2}{(x + 1)^2 (x + 3)} \;dx = \frac{11}{4} \int \frac{1}{x + 1} \;dx - \frac{5}{2} \int (x + 1)^{-2} \;dx - \frac{11}{4} \int \frac{1}{x + 3} \;dx$

We know that $\int \frac{1}{u} du = \log_e |u|$ and $\int u^n du = \frac{u^{n+1}}{n+1}$ for $n \neq -1$.

$\int \frac{1}{x + 1} \;dx = \log_e |x + 1|$

$\int (x + 1)^{-2} \;dx = \frac{(x + 1)^{-1}}{-1} = -\frac{1}{x + 1}$

$\int \frac{1}{x + 3} \;dx = \log_e |x + 3|$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{3x − 2}{(x + 1)^2 (x + 3)} \;dx = \frac{11}{4} \log_e |x + 1| - \frac{5}{2} \left( -\frac{1}{x + 1} \right) - \frac{11}{4} \log_e |x + 3| + C$

$\int \frac{3x − 2}{(x + 1)^2 (x + 3)} \;dx = \frac{11}{4} \log_e |x + 1| + \frac{5}{2(x + 1)} - \frac{11}{4} \log_e |x + 3| + C$

Using the logarithm property $\log_e a - \log_e b = \log_e \frac{a}{b}$, we can combine the logarithm terms:

$\int \frac{3x − 2}{(x + 1)^2 (x + 3)} \;dx = \frac{11}{4} (\log_e |x + 1| - \log_e |x + 3|) + \frac{5}{2(x + 1)} + C$

$\int \frac{3x − 2}{(x + 1)^2 (x + 3)} \;dx = \frac{11}{4} \log_e \left|\frac{x + 1}{x + 3}\right| + \frac{5}{2(x + 1)} + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{3x − 2}{(x + 1)^2 (x + 3)} \;dx = \frac{11}{4} \log_e \left|\frac{x + 1}{x + 3}\right| + \frac{5}{2(x + 1)} + C$.

Given:

We are asked to find the integral $\int \frac{x^2}{(x^2 + 1) (x^2 + 4)} \;dx$.

Solution:

The integrand is a rational function $\frac{x^2}{(x^2 + 1) (x^2 + 4)}$. The denominator consists of distinct irreducible quadratic factors. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x^2 + 1)(x^2 + 4)$, we get:

Expanding the right side:

$x^2 = (Ax^3 + 4Ax + Bx^2 + 4B) + (Cx^3 + Cx + Dx^2 + D)$

$x^2 = Ax^3 + Bx^2 + 4Ax + 4B + Cx^3 + Dx^2 + Cx + D$

Grouping terms by powers of $x$:

$x^2 = (A + C)x^3 + (B + D)x^2 + (4A + C)x + (4B + D)$

Comparing the coefficients of the powers of $x$ on both sides:

Coefficient of $x^3$: $A + C = 0$

Coefficient of $x^2$: $B + D = 1$

Coefficient of $x$: $4A + C = 0$

Constant term: $4B + D = 0$

From the equations for the coefficients of $x^3$ and $x$:
$A + C = 0$
$4A + C = 0$

Subtracting the first equation from the second gives $(4A + C) - (A + C) = 0 - 0$, which simplifies to $3A = 0$. Thus, $\mathbf{A = 0}$.

Substituting $A = 0$ into $A + C = 0$, we get $0 + C = 0$. Thus, $\mathbf{C = 0}$.

From the equations for the coefficients of $x^2$ and the constant term:
$B + D = 1$
$4B + D = 0$

Subtracting the first equation from the second gives $(4B + D) - (B + D) = 0 - 1$, which simplifies to $3B = -1$. Thus, $\mathbf{B = -\frac{1}{3}}$.

Substituting $B = -\frac{1}{3}$ into $B + D = 1$, we get $-\frac{1}{3} + D = 1$. Thus, $D = 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3}$. Thus, $\mathbf{D = \frac{4}{3}}$.

Substituting the values of A, B, C, and D back into equation (1), we get:

Now, we integrate each term:

$\int \frac{x^2}{(x^2 + 1) (x^2 + 4)} \;dx = \int \left( -\frac{1}{3(x^2 + 1)} + \frac{4}{3(x^2 + 4)} \right) dx$

Using the linearity of the integral:

$\int \frac{x^2}{(x^2 + 1) (x^2 + 4)} \;dx = -\frac{1}{3} \int \frac{1}{x^2 + 1} \;dx + \frac{4}{3} \int \frac{1}{x^2 + 4} \;dx$

We use the standard integral formula $\int \frac{1}{x^2 + a^2} \;dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.

For the first integral, $a^2 = 1$, so $a = 1$.

$\int \frac{1}{x^2 + 1} \;dx = \frac{1}{1} \tan^{-1}\left(\frac{x}{1}\right) = \tan^{-1}(x)$

For the second integral, $a^2 = 4$, so $a = 2$.

$\int \frac{1}{x^2 + 4} \;dx = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{x^2}{(x^2 + 1) (x^2 + 4)} \;dx = -\frac{1}{3} \tan^{-1}(x) + \frac{4}{3} \left( \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right) + C$

$\int \frac{x^2}{(x^2 + 1) (x^2 + 4)} \;dx = -\frac{1}{3} \tan^{-1}(x) + \frac{4}{6} \tan^{-1}\left(\frac{x}{2}\right) + C$

$\int \frac{x^2}{(x^2 + 1) (x^2 + 4)} \;dx = -\frac{1}{3} \tan^{-1}(x) + \frac{2}{3} \tan^{-1}\left(\frac{x}{2}\right) + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{x^2}{(x^2 + 1) (x^2 + 4)} \;dx = -\frac{1}{3} \tan^{-1}(x) + \frac{2}{3} \tan^{-1}\left(\frac{x}{2}\right) + C$.

Given:

We are asked to find the integral $\int \frac{(3\sin φ − 2) \cos φ}{5 − \cos^2 φ − 4 \sin φ} \;dφ$.

Solution:

Let's use the substitution method. Let $t = \sin φ$.

Then, differentiating with respect to $φ$, we get $dt = \cos φ \;dφ$.

Also, we can express $\cos^2 φ$ in terms of $t$ using the identity $\sin^2 φ + \cos^2 φ = 1$.

$\cos^2 φ = 1 - \sin^2 φ = 1 - t^2$.

Now substitute $t = \sin φ$, $dt = \cos φ \;dφ$, and $\cos^2 φ = 1 - t^2$ into the integral:

The numerator becomes $(3\sin φ − 2) \cos φ \;dφ = (3t - 2) dt$.

The denominator becomes $5 - \cos^2 φ - 4 \sin φ = 5 - (1 - t^2) - 4t = 5 - 1 + t^2 - 4t = t^2 - 4t + 4$.

The integral transforms to:

The denominator is a perfect square: $t^2 - 4t + 4 = (t - 2)^2$.

So the integral is $\int \frac{3t - 2}{(t - 2)^2} \;dt$.

We use partial fraction decomposition for the integrand $\frac{3t - 2}{(t - 2)^2}$.

We write it in the form:

Multiplying both sides of equation (2) by $(t - 2)^2$, we get:

To find the values of A and B:

Set $t = 2$:

Thus, $\mathbf{B = 4}$.

Comparing the coefficients of $t$ in the equation $3t - 2 = A(t - 2) + B = At - 2A + B$:

Thus, $\mathbf{A = 3}$.

Substitute the values of A and B back into the partial fraction decomposition:

Now, integrate the decomposed expression with respect to $t$ (from equation (1)):

$\int \frac{3t - 2}{(t - 2)^2} \;dt = \int \left( \frac{3}{t - 2} + \frac{4}{(t - 2)^2} \right) dt$

Using the linearity of the integral:

$\int \left( \frac{3}{t - 2} + \frac{4}{(t - 2)^2} \right) dt = \int \frac{3}{t - 2} dt + \int \frac{4}{(t - 2)^2} dt$

We evaluate each integral:

$\int \frac{3}{t - 2} dt = 3 \int \frac{1}{t - 2} dt = 3 \log_e |t - 2|$

$\int \frac{4}{(t - 2)^2} dt = 4 \int (t - 2)^{-2} dt = 4 \frac{(t - 2)^{-2 + 1}}{-2 + 1} + C' = 4 \frac{(t - 2)^{-1}}{-1} + C' = -\frac{4}{t - 2} + C'$

(We combine constants of integration at the end).

So, the integral with respect to $t$ is:

$3 \log_e |t - 2| - \frac{4}{t - 2} + C$

Now, substitute back $t = \sin φ$:

The integral is $3 \log_e |\sin φ - 2| - \frac{4}{\sin φ - 2} + C$.

Since $-1 \leq \sin φ \leq 1$ for real $φ$, the value of $\sin φ - 2$ is always negative (ranging from $-1 - 2 = -3$ to $1 - 2 = -1$).

Therefore, $|\sin φ - 2| = -(\sin φ - 2) = 2 - \sin φ$.

So, the integral can be written as:

$3 \log_e (2 - \sin φ) - \frac{4}{\sin φ - 2} + C$

Alternatively, we can write $-\frac{4}{\sin φ - 2} = \frac{-4}{- (2 - \sin φ)} = \frac{4}{2 - \sin φ}$.

So the result is also $3 \log_e (2 - \sin φ) + \frac{4}{2 - \sin φ} + C$.

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{(3\sin φ − 2) \cos φ}{5 − \cos^2 φ − 4 \sin φ} \;dφ = 3 \log_e (2 - \sin φ) + \frac{4}{2 - \sin φ} + C$.

Given:

We are asked to find the integral $\int \frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} \;dx$.

Solution:

The integrand is a rational function $\frac{x^2 + x + 1}{(x + 2) (x^2 + 1)}$. The denominator consists of a distinct linear factor $(x+2)$ and an irreducible quadratic factor $(x^2+1)$. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x + 2)(x^2 + 1)$, we get:

We can find the values of A, B, and C by substituting a convenient value for $x$ and by comparing coefficients.

Setting $x = -2$ in equation (2):

Thus, $\mathbf{A = \frac{3}{5}}$.

Now, let's expand equation (2) and compare coefficients:

$x^2 + x + 1 = A(x^2 + 1) + (Bx + C)(x + 2)$

$x^2 + x + 1 = Ax^2 + A + Bx^2 + 2Bx + Cx + 2C$

$x^2 + x + 1 = (A + B)x^2 + (2B + C)x + (A + 2C)$

Comparing the coefficients of the powers of $x$ on both sides:

Coefficient of $x^2$: $1 = A + B$

Coefficient of $x$: $1 = 2B + C$

Constant term: $1 = A + 2C$

We have $A = \frac{3}{5}$. Substitute this into the first equation:

Thus, $\mathbf{B = \frac{2}{5}}$.

Substitute $B = \frac{2}{5}$ into the second equation:

Thus, $\mathbf{C = \frac{1}{5}}$.

We can verify this with the third equation using $A = \frac{3}{5}$ and $C = \frac{1}{5}$:

This matches the constant term on the left side, confirming our values for A, B, and C are correct.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we integrate each term:

$\int \frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} \;dx = \int \left( \frac{3}{5(x + 2)} + \frac{2x + 1}{5(x^2 + 1)} \right) dx$

Using the linearity of the integral:

$\int \frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} \;dx = \frac{3}{5} \int \frac{1}{x + 2} \;dx + \frac{1}{5} \int \frac{2x + 1}{x^2 + 1} \;dx$

We evaluate each integral separately.

For the first integral:

$\int \frac{1}{x + 2} \;dx = \log_e |x + 2|$

For the second integral, we split it into two parts:

$\int \frac{2x + 1}{x^2 + 1} \;dx = \int \frac{2x}{x^2 + 1} \;dx + \int \frac{1}{x^2 + 1} \;dx$

For the first part $\int \frac{2x}{x^2 + 1} \;dx$, let $u = x^2 + 1$, then $du = 2x \;dx$.

$\int \frac{2x}{x^2 + 1} \;dx = \int \frac{1}{u} \;du = \log_e |u| = \log_e |x^2 + 1| = \log_e (x^2 + 1)$ (since $x^2+1 > 0$).

For the second part $\int \frac{1}{x^2 + 1} \;dx$, this is a standard integral:

$\int \frac{1}{x^2 + 1} \;dx = \tan^{-1}(x)$

So, $\int \frac{2x + 1}{x^2 + 1} \;dx = \log_e (x^2 + 1) + \tan^{-1}(x)$.

(We add the constant of integration at the end).

Combining all parts of the original integral:

$\int \frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} \;dx = \frac{3}{5} \log_e |x + 2| + \frac{1}{5} (\log_e (x^2 + 1) + \tan^{-1}(x)) + C$

$\int \frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} \;dx = \frac{3}{5} \log_e |x + 2| + \frac{1}{5} \log_e (x^2 + 1) + \frac{1}{5} \tan^{-1}(x) + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{x^2 + x + 1}{(x + 2) (x^2 + 1)} \;dx = \frac{3}{5} \log_e |x + 2| + \frac{1}{5} \log_e (x^2 + 1) + \frac{1}{5} \tan^{-1}(x) + C$.

Given:

We are asked to find the integral $\int \frac{x}{(x + 1) (x + 2)} \;dx$.

Solution:

The integrand is a rational function $\frac{x}{(x + 1) (x + 2)}$. The denominator is a product of distinct linear factors. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x + 1)(x + 2)$, we get:

To find the values of A and B, we substitute convenient values for $x$ in equation (2).

Setting $x = -1$ in equation (2):

Thus, $\mathbf{A = -1}$.

Setting $x = -2$ in equation (2):

Thus, $\mathbf{B = 2}$.

Substituting the values of A and B back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{x}{(x + 1) (x + 2)} \;dx = \int \left( \frac{-1}{x + 1} + \frac{2}{x + 2} \right) \;dx$

Using the linearity of the integral, we have:

$\int \left( \frac{-1}{x + 1} + \frac{2}{x + 2} \right) \;dx = - \int \frac{1}{x + 1} \;dx + 2 \int \frac{1}{x + 2} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u| + C'$.

So, $\int \frac{1}{x + 1} \;dx = \log_e |x + 1|$

and $\int \frac{1}{x + 2} \;dx = \log_e |x + 2|$

(We add the constant of integration at the end).

Therefore, the integral is:

$\int \frac{x}{(x + 1) (x + 2)} \;dx = - \log_e |x + 1| + 2 \log_e |x + 2| + C$

where $C$ is the constant of integration.

Using the logarithm property $n \log_e a = \log_e a^n$ and $\log_e a - \log_e b = \log_e \frac{a}{b}$, we can simplify the result:

$\int \frac{x}{(x + 1) (x + 2)} \;dx = \log_e |(x + 2)^2| - \log_e |x + 1| + C$

$\int \frac{x}{(x + 1) (x + 2)} \;dx = \log_e \left|\frac{(x + 2)^2}{x + 1}\right| + C$

Final Answer:

The integral is $\int \frac{x}{(x + 1) (x + 2)} \;dx = 2 \log_e |x + 2| - \log_e |x + 1| + C$ or $\log_e \left|\frac{(x + 2)^2}{x + 1}\right| + C$.

Given:

We are asked to find the integral $\int \frac{1}{x^2 − 9} \;dx$.

Solution:

The integrand is a rational function $\frac{1}{x^2 − 9}$. The denominator can be factored as a difference of squares: $x^2 − 9 = x^2 - 3^2 = (x - 3)(x + 3)$.

So, the integral is $\int \frac{1}{(x - 3)(x + 3)} \;dx$.

The integrand $\frac{1}{(x - 3)(x + 3)}$ is a rational function with distinct linear factors in the denominator. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x - 3)(x + 3)$, we get:

To find the values of A and B, we substitute convenient values for $x$ in equation (2).

Setting $x = 3$ in equation (2):

Thus, $\mathbf{A = \frac{1}{6}}$.

Setting $x = -3$ in equation (2):

Thus, $\mathbf{B = -\frac{1}{6}}$.

Substituting the values of A and B back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{1}{x^2 - 9} \;dx = \int \left( \frac{1}{6(x - 3)} - \frac{1}{6(x + 3)} \right) \;dx$

Using the linearity of the integral, we have:

$\int \frac{1}{x^2 - 9} \;dx = \frac{1}{6} \int \frac{1}{x - 3} \;dx - \frac{1}{6} \int \frac{1}{x + 3} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u| + C'$.

So, $\int \frac{1}{x - 3} \;dx = \log_e |x - 3|$

and $\int \frac{1}{x + 3} \;dx = \log_e |x + 3|$

(We add the constant of integration at the end).

Therefore, the integral is:

$\int \frac{1}{x^2 - 9} \;dx = \frac{1}{6} \log_e |x - 3| - \frac{1}{6} \log_e |x + 3| + C$

where $C$ is the constant of integration.

Using the logarithm property $\log_e a - \log_e b = \log_e \frac{a}{b}$, we can simplify the result:

$\int \frac{1}{x^2 - 9} \;dx = \frac{1}{6} (\log_e |x - 3| - \log_e |x + 3|) + C$

$\int \frac{1}{x^2 - 9} \;dx = \frac{1}{6} \log_e \left|\frac{x - 3}{x + 3}\right| + C$

Alternate Solution using Standard Formula:

We can directly use the standard integration formula for $\int \frac{1}{x^2 - a^2} \;dx$.

The integral is $\int \frac{1}{x^2 - 9} \;dx$. Here, $a^2 = 9$, so $a = 3$.

The formula is $\int \frac{1}{x^2 - a^2} \;dx = \frac{1}{2a} \log_e \left|\frac{x - a}{x + a}\right| + C$.

Substituting $a = 3$:

$\int \frac{1}{x^2 - 3^2} \;dx = \frac{1}{2(3)} \log_e \left|\frac{x - 3}{x + 3}\right| + C$

$\int \frac{1}{x^2 - 9} \;dx = \frac{1}{6} \log_e \left|\frac{x - 3}{x + 3}\right| + C$

This matches the result obtained using partial fractions.

Final Answer:

The integral is $\int \frac{1}{x^2 − 9} \;dx = \frac{1}{6} \log_e \left|\frac{x - 3}{x + 3}\right| + C$.

Given:

We are asked to find the integral $\int \frac{3x − 1}{(x − 1) (x − 2) (x − 3)} \;dx$.

Solution:

The integrand is a rational function $\frac{3x − 1}{(x − 1) (x − 2) (x − 3)}$. The denominator is a product of distinct linear factors. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x - 1)(x - 2)(x - 3)$, we get:

To find the values of A, B, and C, we substitute convenient values for $x$ in equation (2).

Setting $x = 1$ in equation (2):

Thus, $\mathbf{A = 1}$.

Setting $x = 2$ in equation (2):

Thus, $\mathbf{B = -5}$.

Setting $x = 3$ in equation (2):

Thus, $\mathbf{C = 4}$.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{3x − 1}{(x − 1) (x − 2) (x − 3)} \;dx = \int \left( \frac{1}{x - 1} - \frac{5}{x - 2} + \frac{4}{x - 3} \right) \;dx$

Using the linearity of the integral, we have:

$\int \frac{3x − 1}{(x − 1) (x − 2) (x − 3)} \;dx = \int \frac{1}{x - 1} \;dx - 5 \int \frac{1}{x - 2} \;dx + 4 \int \frac{1}{x - 3} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u| + C'$.

So, $\int \frac{1}{x - 1} \;dx = \log_e |x - 1|$

$\int \frac{1}{x - 2} \;dx = \log_e |x - 2|$

$\int \frac{1}{x - 3} \;dx = \log_e |x - 3|$

(We add the constant of integration at the end).

Therefore, the integral is:

$\int \frac{3x − 1}{(x − 1) (x − 2) (x − 3)} \;dx = \log_e |x - 1| - 5 \log_e |x - 2| + 4 \log_e |x - 3| + C$

where $C$ is the constant of integration.

Using logarithm properties ($n \log_e a = \log_e a^n$ and $\log_e a + \log_e b = \log_e (ab)$), we can also write:

$\int \frac{3x − 1}{(x − 1) (x − 2) (x − 3)} \;dx = \log_e |x - 1| + \log_e |(x - 3)^4| - \log_e |(x - 2)^5| + C$

$\int \frac{3x − 1}{(x − 1) (x − 2) (x − 3)} \;dx = \log_e \left| \frac{(x - 1)(x - 3)^4}{(x - 2)^5} \right| + C$

Final Answer:

The integral is $\int \frac{3x − 1}{(x − 1) (x − 2) (x − 3)} \;dx = \log_e |x - 1| - 5 \log_e |x - 2| + 4 \log_e |x - 3| + C$.

Given:

We are asked to find the integral $\int \frac{x}{(x − 1) (x − 2) (x − 3)} \;dx$.

Solution:

The integrand is a rational function $\frac{x}{(x − 1) (x − 2) (x − 3)}$. The denominator is a product of distinct linear factors. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x - 1)(x - 2)(x - 3)$, we get:

To find the values of A, B, and C, we substitute convenient values for $x$ in equation (2).

Setting $x = 1$ in equation (2):

Thus, $\mathbf{A = \frac{1}{2}}$.

Setting $x = 2$ in equation (2):

Thus, $\mathbf{B = -2}$.

Setting $x = 3$ in equation (2):

Thus, $\mathbf{C = \frac{3}{2}}$.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{x}{(x - 1) (x - 2) (x - 3)} \;dx = \int \left( \frac{1}{2(x - 1)} - \frac{2}{x - 2} + \frac{3}{2(x - 3)} \right) \;dx$

Using the linearity of the integral, we have:

$\int \frac{x}{(x - 1) (x - 2) (x - 3)} \;dx = \frac{1}{2} \int \frac{1}{x - 1} \;dx - 2 \int \frac{1}{x - 2} \;dx + \frac{3}{2} \int \frac{1}{x - 3} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u| + C'$.

So, $\int \frac{1}{x - 1} \;dx = \log_e |x - 1|$

$\int \frac{1}{x - 2} \;dx = \log_e |x - 2|$

$\int \frac{1}{x - 3} \;dx = \log_e |x - 3|$

(We add the constant of integration at the end).

Therefore, the integral is:

$\int \frac{x}{(x - 1) (x - 2) (x - 3)} \;dx = \frac{1}{2} \log_e |x - 1| - 2 \log_e |x - 2| + \frac{3}{2} \log_e |x - 3| + C$

where $C$ is the constant of integration.

Using logarithm properties ($n \log_e a = \log_e a^n$ and $\log_e a + \log_e b = \log_e (ab)$ and $\log_e a - \log_e b = \log_e (a/b)$), we can also write:

$\int \frac{x}{(x - 1) (x - 2) (x - 3)} \;dx = \log_e |x - 1|^{1/2} - \log_e |(x - 2)^2| + \log_e |x - 3|^{3/2} + C$

$\int \frac{x}{(x - 1) (x - 2) (x - 3)} \;dx = \log_e \left| \frac{(x - 1)^{1/2} (x - 3)^{3/2}}{(x - 2)^2} \right| + C$

$\int \frac{x}{(x - 1) (x - 2) (x - 3)} \;dx = \log_e \left| \sqrt{x - 1} \sqrt{(x - 3)^3} \middle/ (x - 2)^2 \right| + C$

Final Answer:

The integral is $\int \frac{x}{(x - 1) (x - 2) (x - 3)} \;dx = \frac{1}{2} \log_e |x - 1| - 2 \log_e |x - 2| + \frac{3}{2} \log_e |x - 3| + C$.

Given:

We are asked to find the integral $\int \frac{2x}{x^2 + 3x + 2} \;dx$.

Solution:

The integrand is a rational function $\frac{2x}{x^2 + 3x + 2}$. First, we factor the denominator.

$x^2 + 3x + 2 = (x + 1)(x + 2)$

So the integral is $\int \frac{2x}{(x + 1) (x + 2)} \;dx$.

The denominator is a product of distinct linear factors. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x + 1)(x + 2)$, we get:

To find the values of A and B, we substitute convenient values for $x$ in equation (2).

Setting $x = -1$ in equation (2):

Thus, $\mathbf{A = -2}$.

Setting $x = -2$ in equation (2):

Thus, $\mathbf{B = 4}$.

Substituting the values of A and B back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{2x}{(x + 1) (x + 2)} \;dx = \int \left( \frac{-2}{x + 1} + \frac{4}{x + 2} \right) \;dx$

Using the linearity of the integral, we have:

$\int \frac{2x}{(x + 1) (x + 2)} \;dx = -2 \int \frac{1}{x + 1} \;dx + 4 \int \frac{1}{x + 2} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u| + C'$.

So, $\int \frac{1}{x + 1} \;dx = \log_e |x + 1|$

and $\int \frac{1}{x + 2} \;dx = \log_e |x + 2|$

(We add the constant of integration at the end).

Therefore, the integral is:

$\int \frac{2x}{x^2 + 3x + 2} \;dx = -2 \log_e |x + 1| + 4 \log_e |x + 2| + C$

where $C$ is the constant of integration.

Using the logarithm property $n \log_e a = \log_e a^n$ and $\log_e a - \log_e b = \log_e \frac{a}{b}$, we can simplify the result:

$\int \frac{2x}{x^2 + 3x + 2} \;dx = \log_e |(x + 2)^4| - \log_e |(x + 1)^2| + C$

$\int \frac{2x}{x^2 + 3x + 2} \;dx = \log_e \left|\frac{(x + 2)^4}{(x + 1)^2}\right| + C$

Final Answer:

The integral is $\int \frac{2x}{x^2 + 3x + 2} \;dx = 4 \log_e |x + 2| - 2 \log_e |x + 1| + C$.

Given:

We are asked to find the integral $\int \frac{1 − x^2}{x (1 − 2x)} \;dx$.

Solution:

The integrand is a rational function $\frac{1 - x^2}{x (1 - 2x)} = \frac{-x^2 + 1}{-2x^2 + x}$. The degree of the numerator (2) is equal to the degree of the denominator (2). This is an improper rational function, so we must perform polynomial long division first.

Divide $-x^2 + 1$ by $-2x^2 + x$:

So, $\frac{-x^2 + 1}{-2x^2 + x} = \frac{1}{2} + \frac{-x/2 + 1}{-2x^2 + x}$.

We can rewrite the remainder term as $\frac{1 - x/2}{-x(2x - 1)} = \frac{\frac{2-x}{2}}{x(2x - 1)} = \frac{2-x}{2x(2x - 1)}$.

Thus, the integrand can be written as:

Now, we need to integrate $\int \left( \frac{1}{2} + \frac{2-x}{2x(1-2x)} \right) dx$. We can integrate the first term easily: $\int \frac{1}{2} dx = \frac{1}{2}x$.

For the second term, we use partial fraction decomposition for $\frac{2-x}{x(1-2x)}$. Note that we can pull out the constant factor $\frac{1}{2}$ from the integral later.

Let's decompose $\frac{2-x}{x(1-2x)}$. The denominator has distinct linear factors $x$ and $1-2x$.

Multiplying both sides of equation (2) by $x(1 - 2x)$, we get:

To find the values of A and B, we substitute convenient values for $x$.

Setting $x = 0$:

Thus, $\mathbf{A = 2}$.

Setting $x = \frac{1}{2}$:

Thus, $\mathbf{B = 3}$.

Substituting the values of A and B back into equation (2), we get:

Now, substitute this back into the original integrand expression (1):

Now, we integrate each term:

$\int \frac{1 − x^2}{x (1 − 2x)} \;dx = \int \left( \frac{1}{2} + \frac{1}{x} + \frac{3}{2(1 - 2x)} \right) \;dx$

Using the linearity of the integral:

$\int \frac{1 − x^2}{x (1 − 2x)} \;dx = \int \frac{1}{2} \;dx + \int \frac{1}{x} \;dx + \frac{3}{2} \int \frac{1}{1 - 2x} \;dx$

We know that $\int c \;dx = cx$, $\int \frac{1}{x} \;dx = \log_e |x|$, and $\int \frac{1}{ax+b} \;dx = \frac{1}{a} \log_e |ax+b|$.

$\int \frac{1}{2} \;dx = \frac{1}{2}x$

$\int \frac{1}{x} \;dx = \log_e |x|$

$\int \frac{1}{1 - 2x} \;dx = \frac{1}{-2} \log_e |1 - 2x| = -\frac{1}{2} \log_e |1 - 2x|$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{1 − x^2}{x (1 − 2x)} \;dx = \frac{1}{2}x + \log_e |x| + \frac{3}{2} \left( -\frac{1}{2} \log_e |1 - 2x| \right) + C$

$\int \frac{1 − x^2}{x (1 − 2x)} \;dx = \frac{1}{2}x + \log_e |x| - \frac{3}{4} \log_e |1 - 2x| + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{1 − x^2}{x (1 − 2x)} \;dx = \frac{x}{2} + \log_e |x| - \frac{3}{4} \log_e |1 - 2x| + C$.

Given:

We are asked to find the integral $\int \frac{x}{(x^2 + 1) (x − 1)} \;dx$.

Solution:

The integrand is a rational function $\frac{x}{(x^2 + 1) (x − 1)}$. The denominator consists of a distinct linear factor $(x-1)$ and an irreducible quadratic factor $(x^2+1)$. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x - 1)(x^2 + 1)$, we get:

We can find the values of A, B, and C by substituting a convenient value for $x$ and by comparing coefficients.

Setting $x = 1$ in equation (2):

Thus, $\mathbf{A = \frac{1}{2}}$.

Now, let's expand equation (2) and compare coefficients:

$x = A(x^2 + 1) + (Bx + C)(x - 1)$

$x = Ax^2 + A + Bx^2 - Bx + Cx - C$

$x = (A + B)x^2 + (-B + C)x + (A - C)$

Comparing the coefficients of the powers of $x$ on both sides:

Coefficient of $x^2$: $0 = A + B$

Coefficient of $x$: $1 = -B + C$

Constant term: $0 = A - C$

We have $A = \frac{1}{2}$. Substitute this into the first equation:

Thus, $\mathbf{B = -\frac{1}{2}}$.

Substitute $B = -\frac{1}{2}$ into the second equation:

Thus, $\mathbf{C = \frac{1}{2}}$.

We can verify this with the third equation using $A = \frac{1}{2}$ and $C = \frac{1}{2}$:

This matches the constant term on the left side, confirming our values for A, B, and C are correct.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we integrate each term:

$\int \frac{x}{(x^2 + 1) (x − 1)} \;dx = \int \left( \frac{1}{2(x - 1)} + \frac{1 - x}{2(x^2 + 1)} \right) dx$

Using the linearity of the integral:

$\int \frac{x}{(x^2 + 1) (x − 1)} \;dx = \frac{1}{2} \int \frac{1}{x - 1} \;dx + \frac{1}{2} \int \frac{1 - x}{x^2 + 1} \;dx$

We evaluate each integral separately.

For the first integral:

$\int \frac{1}{x - 1} \;dx = \log_e |x - 1|$

For the second integral, we split it into two parts:

$\int \frac{1 - x}{x^2 + 1} \;dx = \int \frac{1}{x^2 + 1} \;dx - \int \frac{x}{x^2 + 1} \;dx$

For the first part $\int \frac{1}{x^2 + 1} \;dx$, this is a standard integral:

$\int \frac{1}{x^2 + 1} \;dx = \tan^{-1}(x)$

For the second part $\int \frac{x}{x^2 + 1} \;dx$, let $u = x^2 + 1$, then $du = 2x \;dx$, so $x \;dx = \frac{1}{2} du$.

$\int \frac{x}{x^2 + 1} \;dx = \int \frac{1}{u} \frac{1}{2} \;du = \frac{1}{2} \int \frac{1}{u} \;du = \frac{1}{2} \log_e |u| = \frac{1}{2} \log_e |x^2 + 1| = \frac{1}{2} \log_e (x^2 + 1)$ (since $x^2+1 > 0$).

So, $\int \frac{1 - x}{x^2 + 1} \;dx = \tan^{-1}(x) - \frac{1}{2} \log_e (x^2 + 1)$.

(We add the constant of integration at the end).

Combining all parts of the original integral:

$\int \frac{x}{(x^2 + 1) (x − 1)} \;dx = \frac{1}{2} \log_e |x - 1| + \frac{1}{2} \left( \tan^{-1}(x) - \frac{1}{2} \log_e (x^2 + 1) \right) + C$

$\int \frac{x}{(x^2 + 1) (x − 1)} \;dx = \frac{1}{2} \log_e |x - 1| + \frac{1}{2} \tan^{-1}(x) - \frac{1}{4} \log_e (x^2 + 1) + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{x}{(x^2 + 1) (x − 1)} \;dx = \frac{1}{2} \log_e |x - 1| - \frac{1}{4} \log_e (x^2 + 1) + \frac{1}{2} \tan^{-1}(x) + C$.

Given:

We are asked to find the integral $\int \frac{x}{(x − 1)^2 (x + 2)} \;dx$.

Solution:

The integrand is a rational function $\frac{x}{(x − 1)^2 (x + 2)}$. The denominator consists of a repeated linear factor $(x-1)^2$ and a distinct linear factor $(x+2)$. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x - 1)^2 (x + 2)$, we get:

We can find the values of A, B, and C by substituting convenient values for $x$ or by comparing coefficients.

Setting $x = 1$ in equation (2):

Thus, $\mathbf{B = \frac{1}{3}}$.

Setting $x = -2$ in equation (2):

Thus, $\mathbf{C = -\frac{2}{9}}$.

To find A, compare the coefficient of $x^2$ in equation (2). Expanding the right side:

$x = A(x^2 + x - 2) + B(x + 2) + C(x^2 - 2x + 1)$

$x = Ax^2 + Ax - 2A + Bx + 2B + Cx^2 - 2Cx + C$

$x = (A + C)x^2 + (A + B - 2C)x + (-2A + 2B + C)$

Comparing the coefficients of $x^2$ on both sides:

Since $C = -\frac{2}{9}$, we have:

Thus, $\mathbf{A = \frac{2}{9}}$.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we integrate each term:

$\int \frac{x}{(x - 1)^2 (x + 2)} \;dx = \int \left( \frac{2}{9(x - 1)} + \frac{1}{3(x - 1)^2} - \frac{2}{9(x + 2)} \right) \;dx$

Using the linearity of the integral:

$\int \frac{x}{(x - 1)^2 (x + 2)} \;dx = \frac{2}{9} \int \frac{1}{x - 1} \;dx + \frac{1}{3} \int (x - 1)^{-2} \;dx - \frac{2}{9} \int \frac{1}{x + 2} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u|$ and $\int u^n du = \frac{u^{n+1}}{n+1}$ for $n \neq -1$.

$\int \frac{1}{x - 1} \;dx = \log_e |x - 1|$

$\int (x - 1)^{-2} \;dx = \frac{(x - 1)^{-2 + 1}}{-2 + 1} = \frac{(x - 1)^{-1}}{-1} = -\frac{1}{x - 1}$

$\int \frac{1}{x + 2} \;dx = \log_e |x + 2|$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{x}{(x - 1)^2 (x + 2)} \;dx = \frac{2}{9} \log_e |x - 1| + \frac{1}{3} \left(-\frac{1}{x - 1}\right) - \frac{2}{9} \log_e |x + 2| + C$

$\int \frac{x}{(x - 1)^2 (x + 2)} \;dx = \frac{2}{9} \log_e |x - 1| - \frac{1}{3(x - 1)} - \frac{2}{9} \log_e |x + 2| + C$

where $C$ is the constant of integration.

Using logarithm properties, we can also write:

$\int \frac{x}{(x - 1)^2 (x + 2)} \;dx = \frac{2}{9} (\log_e |x - 1| - \log_e |x + 2|) - \frac{1}{3(x - 1)} + C$

$\int \frac{x}{(x - 1)^2 (x + 2)} \;dx = \frac{2}{9} \log_e \left|\frac{x - 1}{x + 2}\right| - \frac{1}{3(x - 1)} + C$

Final Answer:

The integral is $\int \frac{x}{(x − 1)^2 (x + 2)} \;dx = \frac{2}{9} \log_e |x - 1| - \frac{1}{3(x - 1)} - \frac{2}{9} \log_e |x + 2| + C$.

Given:

We are asked to find the integral $\int \frac{3x + 5}{x^3 − x^2 − x + 1} \;dx$.

Solution:

The integrand is a rational function. First, we need to factor the denominator: $x^3 − x^2 − x + 1$.

We can factor by grouping:

$x^3 − x^2 − x + 1 = x^2(x - 1) - 1(x - 1)$

$= (x^2 - 1)(x - 1)$

Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$, we factor $x^2 - 1$:

$x^2 - 1 = (x - 1)(x + 1)$

So the denominator is $(x - 1)(x + 1)(x - 1) = (x - 1)^2 (x + 1)$.

The integral is $\int \frac{3x + 5}{(x - 1)^2 (x + 1)} \;dx$.

The denominator has a repeated linear factor $(x-1)^2$ and a distinct linear factor $(x+1)$. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x - 1)^2 (x + 1)$, we get:

We can find the values of A, B, and C by substituting convenient values for $x$ or by comparing coefficients.

Setting $x = 1$ in equation (2):

Thus, $\mathbf{B = 4}$.

Setting $x = -1$ in equation (2):

Thus, $\mathbf{C = \frac{1}{2}}$.

To find A, compare the coefficients of $x^2$ in equation (2). Expanding the right side:

$3x + 5 = A(x^2 - 1) + B(x + 1) + C(x^2 - 2x + 1)$

$3x + 5 = Ax^2 - A + Bx + B + Cx^2 - 2Cx + C$

$3x + 5 = (A + C)x^2 + (B - 2C)x + (-A + B + C)$

Comparing the coefficients of $x^2$ on both sides:

Since $C = \frac{1}{2}$, we have:

Thus, $\mathbf{A = -\frac{1}{2}}$.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we integrate each term:

$\int \frac{3x + 5}{(x - 1)^2 (x + 1)} \;dx = \int \left( -\frac{1}{2(x - 1)} + \frac{4}{(x - 1)^2} + \frac{1}{2(x + 1)} \right) \;dx$

Using the linearity of the integral:

$\int \frac{3x + 5}{(x - 1)^2 (x + 1)} \;dx = -\frac{1}{2} \int \frac{1}{x - 1} \;dx + 4 \int (x - 1)^{-2} \;dx + \frac{1}{2} \int \frac{1}{x + 1} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u|$ and $\int u^n du = \frac{u^{n+1}}{n+1}$ for $n \neq -1$.

$\int \frac{1}{x - 1} \;dx = \log_e |x - 1|$

$\int (x - 1)^{-2} \;dx = \frac{(x - 1)^{-2 + 1}}{-2 + 1} = \frac{(x - 1)^{-1}}{-1} = -\frac{1}{x - 1}$

$\int \frac{1}{x + 1} \;dx = \log_e |x + 1|$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{3x + 5}{(x - 1)^2 (x + 1)} \;dx = -\frac{1}{2} \log_e |x - 1| + 4 \left(-\frac{1}{x - 1}\right) + \frac{1}{2} \log_e |x + 1| + C$

$\int \frac{3x + 5}{(x - 1)^2 (x + 1)} \;dx = -\frac{1}{2} \log_e |x - 1| - \frac{4}{x - 1} + \frac{1}{2} \log_e |x + 1| + C$

where $C$ is the constant of integration.

Using logarithm properties, we can rearrange the terms:

$\int \frac{3x + 5}{(x - 1)^2 (x + 1)} \;dx = \frac{1}{2} \log_e |x + 1| - \frac{1}{2} \log_e |x - 1| - \frac{4}{x - 1} + C$

$\int \frac{3x + 5}{(x - 1)^2 (x + 1)} \;dx = \frac{1}{2} (\log_e |x + 1| - \log_e |x - 1|) - \frac{4}{x - 1} + C$

$\int \frac{3x + 5}{(x - 1)^2 (x + 1)} \;dx = \frac{1}{2} \log_e \left|\frac{x + 1}{x - 1}\right| - \frac{4}{x - 1} + C$

Final Answer:

The integral is $\int \frac{3x + 5}{x^3 − x^2 − x + 1} \;dx = \frac{1}{2} \log_e \left|\frac{x + 1}{x - 1}\right| - \frac{4}{x - 1} + C$.

Given:

We are asked to find the integral $\int \frac{2x − 3}{(x^2 − 1) (2x + 3)} \;dx$.

Solution:

The integrand is a rational function. First, we factor the denominator: $(x^2 − 1) (2x + 3)$.

The term $x^2 - 1$ is a difference of squares: $x^2 - 1 = (x - 1)(x + 1)$.

So the denominator is $(x - 1)(x + 1)(2x + 3)$.

The integral is $\int \frac{2x − 3}{(x - 1) (x + 1) (2x + 3)} \;dx$.

The denominator is a product of distinct linear factors. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x - 1)(x + 1)(2x + 3)$, we get:

We can find the values of A, B, and C by substituting convenient values for $x$ in equation (2).

Setting $x = 1$ in equation (2):

Thus, $\mathbf{A = -\frac{1}{10}}$.

Setting $x = -1$ in equation (2):

Thus, $\mathbf{B = \frac{5}{2}}$.

Setting $x = -\frac{3}{2}$ in equation (2):

Thus, $\mathbf{C = -\frac{24}{5}}$.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{2x − 3}{(x^2 − 1) (2x + 3)} \;dx = \int \left( -\frac{1}{10(x - 1)} + \frac{5}{2(x + 1)} - \frac{24}{5(2x + 3)} \right) \;dx$

Using the linearity of the integral:

$\int \frac{2x − 3}{(x^2 − 1) (2x + 3)} \;dx = -\frac{1}{10} \int \frac{1}{x - 1} \;dx + \frac{5}{2} \int \frac{1}{x + 1} \;dx - \frac{24}{5} \int \frac{1}{2x + 3} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u|$ and $\int \frac{1}{ax+b} \;dx = \frac{1}{a} \log_e |ax+b|$.

$\int \frac{1}{x - 1} \;dx = \log_e |x - 1|$

$\int \frac{1}{x + 1} \;dx = \log_e |x + 1|$

$\int \frac{1}{2x + 3} \;dx = \frac{1}{2} \log_e |2x + 3|$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{2x − 3}{(x^2 − 1) (2x + 3)} \;dx = -\frac{1}{10} \log_e |x - 1| + \frac{5}{2} \log_e |x + 1| - \frac{24}{5} \left(\frac{1}{2} \log_e |2x + 3|\right) + C$

$\int \frac{2x − 3}{(x^2 − 1) (2x + 3)} \;dx = -\frac{1}{10} \log_e |x - 1| + \frac{5}{2} \log_e |x + 1| - \frac{12}{5} \log_e |2x + 3| + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{2x − 3}{(x^2 − 1) (2x + 3)} \;dx = -\frac{1}{10} \log_e |x - 1| + \frac{5}{2} \log_e |x + 1| - \frac{12}{5} \log_e |2x + 3| + C$.

Given:

We are asked to find the integral $\int \frac{5x}{(x + 1) (x^2 − 4)} \;dx$.

Solution:

The integrand is a rational function. First, we need to factor the denominator: $(x + 1) (x^2 − 4)$.

The term $x^2 - 4$ is a difference of squares: $x^2 - 4 = x^2 - 2^2 = (x - 2)(x + 2)$.

So the denominator is $(x + 1)(x - 2)(x + 2)$.

The integral is $\int \frac{5x}{(x + 1) (x - 2) (x + 2)} \;dx$.

The denominator is a product of distinct linear factors. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x + 1)(x - 2)(x + 2)$, we get:

We can find the values of A, B, and C by substituting convenient values for $x$ in equation (2).

Setting $x = -1$ in equation (2):

Thus, $\mathbf{A = \frac{5}{3}}$.

Setting $x = 2$ in equation (2):

Thus, $\mathbf{B = \frac{5}{6}}$.

Setting $x = -2$ in equation (2):

Thus, $\mathbf{C = -\frac{5}{2}}$.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{5x}{(x + 1) (x^2 − 4)} \;dx = \int \left( \frac{5}{3(x + 1)} + \frac{5}{6(x - 2)} - \frac{5}{2(x + 2)} \right) \;dx$

Using the linearity of the integral:

$\int \frac{5x}{(x + 1) (x^2 − 4)} \;dx = \frac{5}{3} \int \frac{1}{x + 1} \;dx + \frac{5}{6} \int \frac{1}{x - 2} \;dx - \frac{5}{2} \int \frac{1}{x + 2} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u| + C'$.

So, $\int \frac{1}{x + 1} \;dx = \log_e |x + 1|$

$\int \frac{1}{x - 2} \;dx = \log_e |x - 2|$

$\int \frac{1}{x + 2} \;dx = \log_e |x + 2|$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{5x}{(x + 1) (x^2 − 4)} \;dx = \frac{5}{3} \log_e |x + 1| + \frac{5}{6} \log_e |x - 2| - \frac{5}{2} \log_e |x + 2| + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{5x}{(x + 1) (x^2 − 4)} \;dx = \frac{5}{3} \log_e |x + 1| + \frac{5}{6} \log_e |x - 2| - \frac{5}{2} \log_e |x + 2| + C$.

Given:

We are asked to find the integral $\int \frac{x^3 + x + 1}{x^2 − 1} \;dx$.

Solution:

The integrand is a rational function where the degree of the numerator (3) is greater than the degree of the denominator (2). This is an improper rational function, so we first perform polynomial long division.

Divide $x^3 + x + 1$ by $x^2 - 1$:

From the long division, we can write the integrand as the sum of the quotient and the remainder over the divisor:

Now, we need to integrate $x + \frac{2x + 1}{x^2 - 1}$. The integral of $x$ is $\int x \;dx = \frac{x^2}{2}$.

For the term $\frac{2x + 1}{x^2 - 1}$, we factor the denominator: $x^2 - 1 = (x - 1)(x + 1)$.

So we have the term $\frac{2x + 1}{(x - 1)(x + 1)}$. We use partial fraction decomposition for this proper rational function.

We write it in the form:

Multiplying both sides of equation (2) by $(x - 1)(x + 1)$, we get:

To find the values of A and B, we substitute convenient values for $x$.

Setting $x = 1$:

Thus, $\mathbf{A = \frac{3}{2}}$.

Setting $x = -1$:

Thus, $\mathbf{B = \frac{1}{2}}$.

Substituting the values of A and B back into equation (2), we get:

Now, substitute this back into the integral expression from equation (1):

$\int \frac{x^3 + x + 1}{x^2 − 1} \;dx = \int \left( x + \frac{3}{2(x - 1)} + \frac{1}{2(x + 1)} \right) \;dx$

Using the linearity of the integral:

$\int \frac{x^3 + x + 1}{x^2 − 1} \;dx = \int x \;dx + \frac{3}{2} \int \frac{1}{x - 1} \;dx + \frac{1}{2} \int \frac{1}{x + 1} \;dx$

We evaluate each integral:

$\int x \;dx = \frac{x^2}{2}$

$\int \frac{1}{x - 1} \;dx = \log_e |x - 1|$

$\int \frac{1}{x + 1} \;dx = \log_e |x + 1|$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{x^3 + x + 1}{x^2 − 1} \;dx = \frac{x^2}{2} + \frac{3}{2} \log_e |x - 1| + \frac{1}{2} \log_e |x + 1| + C$

where $C$ is the constant of integration.

Using logarithm properties, we can also write:

$\int \frac{x^3 + x + 1}{x^2 − 1} \;dx = \frac{x^2}{2} + \frac{1}{2} (3 \log_e |x - 1| + \log_e |x + 1|) + C$

$\int \frac{x^3 + x + 1}{x^2 − 1} \;dx = \frac{x^2}{2} + \frac{1}{2} (\log_e |(x - 1)^3| + \log_e |x + 1|) + C$

$\int \frac{x^3 + x + 1}{x^2 − 1} \;dx = \frac{x^2}{2} + \frac{1}{2} \log_e \left|(x - 1)^3 (x + 1)\right| + C$

Final Answer:

The integral is $\int \frac{x^3 + x + 1}{x^2 − 1} \;dx = \frac{x^2}{2} + \frac{3}{2} \log_e |x - 1| + \frac{1}{2} \log_e |x + 1| + C$.

Given:

We are asked to find the integral $\int \frac{2}{(1 − x) (1 + x^2)} \;dx$.

Solution:

The integrand is a rational function $\frac{2}{(1 − x) (1 + x^2)}$. The denominator consists of a distinct linear factor $(1-x)$ and an irreducible quadratic factor $(1+x^2)$. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(1 - x)(1 + x^2)$, we get:

We can find the values of A, B, and C by substituting a convenient value for $x$ and by comparing coefficients.

Setting $x = 1$ in equation (2):

Thus, $\mathbf{A = 1}$.

Now, let's expand equation (2) and compare coefficients:

$2 = A(1 + x^2) + (Bx + C)(1 - x)$

$2 = A + Ax^2 + Bx - Bx^2 + C - Cx$

$2 = (A - B)x^2 + (B - C)x + (A + C)$

Comparing the coefficients of the powers of $x$ on both sides:

Coefficient of $x^2$: $0 = A - B$

Coefficient of $x$: $0 = B - C$

Constant term: $2 = A + C$

We have $A = 1$. Substitute this into the first equation:

Thus, $\mathbf{B = 1}$.

Substitute $B = 1$ into the second equation:

Thus, $\mathbf{C = 1}$.

We can verify this with the third equation using $A = 1$ and $C = 1$:

This matches the constant term on the left side, confirming our values for A, B, and C are correct.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we integrate each term:

$\int \frac{2}{(1 - x) (1 + x^2)} \;dx = \int \left( \frac{1}{1 - x} + \frac{x + 1}{x^2 + 1} \right) dx$

Using the linearity of the integral:

$\int \frac{2}{(1 - x) (1 + x^2)} \;dx = \int \frac{1}{1 - x} \;dx + \int \frac{x + 1}{x^2 + 1} \;dx$

We evaluate each integral separately.

For the first integral, let $u = 1 - x$, then $du = -dx$:

$\int \frac{1}{1 - x} \;dx = \int \frac{1}{u} (-du) = - \int \frac{1}{u} \;du = -\log_e |u| = -\log_e |1 - x|$.

For the second integral, we split it into two parts:

$\int \frac{x + 1}{x^2 + 1} \;dx = \int \frac{x}{x^2 + 1} \;dx + \int \frac{1}{x^2 + 1} \;dx$

For the first part $\int \frac{x}{x^2 + 1} \;dx$, let $v = x^2 + 1$, then $dv = 2x \;dx$:

$\int \frac{x}{x^2 + 1} \;dx = \int \frac{1}{v} \frac{1}{2} \;dv = \frac{1}{2} \int \frac{1}{v} \;dv = \frac{1}{2} \log_e |v| = \frac{1}{2} \log_e |x^2 + 1| = \frac{1}{2} \log_e (x^2 + 1)$ (since $x^2+1 > 0$).

For the second part $\int \frac{1}{x^2 + 1} \;dx$, this is a standard integral:

$\int \frac{1}{x^2 + 1} \;dx = \tan^{-1}(x)$

So, $\int \frac{x + 1}{x^2 + 1} \;dx = \frac{1}{2} \log_e (x^2 + 1) + \tan^{-1}(x)$.

(We add the constant of integration at the end).

Combining all parts of the original integral:

$\int \frac{2}{(1 - x) (1 + x^2)} \;dx = -\log_e |1 - x| + \frac{1}{2} \log_e (x^2 + 1) + \tan^{-1}(x) + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{2}{(1 − x) (1 + x^2)} \;dx = -\log_e |1 - x| + \frac{1}{2} \log_e (x^2 + 1) + \tan^{-1}(x) + C$.

Given:

We are asked to find the integral $\int \frac{3x − 1}{(x + 2)^2} \;dx$.

Solution:

The integrand is a rational function $\frac{3x − 1}{(x + 2)^2}$. The denominator has a repeated linear factor $(x+2)^2$. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x + 2)^2$, we get:

To find the values of A and B, we substitute a convenient value for $x$ or compare coefficients.

Setting $x = -2$:

Thus, $\mathbf{B = -7}$.

Compare the coefficients of $x$ in the equation $3x - 1 = A(x + 2) + B = Ax + 2A + B$:

Thus, $\mathbf{A = 3}$.

Substituting the values of A and B back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{3x − 1}{(x + 2)^2} \;dx = \int \left( \frac{3}{x + 2} - \frac{7}{(x + 2)^2} \right) \;dx$

Using the linearity of the integral:

$\int \frac{3x − 1}{(x + 2)^2} \;dx = 3 \int \frac{1}{x + 2} \;dx - 7 \int (x + 2)^{-2} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u|$ and $\int u^n du = \frac{u^{n+1}}{n+1}$ for $n \neq -1$.

$\int \frac{1}{x + 2} \;dx = \log_e |x + 2|$

$\int (x + 2)^{-2} \;dx = \frac{(x + 2)^{-2 + 1}}{-2 + 1} = \frac{(x + 2)^{-1}}{-1} = -\frac{1}{x + 2}$

(We add the constant of integration at the end).

Combining these results:

$\int \frac{3x − 1}{(x + 2)^2} \;dx = 3 \log_e |x + 2| - 7 \left(-\frac{1}{x + 2}\right) + C$

$\int \frac{3x − 1}{(x + 2)^2} \;dx = 3 \log_e |x + 2| + \frac{7}{x + 2} + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{3x − 1}{(x + 2)^2} \;dx = 3 \log_e |x + 2| + \frac{7}{x + 2} + C$.

Given:

We are asked to find the integral $\int \frac{1}{x^4 − 1} \;dx$.

Solution:

The integrand is a rational function $\frac{1}{x^4 − 1}$. First, we need to factor the denominator $x^4 - 1$.

We can factor it as a difference of squares: $x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)$.

The term $x^2 - 1$ is also a difference of squares: $x^2 - 1 = (x - 1)(x + 1)$.

So, the complete factorization of the denominator is $x^4 - 1 = (x - 1)(x + 1)(x^2 + 1)$.

The integral is $\int \frac{1}{(x - 1)(x + 1)(x^2 + 1)} \;dx$.

The denominator consists of distinct linear factors $(x-1)$ and $(x+1)$, and an irreducible quadratic factor $(x^2+1)$. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x - 1)(x + 1)(x^2 + 1)$, we get:

We can find the values of A, B, C, and D by substituting convenient values for $x$ or by comparing coefficients.

Setting $x = 1$ in equation (2):

Thus, $\mathbf{A = \frac{1}{4}}$.

Setting $x = -1$ in equation (2):

Thus, $\mathbf{B = -\frac{1}{4}}$.

Now, let's expand equation (2) and compare coefficients:

$1 = A(x^3 + x^2 + x + 1) + B(x^3 - x^2 + x - 1) + (Cx + D)(x^2 - 1)$

$1 = Ax^3 + Ax^2 + Ax + A + Bx^3 - Bx^2 + Bx - B + Cx^3 - Cx + Dx^2 - D$

$1 = (A + B + C)x^3 + (A - B + D)x^2 + (A + B - C)x + (A - B - D)$

Comparing the coefficients of the powers of $x$ on both sides:

Coefficient of $x^3$: $0 = A + B + C$

Coefficient of $x^2$: $0 = A - B + D$

Coefficient of $x$: $0 = A + B - C$

Constant term: $1 = A - B - D$

We have $A = \frac{1}{4}$ and $B = -\frac{1}{4}$. Substitute these into the equation for the coefficient of $x^3$:

Thus, $\mathbf{C = 0}$.

Now substitute $A = \frac{1}{4}$ and $B = -\frac{1}{4}$ into the equation for the coefficient of $x^2$:

Thus, $\mathbf{D = -\frac{1}{2}}$.

We can verify these values with the constant term equation using $A = \frac{1}{4}$, $B = -\frac{1}{4}$, and $D = -\frac{1}{2}$:

This matches the constant term on the left side, confirming our values for A, B, C, and D are correct.

Substituting the values of A, B, C, and D back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{1}{x^4 - 1} \;dx = \int \left( \frac{1}{4(x - 1)} - \frac{1}{4(x + 1)} - \frac{1}{2(x^2 + 1)} \right) \;dx$

Using the linearity of the integral:

$\int \frac{1}{x^4 - 1} \;dx = \frac{1}{4} \int \frac{1}{x - 1} \;dx - \frac{1}{4} \int \frac{1}{x + 1} \;dx - \frac{1}{2} \int \frac{1}{x^2 + 1} \;dx$

We use the standard integral formulas:

$\int \frac{1}{u} \;du = \log_e |u|$

$\int \frac{1}{x^2 + a^2} \;dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$

Applying these formulas:

$\int \frac{1}{x - 1} \;dx = \log_e |x - 1|$

$\int \frac{1}{x + 1} \;dx = \log_e |x + 1|$

$\int \frac{1}{x^2 + 1} \;dx = \tan^{-1}(x)$ (Here $a=1$).

(We add the constant of integration at the end).

Combining these results:

$\int \frac{1}{x^4 - 1} \;dx = \frac{1}{4} \log_e |x - 1| - \frac{1}{4} \log_e |x + 1| - \frac{1}{2} \tan^{-1}(x) + C$

where $C$ is the constant of integration.

Using the logarithm property $\log_e a - \log_e b = \log_e \frac{a}{b}$, we can simplify the result:

$\int \frac{1}{x^4 - 1} \;dx = \frac{1}{4} (\log_e |x - 1| - \log_e |x + 1|) - \frac{1}{2} \tan^{-1}(x) + C$

$\int \frac{1}{x^4 - 1} \;dx = \frac{1}{4} \log_e \left|\frac{x - 1}{x + 1}\right| - \frac{1}{2} \tan^{-1}(x) + C$

Final Answer:

The integral is $\int \frac{1}{x^4 − 1} \;dx = \frac{1}{4} \log_e \left|\frac{x - 1}{x + 1}\right| - \frac{1}{2} \tan^{-1}(x) + C$.

Given:

We are asked to find the integral $\int \frac{1}{x(x^n + 1)} \;dx$.

Solution:

We follow the hint provided.

Multiply the numerator and the denominator by $x^{n-1}$.

$\frac{1}{x(x^n + 1)} = \frac{1}{x(x^n + 1)} \times \frac{x^{n-1}}{x^{n-1}} = \frac{x^{n-1}}{x \cdot x^{n-1} (x^n + 1)} = \frac{x^{n-1}}{x^n (x^n + 1)}$

So, the integral becomes:

Now, let's use the substitution suggested: $t = x^n$.

Differentiate $t$ with respect to $x$:

This implies $dt = nx^{n-1} \;dx$.

So, $x^{n-1} \;dx = \frac{1}{n} \;dt$.

Substitute $t = x^n$ and $x^{n-1} \;dx = \frac{1}{n} \;dt$ into integral (1):

Now, we need to evaluate the integral $\int \frac{1}{t(t + 1)} \;dt$. We use partial fraction decomposition for the integrand $\frac{1}{t(t + 1)}$.

We write it in the form:

Multiply both sides of equation (3) by $t(t + 1)$:

To find the values of A and B:

Set $t = 0$:

Thus, $\mathbf{A = 1}$.

Set $t = -1$:

Thus, $\mathbf{B = -1}$.

Substitute the values of A and B back into equation (3):

Now, substitute this back into integral (2):

Using the linearity of the integral:

We know that $\int \frac{1}{u} \;du = \log_e |u| + C'$.

(We add the constant of integration at the end).

Combining these results for the integral with respect to $t$:

Using the logarithm property $\log_e a - \log_e b = \log_e \frac{a}{b}$:

Finally, substitute back $t = x^n$:

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{1}{x(x^n + 1)} \;dx = \frac{1}{n} \log_e \left|\frac{x^n}{x^n + 1}\right| + C$.

Given:

We are asked to find the integral $\int \frac{\cos x}{(1 − \sin x) (2 − \sin x)} \;dx$.

Solution:

We use the substitution suggested in the hint. Let $t = \sin x$.

Differentiating both sides with respect to $x$, we get:

This implies $dt = \cos x \;dx$.

Substitute $t = \sin x$ and $dt = \cos x \;dx$ into the given integral:

Now, we evaluate the integral $\int \frac{1}{(1 - t)(2 - t)} \;dt$ using partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (2) by $(1 - t)(2 - t)$, we get:

To find the values of A and B, we substitute convenient values for $t$:

Setting $t = 1$:

Thus, $\mathbf{A = 1}$.

Setting $t = 2$:

Thus, $\mathbf{B = 1}$.

Substituting the values of A and B back into equation (2), we get:

Now, we integrate this expression with respect to $t$ (from integral (1)):

$\int \frac{dt}{(1 - t)(2 - t)} = \int \left( \frac{1}{1 - t} + \frac{1}{2 - t} \right) \;dt$

Using the linearity of the integral:

$\int \left( \frac{1}{1 - t} + \frac{1}{2 - t} \right) \;dt = \int \frac{1}{1 - t} \;dt + \int \frac{1}{2 - t} \;dt$

We know that $\int \frac{1}{ax+b} \;dx = \frac{1}{a} \log_e |ax+b| + C'$.

For the first integral, let $u = 1 - t$, then $du = -dt$, so $\int \frac{1}{1 - t} \;dt = - \int \frac{1}{u} \;du = -\log_e |1 - t|$.

For the second integral, let $v = 2 - t$, then $dv = -dt$, so $\int \frac{1}{2 - t} \;dt = - \int \frac{1}{v} \;dv = -\log_e |2 - t|$.

(We add the constant of integration at the end).

So, the integral with respect to $t$ is:

$-\log_e |1 - t| - \log_e |2 - t| + C$

This simplifies to $-\log_e (|1 - t| |2 - t|) + C$.

Let's double check the partial fraction calculation. $1 = A(2 - t) + B(1 - t)$ If $t=1$, $1 = A(1) + B(0) \implies A = 1$. If $t=2$, $1 = A(0) + B(-1) \implies B = -1$. The decomposition should be $\frac{1}{1 - t} - \frac{1}{2 - t}$.

Let's recalculate the integral using the correct decomposition:

$\int \left( \frac{1}{1 - t} - \frac{1}{2 - t} \right) \;dt = \int \frac{1}{1 - t} \;dt - \int \frac{1}{2 - t} \;dt$

Using $\int \frac{1}{ax+b} \;dx = \frac{1}{a} \log_e |ax+b| + C'$:

$\int \frac{1}{1 - t} \;dt = \frac{1}{-1} \log_e |1 - t| = -\log_e |1 - t|$

$\int \frac{1}{2 - t} \;dt = \frac{1}{-1} \log_e |2 - t| = -\log_e |2 - t|$

So, the integral is $-\log_e |1 - t| - (-\log_e |2 - t|) + C = -\log_e |1 - t| + \log_e |2 - t| + C$.

Using the logarithm property $\log_e a - \log_e b = \log_e \frac{a}{b}$:

$\int \frac{dt}{(1 - t)(2 - t)} = \log_e \left|\frac{2 - t}{1 - t}\right| + C$

Finally, substitute back $t = \sin x$:

$\int \frac{\cos x}{(1 − \sin x) (2 − \sin x)} \;dx = \log_e \left|\frac{2 - \sin x}{1 - \sin x}\right| + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{\cos x}{(1 − \sin x) (2 − \sin x)} \;dx = \log_e \left|\frac{2 - \sin x}{1 - \sin x}\right| + C$.

Given:

We are asked to find the integral $\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} \;dx$.

Solution:

The integrand is a rational function. First, let's expand the numerator and the denominator:

Numerator = $(x^2 + 1)(x^2 + 2) = x^4 + 2x^2 + x^2 + 2 = x^4 + 3x^2 + 2$.

Denominator = $(x^2 + 3)(x^2 + 4) = x^4 + 4x^2 + 3x^2 + 12 = x^4 + 7x^2 + 12$.

The integrand is $\frac{x^4 + 3x^2 + 2}{x^4 + 7x^2 + 12}$. Since the degree of the numerator is equal to the degree of the denominator, this is an improper rational function. We can perform algebraic manipulation or long division.

Let's rewrite the numerator in terms of the denominator:

$x^4 + 3x^2 + 2 = (x^4 + 7x^2 + 12) - 4x^2 - 10$.

So, the integrand becomes:

$\frac{x^4 + 3x^2 + 2}{x^4 + 7x^2 + 12} = \frac{(x^4 + 7x^2 + 12) - (4x^2 + 10)}{x^4 + 7x^2 + 12} = 1 - \frac{4x^2 + 10}{x^4 + 7x^2 + 12}$.

The integral is $\int \left( 1 - \frac{4x^2 + 10}{(x^2 + 3)(x^2 + 4)} \right) dx = \int 1 \;dx - \int \frac{4x^2 + 10}{(x^2 + 3)(x^2 + 4)} \;dx$.

Now we need to evaluate the integral of the second term: $\int \frac{4x^2 + 10}{(x^2 + 3)(x^2 + 4)} \;dx$.

The integrand $\frac{4x^2 + 10}{(x^2 + 3)(x^2 + 4)}$ is a proper rational function with irreducible quadratic factors in the denominator.

To perform partial fraction decomposition, we can treat $x^2$ as a variable temporarily. Let $y = x^2$.

$\frac{4y + 10}{(y + 3)(y + 4)} = \frac{A}{y + 3} + \frac{B}{y + 4}$.

Multiplying both sides by $(y + 3)(y + 4)$, we get:

Set $y = -3$:

Thus, $\mathbf{A = -2}$.

Set $y = -4$:

Thus, $\mathbf{B = 6}$.

Substituting the values of A and B back into the partial fraction decomposition with $y = x^2$:

Now substitute this back into the original integral:

$\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} \;dx = \int \left( 1 - \left( \frac{-2}{x^2 + 3} + \frac{6}{x^2 + 4} \right) \right) dx$

$\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} \;dx = \int \left( 1 + \frac{2}{x^2 + 3} - \frac{6}{x^2 + 4} \right) dx$

Using the linearity of the integral:

$\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} \;dx = \int 1 \;dx + 2 \int \frac{1}{x^2 + 3} \;dx - 6 \int \frac{1}{x^2 + 4} \;dx$

We use the standard integral formula $\int \frac{1}{x^2 + a^2} \;dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C'$.

For $\int \frac{1}{x^2 + 3} \;dx$, we have $a^2 = 3$, so $a = \sqrt{3}$.

For $\int \frac{1}{x^2 + 4} \;dx$, we have $a^2 = 4$, so $a = 2$.

Also, $\int 1 \;dx = x$.

(We add the constant of integration at the end).

Combining these results:

$\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} \;dx = x + 2 \left( \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) \right) - 6 \left( \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right) + C$

$\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} \;dx = x + \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) - 3 \tan^{-1}\left(\frac{x}{2}\right) + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{(x^2 + 1) (x^2 + 2)}{(x^2 + 3) (x^2 + 4)} \;dx = x + \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) - 3 \tan^{-1}\left(\frac{x}{2}\right) + C$.

Given:

We are asked to find the integral $\int \frac{2x}{(x^2 + 1) (x^2 + 3)} \;dx$.

Solution:

Let's use the method of substitution to simplify the integral. Let $u = x^2$.

Differentiating $u$ with respect to $x$, we get:

This implies $du = 2x \;dx$.

Now, substitute $u = x^2$ and $du = 2x \;dx$ into the given integral:

This is a proper rational function in terms of $u$ with distinct linear factors in the denominator. We use partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (2) by $(u + 1)(u + 3)$, we get:

To find the values of A and B, we substitute convenient values for $u$:

Setting $u = -1$:

Thus, $\mathbf{A = \frac{1}{2}}$.

Setting $u = -3$:

Thus, $\mathbf{B = -\frac{1}{2}}$.

Substituting the values of A and B back into equation (2), we get:

Now, integrate this expression with respect to $u$ (from integral (1)):

$\int \frac{du}{(u + 1)(u + 3)} = \int \left( \frac{1}{2(u + 1)} - \frac{1}{2(u + 3)} \right) \;du$

Using the linearity of the integral:

$\int \left( \frac{1}{2(u + 1)} - \frac{1}{2(u + 3)} \right) \;du = \frac{1}{2} \int \frac{1}{u + 1} \;du - \frac{1}{2} \int \frac{1}{u + 3} \;du$

We know that $\int \frac{1}{v} \;dv = \log_e |v| + C'$.

(We add the constant of integration at the end).

So, the integral with respect to $u$ is:

$\frac{1}{2} \log_e |u + 1| - \frac{1}{2} \log_e |u + 3| + C$

Using the logarithm property $\log_e a - \log_e b = \log_e \frac{a}{b}$:

$\frac{1}{2} (\log_e |u + 1| - \log_e |u + 3|) + C = \frac{1}{2} \log_e \left|\frac{u + 1}{u + 3}\right| + C$

Finally, substitute back $u = x^2$:

$\int \frac{2x}{(x^2 + 1) (x^2 + 3)} \;dx = \frac{1}{2} \log_e \left|\frac{x^2 + 1}{x^2 + 3}\right| + C$

Since $x^2 + 1 > 0$ and $x^2 + 3 > 0$ for all real $x$, we can remove the absolute value signs from the fraction inside the logarithm:

$\int \frac{2x}{(x^2 + 1) (x^2 + 3)} \;dx = \frac{1}{2} \log_e \left(\frac{x^2 + 1}{x^2 + 3}\right) + C$

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{2x}{(x^2 + 1) (x^2 + 3)} \;dx = \frac{1}{2} \log_e \left(\frac{x^2 + 1}{x^2 + 3}\right) + C$.

Given:

We are asked to find the integral $\int \frac{1}{x(x^4 − 1)} \;dx$.

Solution:

The integrand is a rational function $\frac{1}{x(x^4 − 1)}$. First, we need to factor the denominator $x(x^4 - 1)$.

The term $x^4 - 1$ can be factored as a difference of squares: $x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)$.

The term $x^2 - 1$ can be further factored as a difference of squares: $x^2 - 1 = (x - 1)(x + 1)$.

So, the complete factorization of the denominator is $x(x - 1)(x + 1)(x^2 + 1)$.

The integral is $\int \frac{1}{x(x - 1)(x + 1)(x^2 + 1)} \;dx$.

The denominator consists of distinct linear factors ($x$, $x-1$, $x+1$) and an irreducible quadratic factor ($x^2+1$). We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $x(x - 1)(x + 1)(x^2 + 1)$, we get:

We can find the values of A, B, C, D, and E by substituting convenient values for $x$ and by comparing coefficients.

Setting $x = 0$ in equation (2):

Thus, $\mathbf{A = -1}$.

Setting $x = 1$ in equation (2):

Thus, $\mathbf{B = \frac{1}{4}}$.

Setting $x = -1$ in equation (2):

Thus, $\mathbf{C = \frac{1}{4}}$.

Expanding equation (2) and comparing coefficients of $x^3$:
$1 = A(x^4 - 1) + Bx(x^3 + x^2 + x + 1) + Cx(x^3 - x^2 + x - 1) + (Dx + E)(x^2 - 1)x$
$1 = A(x^4 - 1) + B(x^4 + x^3 + x^2 + x) + C(x^4 - x^3 + x^2 - x) + (Dx + E)(x^3 - x)$
$1 = Ax^4 - A + Bx^4 + Bx^3 + Bx^2 + Bx + Cx^4 - Cx^3 + Cx^2 - Cx + Dx^4 + Ex^3 - Dx^2 - Ex$
$1 = (A + B + C + D)x^4 + (B - C + E)x^3 + (B + C - D)x^2 + (B - C - E)x - A$

Comparing the coefficients of $x^3$ on both sides (left side has $0x^3$):

Substitute $B = \frac{1}{4}$ and $C = \frac{1}{4}$:

Thus, $\mathbf{E = 0}$.

Comparing the coefficients of $x^2$ on both sides (left side has $0x^2$):

Substitute $B = \frac{1}{4}$ and $C = \frac{1}{4}$:

Thus, $\mathbf{D = \frac{1}{2}}$.

Substituting the values of A, B, C, D, and E back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{1}{x(x^4 - 1)} \;dx = \int \left( -\frac{1}{x} + \frac{1}{4(x - 1)} + \frac{1}{4(x + 1)} + \frac{x}{2(x^2 + 1)} \right) \;dx$

Using the linearity of the integral:

$\int \frac{1}{x(x^4 - 1)} \;dx = -\int \frac{1}{x} \;dx + \frac{1}{4} \int \frac{1}{x - 1} \;dx + \frac{1}{4} \int \frac{1}{x + 1} \;dx + \frac{1}{2} \int \frac{x}{x^2 + 1} \;dx$

We use the standard integral formulas:

$\int \frac{1}{u} \;du = \log_e |u|$

For the last integral $\int \frac{x}{x^2 + 1} \;dx$, let $u = x^2 + 1$, then $du = 2x \;dx$, so $x \;dx = \frac{1}{2} du$.

$\int \frac{x}{x^2 + 1} \;dx = \int \frac{1}{u} \frac{1}{2} \;du = \frac{1}{2} \int \frac{1}{u} \;du = \frac{1}{2} \log_e |u| = \frac{1}{2} \log_e |x^2 + 1| = \frac{1}{2} \log_e (x^2 + 1)$ (since $x^2+1 > 0$).

(We add the constant of integration at the end).

Combining these results:

$\int \frac{1}{x(x^4 - 1)} \;dx = -\log_e |x| + \frac{1}{4} \log_e |x - 1| + \frac{1}{4} \log_e |x + 1| + \frac{1}{2} \left(\frac{1}{2} \log_e (x^2 + 1)\right) + C$

$\int \frac{1}{x(x^4 - 1)} \;dx = -\log_e |x| + \frac{1}{4} \log_e |x - 1| + \frac{1}{4} \log_e |x + 1| + \frac{1}{4} \log_e (x^2 + 1) + C$

where $C$ is the constant of integration.

Using logarithm properties ($n \log_e a = \log_e a^n$ and $\log_e a + \log_e b = \log_e (ab)$ and $\log_e a - \log_e b = \log_e (a/b)$), we can simplify the result:

$\int \frac{1}{x(x^4 - 1)} \;dx = \frac{1}{4} (\log_e |x - 1| + \log_e |x + 1| + \log_e (x^2 + 1)) - \log_e |x| + C$

$\int \frac{1}{x(x^4 - 1)} \;dx = \frac{1}{4} \log_e (|x - 1||x + 1|(x^2 + 1)) - \frac{4}{4}\log_e |x| + C$

$\int \frac{1}{x(x^4 - 1)} \;dx = \frac{1}{4} \log_e |(x^2 - 1)(x^2 + 1)| - \frac{1}{4}\log_e |x^4| + C$

$\int \frac{1}{x(x^4 - 1)} \;dx = \frac{1}{4} \log_e |x^4 - 1| - \frac{1}{4}\log_e |x^4| + C$

$\int \frac{1}{x(x^4 - 1)} \;dx = \frac{1}{4} \log_e \left|\frac{x^4 - 1}{x^4}\right| + C$

Final Answer:

The integral is $\int \frac{1}{x(x^4 − 1)} \;dx = \frac{1}{4} \log_e \left|\frac{x^4 - 1}{x^4}\right| + C$.

Given:

We are asked to find the integral $\int \frac{1}{e^x - 1} \;dx$.

Solution:

We use the substitution suggested in the hint. Let $t = e^x$.

Differentiating both sides with respect to $x$, we get:

So, $dt = e^x \;dx$.

Since $t = e^x$, we have $dx = \frac{dt}{e^x} = \frac{dt}{t}$.

Now, substitute $t = e^x$ and $dx = \frac{dt}{t}$ into the given integral:

Now, we evaluate the integral $\int \frac{1}{t(t - 1)} \;dt$ using partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (2) by $t(t - 1)$, we get:

To find the values of A and B, we substitute convenient values for $t$:

Setting $t = 0$:

Thus, $\mathbf{A = -1}$.

Setting $t = 1$:

Thus, $\mathbf{B = 1}$.

Substituting the values of A and B back into equation (2), we get:

Now, integrate this expression with respect to $t$ (from integral (1)):

$\int \frac{1}{t(t - 1)} \;dt = \int \left( \frac{1}{t - 1} - \frac{1}{t} \right) \;dt$

Using the linearity of the integral:

$\int \left( \frac{1}{t - 1} - \frac{1}{t} \right) \;dt = \int \frac{1}{t - 1} \;dt - \int \frac{1}{t} \;dt$

We know that $\int \frac{1}{u} \;du = \log_e |u| + C'$.

(We add the constant of integration at the end).

So, the integral with respect to $t$ is:

$\log_e |t - 1| - \log_e |t| + C$

Using the logarithm property $\log_e a - \log_e b = \log_e \frac{a}{b}$:

Finally, substitute back $t = e^x$:

$\int \frac{1}{e^x - 1} \;dx = \log_e \left|\frac{e^x - 1}{e^x}\right| + C$

Since $e^x > 0$ for all real $x$, we have $|e^x| = e^x$. The term $e^x - 1$ can be positive, negative, or zero (when $x=0$). The absolute value around $e^x - 1$ is necessary. The absolute value around $\frac{e^x-1}{e^x}$ is equivalent to $\frac{|e^x - 1|}{e^x}$.

The integral can also be written as $\log_e |e^x - 1| - \log_e |e^x| + C = \log_e |e^x - 1| - x + C$.

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{1}{e^x - 1} \;dx = \log_e \left|\frac{e^x - 1}{e^x}\right| + C$ or $\log_e |e^x - 1| - x + C$.

Given:

We are asked to find the integral $\int \frac{x}{(x − 1) (x − 2)} \;dx$ and choose the correct option.

Solution:

The integrand is a rational function $\frac{x}{(x − 1) (x − 2)}$. The denominator is a product of distinct linear factors. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $(x - 1)(x - 2)$, we get:

To find the values of A and B, we substitute convenient values for $x$ in equation (2).

Setting $x = 1$ in equation (2):

Thus, $\mathbf{A = -1}$.

Setting $x = 2$ in equation (2):

Thus, $\mathbf{B = 2}$.

Substituting the values of A and B back into equation (1), we get:

Now, we can integrate the expression:

$\int \frac{x}{(x - 1)(x - 2)} \;dx = \int \left( \frac{-1}{x - 1} + \frac{2}{x - 2} \right) \;dx$

Using the linearity of the integral, we have:

$\int \frac{x}{(x - 1)(x - 2)} \;dx = - \int \frac{1}{x - 1} \;dx + 2 \int \frac{1}{x - 2} \;dx$

We know that $\int \frac{1}{u} \;du = \log_e |u| + C'$.

So, $\int \frac{1}{x - 1} \;dx = \log_e |x - 1|$

and $\int \frac{1}{x - 2} \;dx = \log_e |x - 2|$

(We add the constant of integration at the end).

Therefore, the integral is:

$\int \frac{x}{(x - 1)(x - 2)} \;dx = - \log_e |x - 1| + 2 \log_e |x - 2| + C$

where $C$ is the constant of integration.

Using the logarithm property $n \log_e a = \log_e a^n$ and $\log_e a - \log_e b = \log_e \frac{a}{b}$, we can simplify the result:

$\int \frac{x}{(x - 1)(x - 2)} \;dx = \log_e |(x - 2)^2| - \log_e |x - 1| + C$

$\int \frac{x}{(x - 1)(x - 2)} \;dx = \log_e \left|\frac{(x - 2)^2}{x - 1}\right| + C$

Comparing this result with the given options, we see that it matches option (B).

Option (A) is $\log_e \left|\frac{(x - 1)^2}{x - 2}\right| + C = 2\log_e |x - 1| - \log_e |x - 2| + C$. This is incorrect.

Option (C) is $\log_e \left|\left(\frac{x - 1}{x - 2}\right)^2\right| + C = 2\log_e \left|\frac{x - 1}{x - 2}\right| + C = 2(\log_e |x - 1| - \log_e |x - 2|) + C = 2\log_e |x - 1| - 2\log_e |x - 2| + C$. This is incorrect.

Option (D) is $\log_e |(x - 1)(x - 2)| + C = \log_e |x - 1| + \log_e |x - 2| + C$. This is incorrect.

Final Answer:

The correct answer is (B).

Given:

We are asked to find the integral $\int \frac{1}{x (x^2 + 1)} \;dx$ and choose the correct option.

Solution:

The integrand is a rational function $\frac{1}{x (x^2 + 1)}$. The denominator consists of a distinct linear factor $x$ and an irreducible quadratic factor $x^2 + 1$. We use the method of partial fraction decomposition.

We write the integrand in the form:

Multiplying both sides of equation (1) by $x(x^2 + 1)$, we get:

Expanding equation (2):

$1 = Ax^2 + A + Bx^2 + Cx$

$1 = (A + B)x^2 + Cx + A$

Comparing the coefficients of the powers of $x$ on both sides:

Coefficient of $x^2$: $A + B = 0$

Coefficient of $x$: $C = 0$

Constant term: $A = 1$

From the constant term, we have $\mathbf{A = 1}$.

From the coefficient of $x$, we have $\mathbf{C = 0}$.

Substitute $A = 1$ into the coefficient of $x^2$ equation:

Thus, $\mathbf{B = -1}$.

Substituting the values of A, B, and C back into equation (1), we get:

Now, we integrate the expression:

$\int \frac{1}{x(x^2 + 1)} \;dx = \int \left( \frac{1}{x} - \frac{x}{x^2 + 1} \right) \;dx$

Using the linearity of the integral:

$\int \frac{1}{x(x^2 + 1)} \;dx = \int \frac{1}{x} \;dx - \int \frac{x}{x^2 + 1} \;dx$

We evaluate each integral:

$\int \frac{1}{x} \;dx = \log_e |x|$

For the second integral $\int \frac{x}{x^2 + 1} \;dx$, let $u = x^2 + 1$. Then $du = 2x \;dx$. So $x \;dx = \frac{1}{2} du$.

Since $x^2 + 1 > 0$ for all real $x$, $|x^2 + 1| = x^2 + 1$. So, $\int \frac{x}{x^2 + 1} \;dx = \frac{1}{2} \log_e (x^2 + 1)$.

(We add the constant of integration at the end).

Combining these results:

$\int \frac{1}{x(x^2 + 1)} \;dx = \log_e |x| - \frac{1}{2} \log_e (x^2 + 1) + C$

where $C$ is the constant of integration.

Comparing this result with the given options, we see that it matches option (A).

Final Answer:

The correct answer is (A).

Given:

We are asked to find the integral $\int x \cos x \;dx$.

Solution:

The integral involves a product of two different types of functions: an algebraic function ($x$) and a trigonometric function ($\cos x$). We can solve this integral using the method of integration by parts.

The formula for integration by parts is:

We need to choose which part of the integrand will be $u$ and which will be $dv$. We can use the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to help make this choice. We generally choose $u$ as the function that comes first in the LIATE order.

In this case, we have $x$ (Algebraic) and $\cos x$ (Trigonometric). 'A' comes before 'T' in LIATE. So we choose $u = x$ and $dv = \cos x \;dx$.

Let $u = x$

Differentiating with respect to $x$ to find $du$:

Let $dv = \cos x \;dx$

Integrating with respect to $x$ to find $v$:

(We omit the constant of integration for $v$ at this step, as we will add a single constant at the end of the entire integral).

Now, substitute $u=x$, $dv=\cos x \;dx$, $du=dx$, and $v=\sin x$ into the integration by parts formula (1):

Now, we evaluate the remaining integral $\int \sin x \;dx$:

(We add the constant of integration here for the overall result).

Substitute this back into the equation for the main integral:

where $C$ is the constant of integration.

Final Answer:

The integral is $\int x \cos x \;dx = x \sin x + \cos x + C$.

Given:

We are asked to find the integral $\int \log x \;dx$. We assume $\log x$ refers to the natural logarithm, $\log_e x$. So, we need to find $\int \log_e x \;dx$.

Solution:

This integral can be evaluated using the method of integration by parts by considering the integrand as a product of two functions: $\log_e x$ and $1$.

The formula for integration by parts is:

We choose $u = \log_e x$ (as it comes first in the LIATE rule) and $dv = 1 \;dx$.

Let $u = \log_e x$

Differentiating with respect to $x$ to find $du$:

Let $dv = 1 \;dx$

Integrating with respect to $x$ to find $v$:

(We omit the constant of integration for $v$ at this step).

Now, substitute $u=\log_e x$, $dv=1 \;dx$, $du=\frac{1}{x} \;dx$, and $v=x$ into the integration by parts formula (1):

Now, we evaluate the remaining integral $\int 1 \;dx$:

(We add the constant of integration here for the overall result).

Substitute this back into the equation for the main integral:

This can also be written by factoring out $x$:

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \log x \;dx = x \log x - x + C$.

Given:

We are asked to find the integral $\int x e^x \;dx$.

Solution:

The integral involves a product of two different types of functions: an algebraic function ($x$) and an exponential function ($e^x$). We can solve this integral using the method of integration by parts.

The formula for integration by parts is:

We use the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to choose $u$ and $dv$. 'A' (Algebraic) comes before 'E' (Exponential).

So we choose $u = x$ and $dv = e^x \;dx$.

Let $u = x$

Differentiating with respect to $x$ to find $du$:

Let $dv = e^x \;dx$

Integrating with respect to $x$ to find $v$:

(We omit the constant of integration for $v$ at this step).

Now, substitute $u=x$, $dv=e^x \;dx$, $du=dx$, and $v=e^x$ into the integration by parts formula (1):

Now, we evaluate the remaining integral $\int e^x \;dx$:

(We add the constant of integration here for the overall result).

Substitute this back into the equation for the main integral:

This can also be written by factoring out $e^x$:

where $C$ is the constant of integration.

Final Answer:

The integral is $\int x e^x \;dx = x e^x - e^x + C$.

Given:

We are asked to find the integral $\int \frac{x \sin^{−1} x}{\sqrt{1 − x^2}} \;dx$.

Solution:

We can use the method of substitution to simplify the integral.

Let $t = \sin^{-1} x$.

Differentiating both sides with respect to $x$, we get:

This implies $dt = \frac{dx}{\sqrt{1 - x^2}}$.

Also, from $t = \sin^{-1} x$, we have $x = \sin t$.

Now, substitute $t = \sin^{-1} x$, $x = \sin t$, and $dt = \frac{dx}{\sqrt{1 - x^2}}$ into the given integral:

Now, we need to evaluate the integral $\int t \sin t \;dt$. This is a product of an algebraic function ($t$) and a trigonometric function ($\sin t$). We use the method of integration by parts.

The formula for integration by parts is:

Using the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we choose $u = t$ and $dv = \sin t \;dt$.

Let $u = t$

Differentiating to find $du$:

Let $dv = \sin t \;dt$

Integrating to find $v$:

(We omit the constant of integration for $v$ here).

Substitute $u=t$, $dv=\sin t \;dt$, $du=dt$, and $v=-\cos t$ into the integration by parts formula (2):

Now, evaluate the remaining integral $\int \cos t \;dt$:

(We add the constant of integration here for the overall result).

Combine the results for the integral with respect to $t$:

Finally, substitute back to the original variable $x$. Recall $t = \sin^{-1} x$ and $\sin t = x$.

We also need to express $\cos t$ in terms of $x$. Since $t = \sin^{-1} x$, we have $\sin t = x$. Consider a right-angled triangle where the angle is $t$. The opposite side is $x$ and the hypotenuse is $1$. By the Pythagorean theorem, the adjacent side is $\sqrt{1^2 - x^2} = \sqrt{1 - x^2}$.

Substitute $t = \sin^{-1} x$, $\cos t = \sqrt{1 - x^2}$, and $\sin t = x$ back into the result:

We can rearrange the terms:

where $C$ is the constant of integration.

Final Answer:

The integral is $\int \frac{x \sin^{−1} x}{\sqrt{1 − x^2}} \;dx = x - \sqrt{1 - x^2} \sin^{-1} x + C$.

Given:

We are asked to find the integral $\int e^x \sin x \;dx$.

Solution:

The integral involves a product of an exponential function ($e^x$) and a trigonometric function ($\sin x$). We can solve this integral using the method of integration by parts.

The formula for integration by parts is:

We choose $u = \sin x$ and $dv = e^x \;dx$ based on the LIATE rule (though the choice between exponential and trigonometric is less critical as long as it's consistent in subsequent steps). Choosing the trigonometric function as $u$ often works well here.

Let $u = \sin x$.

Differentiating with respect to $x$ to find $du$:

Let $dv = e^x \;dx$.

Integrating with respect to $x$ to find $v$:

(We omit the constant of integration for $v$ at this step).

Apply the integration by parts formula (1) with these choices:

Now, we need to evaluate the new integral $\int e^x \cos x \;dx$. This is also a product of an exponential and a trigonometric function, so we apply integration by parts again.

For $\int e^x \cos x \;dx$, we must make a consistent choice for $u'$ and $dv'$. Since we chose the trigonometric function ($\sin x$) as $u$ and the exponential function ($e^x$) as $dv$ in the first step, we do the same for $\cos x$ and $e^x$.

Let $u' = \cos x$.

Differentiating to find $du'$:

Let $dv' = e^x \;dx$.

Integrating to find $v'$:

Apply the integration by parts formula to $\int e^x \cos x \;dx$ with $u'=\cos x$, $dv'=e^x \;dx$, $du'=-\sin x \;dx$, and $v'=e^x$:

Substitute the result from equation (3) back into equation (2):

Let $I = \int e^x \sin x \;dx$. The equation becomes:

Now, we solve this equation for $I$ by adding $I$ to both sides:

Divide by 2 to find $I$:

Finally, add the constant of integration $C$ to the result:

where $C$ is the constant of integration.

Final Answer:

The integral is $\int e^x \sin x \;dx = \frac{1}{2} e^x (\sin x - \cos x) + C$.

Part (i)

Given:

We are asked to find the integral $\int e^x \left( \tan^{−1} x + \frac{1}{1 + x^2} \right) \;dx$.

Solution:

We observe that the integrand is in the form $e^x [f(x) + f'(x)]$.

We know the standard integral formula:

In the given integral $\int e^x \left( \tan^{−1} x + \frac{1}{1 + x^2} \right) \;dx$, let's identify $f(x)$ and $f'(x)$.

Let $f(x) = \tan^{-1} x$.

Then, the derivative of $f(x)$ with respect to $x$ is:

We see that the term $\frac{1}{1 + x^2}$ is exactly the other term inside the parenthesis multiplied by $e^x$.

So, the integrand is indeed of the form $e^x [f(x) + f'(x)]$ with $f(x) = \tan^{-1} x$ and $f'(x) = \frac{1}{1 + x^2}$.

Applying the formula (1):

where $C$ is the constant of integration.

Final Answer (i):

The integral is $\int e^x \left( \tan^{−1} x + \frac{1}{1 + x^2} \right) \;dx = e^x \tan^{-1} x + C$.

Part (ii)

Given:

We are asked to find the integral $\int \frac{(x^2 + 1) e^x}{(x + 1)^2} \;dx$.

Solution:

We want to see if this integral can also be expressed in the form $\int e^x [f(x) + f'(x)] \;dx$.

We need to manipulate the fraction $\frac{x^2 + 1}{(x + 1)^2}$. The denominator is $(x + 1)^2 = x^2 + 2x + 1$.

Let's rewrite the numerator $x^2 + 1$ in terms of $(x+1)^2$ or $x+1$.

We can write $x^2 + 1 = (x^2 + 2x + 1) - 2x = (x + 1)^2 - 2x$.

So, the fraction is $\frac{(x + 1)^2 - 2x}{(x + 1)^2} = \frac{(x + 1)^2}{(x + 1)^2} - \frac{2x}{(x + 1)^2} = 1 - \frac{2x}{(x + 1)^2}$.

This doesn't seem to directly fit the form $f(x) + f'(x)$. Let's try another manipulation.

Consider $f(x) = \frac{x-1}{x+1}$.

Then the derivative $f'(x)$ is calculated using the quotient rule:

Now let's consider the sum $f(x) + f'(x)$:

To combine these, we find a common denominator $(x+1)^2$:

This is exactly the fractional part of our integrand.

So, the integrand is in the form $e^x [f(x) + f'(x)]$ with $f(x) = \frac{x-1}{x+1}$ and $f'(x) = \frac{2}{(x+1)^2}$.

Applying the standard integral formula (1):

where $C$ is the constant of integration.

Final Answer (ii):

The integral is $\int \frac{(x^2 + 1) e^x}{(x + 1)^2} \;dx = e^x \frac{x-1}{x+1} + C$.

We need to evaluate the integral $\int x \sin x \, dx$.