Integration by Parts Questions in English

(B) Given that $\int f(x) \, dx = g(x)$.
We need to evaluate $\int f^{-1}(x) \, dx$.
Using integration by parts,let $u = f^{-1}(x)$ and $dv = dx$. Then $du = \frac{d}{dx} f^{-1}(x) \, dx$ and $v = x$.
$\int f^{-1}(x) \, dx = x f^{-1}(x) - \int x \frac{d}{dx} f^{-1}(x) \, dx$.
Let $f^{-1}(x) = t$,then $x = f(t)$ and $dx = f'(t) \, dt$.
Also,$\frac{d}{dx} f^{-1}(x) \, dx = dt$.
Substituting these into the integral:
$\int f^{-1}(x) \, dx = x f^{-1}(x) - \int f(t) \, dt$.
Since $\int f(t) \, dt = g(t)$,we have:
$\int f^{-1}(x) \, dx = x f^{-1}(x) - g(t) = x f^{-1}(x) - g(f^{-1}(x))$.
Thus,the correct option is $B$.

(D) To evaluate the integral $\int x \sec^2 x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x$ and $dv = \sec^2 x \, dx$.
Then $du = dx$ and $v = \tan x$.
Applying the formula:
$\int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx$.
Since $\int \tan x \, dx = \log |\sec x| = -\log |\cos x|$,we have:
$\int x \sec^2 x \, dx = x \tan x - (-\log |\cos x|) + c = x \tan x + \log |\cos x| + c$.

(C) Let $I = \int \sin(\log x) \, dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t \, dt$.
Then,$I = \int e^t \sin t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,let $u = \sin t$ and $dv = e^t \, dt$.
Then $du = \cos t \, dt$ and $v = e^t$.
$I = e^t \sin t - \int e^t \cos t \, dt$.
Applying integration by parts again to $\int e^t \cos t \, dt$:
$I = e^t \sin t - [e^t \cos t - \int e^t (-\sin t) \, dt]$.
$I = e^t \sin t - e^t \cos t - I$.
$2I = e^t(\sin t - \cos t)$.
$I = \frac{1}{2} e^t(\sin t - \cos t) + C$.
Substituting $t = \log x$ and $e^t = x$:
$I = \frac{1}{2} x[\sin(\log x) - \cos(\log x)] + C$.

(A) We use the method of integration by parts for $\int x \sin x \, dx$.
Let $u = x$ and $dv = \sin x \, dx$.
Then $du = dx$ and $v = -\cos x$.
Using the formula $\int u \, dv = uv - \int v \, du$,we get:
$\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx$
$= -x \cos x + \int \cos x \, dx$
$= -x \cos x + \sin x + C$,where $C$ is the constant of integration.
Comparing this with the given expression $-x \cos x + A$,we find that $A = \sin x + C$.

(B) To evaluate the integral $\int x \log x \, dx$,we use the integration by parts formula: $\int u \, dv = uv - \int v \, du$.
Let $u = \log x$ and $dv = x \, dx$.
Then $du = \frac{1}{x} \, dx$ and $v = \int x \, dx = \frac{x^2}{2}$.
Applying the formula:
$\int x \log x \, dx = (\log x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx + c$
$= \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx + c$
$= \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} + c$
$= \frac{x^2}{2} \log x - \frac{x^2}{4} + c$.

(A) To solve the integral $\int x \cos x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x$ and $dv = \cos x \, dx$.
Then $du = dx$ and $v = \int \cos x \, dx = \sin x$.
Applying the formula: $\int x \cos x \, dx = x \sin x - \int \sin x \, dx$.
Since $\int \sin x \, dx = -\cos x$,we get:
$\int x \cos x \, dx = x \sin x - (-\cos x) + C = x \sin x + \cos x + C$.

(B) To evaluate the integral $\int \tan^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \tan^{-1} x$ and $dv = dx$.
Then $du = \frac{1}{1 + x^2} \, dx$ and $v = x$.
Applying the formula:
$\int \tan^{-1} x \, dx = x \tan^{-1} x - \int x \cdot \frac{1}{1 + x^2} \, dx$
$= x \tan^{-1} x - \int \frac{x}{1 + x^2} \, dx$
To solve $\int \frac{x}{1 + x^2} \, dx$,let $t = 1 + x^2$,so $dt = 2x \, dx$ or $x \, dx = \frac{1}{2} dt$.
$= x \tan^{-1} x - \frac{1}{2} \int \frac{1}{t} \, dt$
$= x \tan^{-1} x - \frac{1}{2} \log|t| + C$
$= x \tan^{-1} x - \frac{1}{2} \log(1 + x^2) + C$.

(A) To solve the integral $\int x \tan^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \tan^{-1} x$ and $dv = x \, dx$.
Then $du = \frac{1}{1+x^2} \, dx$ and $v = \frac{x^2}{2}$.
Applying the formula:
$\int x \tan^{-1} x \, dx = \frac{x^2}{2} \tan^{-1} x - \int \frac{x^2}{2(1+x^2)} \, dx$
$= \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int \frac{x^2 + 1 - 1}{1+x^2} \, dx$
$= \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} \int (1 - \frac{1}{1+x^2}) \, dx$
$= \frac{x^2}{2} \tan^{-1} x - \frac{1}{2} (x - \tan^{-1} x) + c$
$= \frac{x^2}{2} \tan^{-1} x - \frac{1}{2}x + \frac{1}{2} \tan^{-1} x + c$
$= \frac{1}{2}(x^2 + 1) \tan^{-1} x - \frac{1}{2}x + c$.

(B) To solve the integral $\int \frac{\log x}{x^3} \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \log x$ and $dv = x^{-3} \, dx$.
Then $du = \frac{1}{x} \, dx$ and $v = \int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$.
Applying the formula:
$\int \frac{\log x}{x^3} \, dx = (\log x)\left(-\frac{1}{2x^2}\right) - \int \left(-\frac{1}{2x^2}\right) \left(\frac{1}{x}\right) \, dx$
$= -\frac{\log x}{2x^2} + \frac{1}{2} \int x^{-3} \, dx$
$= -\frac{\log x}{2x^2} + \frac{1}{2} \left(\frac{x^{-2}}{-2}\right) + c$
$= -\frac{\log x}{2x^2} - \frac{1}{4x^2} + c$
$= -\frac{1}{4x^2}(2\log x + 1) + c$.

(D) Let $I = \int e^{-2x} \sin 3x \, dx$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = \sin 3x$ and $dv = e^{-2x} \, dx$.
Then $du = 3 \cos 3x \, dx$ and $v = -\frac{1}{2} e^{-2x}$.
$I = -\frac{1}{2} e^{-2x} \sin 3x - \int (-\frac{1}{2} e^{-2x}) (3 \cos 3x) \, dx$
$I = -\frac{1}{2} e^{-2x} \sin 3x + \frac{3}{2} \int e^{-2x} \cos 3x \, dx$.
Now,integrate $\int e^{-2x} \cos 3x \, dx$ by parts again:
Let $u = \cos 3x$ and $dv = e^{-2x} \, dx$.
Then $du = -3 \sin 3x \, dx$ and $v = -\frac{1}{2} e^{-2x}$.
$\int e^{-2x} \cos 3x \, dx = -\frac{1}{2} e^{-2x} \cos 3x - \int (-\frac{1}{2} e^{-2x}) (-3 \sin 3x) \, dx$
$= -\frac{1}{2} e^{-2x} \cos 3x - \frac{3}{2} \int e^{-2x} \sin 3x \, dx$.
Substituting this back into the expression for $I$:
$I = -\frac{1}{2} e^{-2x} \sin 3x + \frac{3}{2} [-\frac{1}{2} e^{-2x} \cos 3x - \frac{3}{2} I]$
$I = -\frac{1}{2} e^{-2x} \sin 3x - \frac{3}{4} e^{-2x} \cos 3x - \frac{9}{4} I$
$I + \frac{9}{4} I = -\frac{1}{4} e^{-2x} [2 \sin 3x + 3 \cos 3x]$
$\frac{13}{4} I = -\frac{1}{4} e^{-2x} [2 \sin 3x + 3 \cos 3x]$
$I = -\frac{1}{13} e^{-2x} [2 \sin 3x + 3 \cos 3x] + c$.

(B) We have $I = \int x \sin x \sec^3 x \, dx = \int x \tan x \sec^2 x \, dx$.
Using integration by parts,let $u = x$ and $dv = \tan x \sec^2 x \, dx$.
Then $du = dx$ and $v = \int \tan x \sec^2 x \, dx = \frac{\tan^2 x}{2}$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
$I = x \cdot \frac{\tan^2 x}{2} - \int \frac{\tan^2 x}{2} \, dx$.
$I = \frac{x \tan^2 x}{2} - \frac{1}{2} \int (\sec^2 x - 1) \, dx$.
$I = \frac{x \tan^2 x}{2} - \frac{1}{2} (\tan x - x) + c$.
$I = \frac{x(\sec^2 x - 1)}{2} - \frac{1}{2} \tan x + \frac{x}{2} + c$.
$I = \frac{x \sec^2 x}{2} - \frac{x}{2} - \frac{1}{2} \tan x + \frac{x}{2} + c$.
$I = \frac{1}{2} [x \sec^2 x - \tan x] + c$.

(D) We know that $\sin^2 x = \frac{1 - \cos 2x}{2}$.
Substituting this into the integral:
$\int x \sin^2 x \, dx = \int x \left( \frac{1 - \cos 2x}{2} \right) dx$
$= \frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos 2x \, dx$
$= \frac{1}{2} \left( \frac{x^2}{2} \right) - \frac{1}{2} \left[ x \left( \frac{\sin 2x}{2} \right) - \int 1 \cdot \frac{\sin 2x}{2} \, dx \right]$
$= \frac{x^2}{4} - \frac{x}{4} \sin 2x + \frac{1}{2} \int \frac{\sin 2x}{2} \, dx$
$= \frac{x^2}{4} - \frac{x}{4} \sin 2x + \frac{1}{4} \left( -\frac{\cos 2x}{2} \right) + c$
$= \frac{x^2}{4} - \frac{x}{4} \sin 2x - \frac{1}{8} \cos 2x + c$.

(A) We are given the integral $I = \int e^{2x + \log x} dx$.
Using the property of exponents $e^{a+b} = e^a \cdot e^b$ and $e^{\log x} = x$,we get:
$I = \int e^{2x} \cdot e^{\log x} dx = \int x e^{2x} dx$.
Now,apply integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = x$ and $v = e^{2x}$. Then $u' = 1$ and $\int v dx = \frac{e^{2x}}{2}$.
$I = x \cdot \frac{e^{2x}}{2} - \int 1 \cdot \frac{e^{2x}}{2} dx$.
$I = \frac{x e^{2x}}{2} - \frac{1}{2} \cdot \frac{e^{2x}}{2} + c$.
$I = \frac{2x e^{2x} - e^{2x}}{4} + c = \frac{e^{2x}}{4}(2x - 1) + c$.

(A) To evaluate the integral $I = \int \log(x + 1) \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \log(x + 1)$ and $dv = dx$.
Then $du = \frac{1}{x + 1} \, dx$ and $v = x$.
Applying the formula:
$I = x \log(x + 1) - \int \frac{x}{x + 1} \, dx + c$
$I = x \log(x + 1) - \int \frac{x + 1 - 1}{x + 1} \, dx + c$
$I = x \log(x + 1) - \int (1 - \frac{1}{x + 1}) \, dx + c$
$I = x \log(x + 1) - (x - \log(x + 1)) + c$
$I = x \log(x + 1) - x + \log(x + 1) + c$
$I = (x + 1) \log(x + 1) - x + c$.

(D) We are given $\int {\ln ({x^2} + x)dx} = x\ln ({x^2} + x) + A$.
Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = \ln(x^2+x)$ and $dv = dx$.
Then $du = \frac{2x+1}{x^2+x} dx$ and $v = x$.
So,$\int {\ln ({x^2} + x)dx} = x \ln(x^2+x) - \int x \cdot \frac{2x+1}{x^2+x} dx$.
$= x \ln(x^2+x) - \int \frac{2x^2+x}{x^2+x} dx$.
$= x \ln(x^2+x) - \int \frac{2(x^2+x) - x}{x^2+x} dx$.
$= x \ln(x^2+x) - \int (2 - \frac{x}{x(x+1)}) dx$.
$= x \ln(x^2+x) - \int (2 - \frac{1}{x+1}) dx$.
$= x \ln(x^2+x) - (2x - \ln|x+1|) + C$.
$= x \ln(x^2+x) - 2x + \ln|x+1| + C$.
Comparing this with $x \ln(x^2+x) + A$,we get $A = -2x + \ln|x+1| + C$.

(B) To evaluate $I = \int x^2 \sin 2x \, dx$,we use the integration by parts formula: $\int u \, dv = uv - \int v \, du$.
Let $u = x^2$ and $dv = \sin 2x \, dx$.
Then $du = 2x \, dx$ and $v = -\frac{\cos 2x}{2}$.
Applying the formula:
$I = x^2 \left(-\frac{\cos 2x}{2}\right) - \int \left(-\frac{\cos 2x}{2}\right) (2x \, dx)$
$I = -\frac{x^2 \cos 2x}{2} + \int x \cos 2x \, dx$.
Now,apply integration by parts again for $\int x \cos 2x \, dx$:
Let $u = x$ and $dv = \cos 2x \, dx$.
Then $du = dx$ and $v = \frac{\sin 2x}{2}$.
$\int x \cos 2x \, dx = x \left(\frac{\sin 2x}{2}\right) - \int \frac{\sin 2x}{2} \, dx$
$= \frac{x \sin 2x}{2} - \frac{1}{2} \left(-\frac{\cos 2x}{2}\right) = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4}$.
Substituting this back into the expression for $I$:
$I = -\frac{x^2 \cos 2x}{2} + \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} + c$.

(B) To evaluate the integral $\int \log x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \log x$ and $dv = dx$.
Then,$du = \frac{1}{x} \, dx$ and $v = x$.
Applying the formula:
$\int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx + c$
$= x \log x - \int 1 \, dx + c$
$= x \log x - x + c$.
Thus,the correct option is $B$.

(A) Let $I = \int e^x \sin x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \sin x$ and $dv = e^x \, dx$. Then $du = \cos x \, dx$ and $v = e^x$.
$I = e^x \sin x - \int e^x \cos x \, dx$.
Applying integration by parts again to $\int e^x \cos x \, dx$ with $u = \cos x$ and $dv = e^x \, dx$:
$I = e^x \sin x - [e^x \cos x - \int e^x (-\sin x) \, dx]$
$I = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx$
$I = e^x \sin x - e^x \cos x - I$
$2I = e^x (\sin x - \cos x) + c$
$I = \frac{1}{2} e^x (\sin x - \cos x) + c$.
Comparing this with the given expression $\frac{1}{2} e^x \cdot a + c$,we get $a = \sin x - \cos x$.

(D) To evaluate the integral $\int x^n \log x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \log x$ and $dv = x^n \, dx$.
Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^{n+1}}{n+1}$.
Applying the formula:
$\int x^n \log x \, dx = \log x \cdot \frac{x^{n+1}}{n+1} - \int \frac{x^{n+1}}{n+1} \cdot \frac{1}{x} \, dx$
$= \frac{x^{n+1}}{n+1} \log x - \frac{1}{n+1} \int x^n \, dx$
$= \frac{x^{n+1}}{n+1} \log x - \frac{1}{n+1} \cdot \frac{x^{n+1}}{n+1} + c$
$= \frac{x^{n+1}}{n+1} \left( \log x - \frac{1}{n+1} \right) + c$.

(B) Let $I = \int {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx$.
Using integration by parts for the first term $\int \frac{1}{\log x} dx$:
Let $u = \frac{1}{\log x}$ and $dv = dx$.
Then $du = -\frac{1}{x(\log x)^2} dx$ and $v = x$.
So,$\int \frac{1}{\log x} dx = \frac{x}{\log x} - \int x \left( -\frac{1}{x(\log x)^2} \right) dx = \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx$.
Substituting this back into the original integral:
$I = \left( \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx \right) - \int \frac{1}{(\log x)^2} dx + c$.
$I = \frac{x}{\log x} + c$.

(B) Let $I = \int e^x \sin x \, dx$.
Using integration by parts,$\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$.
Taking $u = \sin x$ and $v = e^x$,we get:
$I = \sin x \cdot e^x - \int \cos x \cdot e^x \, dx$.
Applying integration by parts again to $\int e^x \cos x \, dx$ with $u = \cos x$ and $v = e^x$:
$\int e^x \cos x \, dx = \cos x \cdot e^x - \int (-\sin x) \cdot e^x \, dx = e^x \cos x + \int e^x \sin x \, dx$.
Substituting this back into the equation for $I$:
$I = e^x \sin x - (e^x \cos x + I) + c$.
$I = e^x \sin x - e^x \cos x - I + c$.
$2I = e^x (\sin x - \cos x) + c$.
$I = \frac{1}{2} e^x (\sin x - \cos x) + c$.

(A) We need to evaluate the integral $I = \int (1 - x^2) \log x \, dx$.
Using the distributive property,we can write this as $I = \int \log x \, dx - \int x^2 \log x \, dx$.
For the first part,$\int \log x \, dx = x \log x - x$.
For the second part,we use integration by parts $\int u \, dv = uv - \int v \, du$. Let $u = \log x$ and $dv = x^2 \, dx$. Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^3}{3}$.
So,$\int x^2 \log x \, dx = \frac{x^3}{3} \log x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3}{3} \log x - \frac{1}{3} \int x^2 \, dx = \frac{x^3}{3} \log x - \frac{x^3}{9}$.
Combining these results: $I = (x \log x - x) - \left( \frac{x^3}{3} \log x - \frac{x^3}{9} \right) + c$.
Rearranging the terms: $I = \left( x - \frac{x^3}{3} \right) \log x - x + \frac{x^3}{9} + c$.
This can be written as $I = \left( x - \frac{x^3}{3} \right) \log x - \left( x - \frac{x^3}{9} \right) + c$.

(C) We need to evaluate the integral $I = \int [f(x)g''(x) - f''(x)g(x)] dx$.
Using the linearity property of integrals,we can write:
$I = \int f(x)g''(x) dx - \int f''(x)g(x) dx$.
Applying integration by parts to the first term $\int f(x)g''(x) dx$ (taking $u = f(x)$ and $dv = g''(x) dx$):
$\int f(x)g''(x) dx = f(x)g'(x) - \int f'(x)g'(x) dx$.
Applying integration by parts to the second term $\int f''(x)g(x) dx$ (taking $u = g(x)$ and $dv = f''(x) dx$):
$\int f''(x)g(x) dx = g(x)f'(x) - \int g'(x)f'(x) dx$.
Substituting these back into the expression for $I$:
$I = [f(x)g'(x) - \int f'(x)g'(x) dx] - [g(x)f'(x) - \int g'(x)f'(x) dx]$.
The integral terms $\int f'(x)g'(x) dx$ cancel out:
$I = f(x)g'(x) - f'(x)g(x) + C$.
Thus,the correct option is $C$.

(D) To evaluate the integral $I = \int x \sin kx \, dx$,we use the method of integration by parts,which is given by $\int u \, dv = uv - \int v \, du$.
Let $u = x$ and $dv = \sin kx \, dx$.
Then,$du = dx$ and $v = \int \sin kx \, dx = -\frac{\cos kx}{k}$.
Applying the formula:
$I = x \left( -\frac{\cos kx}{k} \right) - \int \left( -\frac{\cos kx}{k} \right) dx$
$I = -\frac{x \cos kx}{k} + \frac{1}{k} \int \cos kx \, dx$
$I = -\frac{x \cos kx}{k} + \frac{1}{k} \left( \frac{\sin kx}{k} \right) + c$
$I = -\frac{x \cos kx}{k} + \frac{\sin kx}{k^2} + c$.

(B) To evaluate the integral $\int x{e^x} dx$,we use the method of integration by parts.
The formula for integration by parts is $\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = x$ and $v = {e^x}$.
Then $u' = 1$ and $\int v dx = {e^x}$.
Applying the formula:
$\int x{e^x} dx = x{e^x} - \int (1 \cdot {e^x}) dx$
$= x{e^x} - \int {e^x} dx$
$= x{e^x} - {e^x} + c$
$= {e^x}(x - 1) + c$.

(C) Let $I = \int x^3 e^{x^2} dx$.
Substitute $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$.
The integral becomes:
$I = \int x^2 e^{x^2} (x dx) = \int t e^t (\frac{1}{2} dt) = \frac{1}{2} \int t e^t dt$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = e^t dt$:
$I = \frac{1}{2} [t e^t - \int e^t dt] = \frac{1}{2} [t e^t - e^t] + c$.
Substituting back $t = x^2$:
$I = \frac{1}{2} e^{x^2} (x^2 - 1) + c$.

(A) We use the method of integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = \log x$ and $dv = (x + 1)^{-2} dx$.
Then $du = \frac{1}{x} dx$ and $v = -(x + 1)^{-1}$.
Applying the formula:
$\int \frac{\log x}{(x + 1)^2} dx = \log x \cdot (-(x + 1)^{-1}) - \int \frac{1}{x} \cdot (-(x + 1)^{-1}) dx$
$= -\frac{\log x}{x + 1} + \int \frac{1}{x(x + 1)} dx$.
Using partial fractions: $\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$.
So,$\int \frac{1}{x(x + 1)} dx = \int \left( \frac{1}{x} - \frac{1}{x + 1} \right) dx = \log |x| - \log |x + 1| + C$.
Therefore,the integral is $-\frac{\log x}{x + 1} + \log x - \log (x + 1) + C$.

(C) To solve the integral $\int {x{e^{2x}}\,dx}$,we use the integration by parts formula: $\int {u\,dv} = uv - \int {v\,du}$.
Let $u = x$ and $dv = {e^{2x}}\,dx$.
Then $du = dx$ and $v = \frac{{{e^{2x}}}}{2}$.
Substituting these into the formula:
$\int {x{e^{2x}}\,dx} = x \cdot \frac{{{e^{2x}}}}{2} - \int {\frac{{{e^{2x}}}}{2}\,dx}$
$= \frac{{x{e^{2x}}}}{2} - \frac{1}{2} \cdot \frac{{{e^{2x}}}}{2} + C$
$= \frac{{x{e^{2x}}}}{2} - \frac{{{e^{2x}}}}{4} + C$
$= {e^{2x}}\left( {\frac{x}{2} - \frac{1}{4}} \right) + C$
$= {e^{2x}}\left( {\frac{{2x - 1}}{4}} \right) + C$
Comparing this with the given form ${e^{2x}}f(x) + C$,we get $f(x) = \frac{{2x - 1}}{4}$.

(C) Given $\frac{d}{dx}f(x) = x\cos x + \sin x$.
Integrating both sides with respect to $x$:
$f(x) = \int (x\cos x + \sin x) dx$.
Using integration by parts for $\int x\cos x dx$:
$\int x\cos x dx = x\sin x - \int (1 \cdot \sin x) dx = x\sin x - (-\cos x) = x\sin x + \cos x$.
Thus,$f(x) = (x\sin x + \cos x) + \int \sin x dx + C$.
$f(x) = x\sin x + \cos x - \cos x + C = x\sin x + C$.
Given $f(0) = 2$,we substitute $x = 0$:
$f(0) = 0 \cdot \sin(0) + C = 2 \Rightarrow C = 2$.
Therefore,$f(x) = x\sin x + 2$.

(B) Let $I = \int {{\cos }^{ - 1}}\left( {\frac{1}{x}} \right)\,dx$.
Since ${\cos }^{ - 1}\left( {\frac{1}{x}} \right) = {\sec ^{ - 1}}x$,we have $I = \int {{\sec ^{ - 1}}x \cdot 1\,dx}$.
Using integration by parts,$\int {u \cdot v\,dx} = u\int {v\,dx} - \int {\left( {\frac{du}{dx} \cdot \int {v\,dx} } \right)dx}$.
Let $u = {\sec ^{ - 1}}x$ and $v = 1$. Then $\frac{du}{dx} = \frac{1}{x\sqrt {{x^2} - 1}}$ and $\int {v\,dx} = x$.
$I = {\sec ^{ - 1}}x \cdot x - \int {\left( {\frac{1}{x\sqrt {{x^2} - 1}} \cdot x} \right)dx}$
$I = x{\sec ^{ - 1}}x - \int {\frac{1}{{\sqrt {{x^2} - 1} }}dx}$
Using the standard integral $\int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}dx} = {\cosh ^{ - 1}}\left( {\frac{x}{a}} \right) + C$,we get:
$I = x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + C$.

(B) To solve the integral $I = \int {{x^3}\log x\,dx}$,we use the method of integration by parts: $\int {u\,dv} = uv - \int {v\,du}$.
Let $u = \log x$ and $dv = x^3\,dx$.
Then $du = \frac{1}{x}\,dx$ and $v = \frac{x^4}{4}$.
Applying the formula:
$I = \left( \log x \cdot \frac{x^4}{4} \right) - \int {\frac{x^4}{4} \cdot \frac{1}{x}\,dx} + c$
$I = \frac{x^4}{4}\log x - \frac{1}{4}\int {x^3\,dx} + c$
$I = \frac{x^4}{4}\log x - \frac{1}{4} \cdot \frac{x^4}{4} + c$
$I = \frac{x^4}{4}\log x - \frac{x^4}{16} + c$
Taking $\frac{1}{16}$ as a common factor:
$I = \frac{1}{16}[4x^4\log x - x^4] + c$.

(A) Let $I = \int \cos (\log_e x) \, dx$.
Using integration by parts,let $u = \cos (\log_e x)$ and $dv = dx$. Then $du = -\sin (\log_e x) \cdot \frac{1}{x} \, dx$ and $v = x$.
$I = x \cos (\log_e x) - \int x \cdot (-\sin (\log_e x)) \cdot \frac{1}{x} \, dx$
$I = x \cos (\log_e x) + \int \sin (\log_e x) \, dx$
Now,apply integration by parts again to $\int \sin (\log_e x) \, dx$ with $u = \sin (\log_e x)$ and $dv = dx$.
$I = x \cos (\log_e x) + [x \sin (\log_e x) - \int x \cdot \cos (\log_e x) \cdot \frac{1}{x} \, dx]$
$I = x \cos (\log_e x) + x \sin (\log_e x) - I$
$2I = x [\cos (\log_e x) + \sin (\log_e x)]$
$I = \frac{1}{2} x [\cos (\log_e x) + \sin (\log_e x)] + C$.

(D) To evaluate $I = \int \sin^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \sin^{-1} x$ and $dv = dx$.
Then $du = \frac{1}{\sqrt{1 - x^2}} \, dx$ and $v = x$.
Applying the formula:
$I = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx$.
To solve the integral $\int \frac{x}{\sqrt{1 - x^2}} \, dx$,let $t = 1 - x^2$,so $dt = -2x \, dx$,which means $x \, dx = -\frac{1}{2} dt$.
Substituting this:
$\int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \cdot (2t^{1/2}) = -\sqrt{t} = -\sqrt{1 - x^2}$.
Therefore,$I = x \sin^{-1} x - (-\sqrt{1 - x^2}) + c = x \sin^{-1} x + \sqrt{1 - x^2} + c$.

(B) We have the integral $I = \int \frac{x + \sin x}{1 + \cos x} \, dx$.
Using the identities $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$,we get:
$I = \int \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \, dx$
$I = \int \left( \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) \, dx$
$I = \int \left( \frac{1}{2} x \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) \, dx$
Now,using integration by parts for the first term $\int \frac{1}{2} x \sec^2 \frac{x}{2} \, dx$:
Let $u = x$ and $dv = \frac{1}{2} \sec^2 \frac{x}{2} \, dx$. Then $du = dx$ and $v = \tan \frac{x}{2}$.
$\int \frac{1}{2} x \sec^2 \frac{x}{2} \, dx = x \tan \frac{x}{2} - \int \tan \frac{x}{2} \, dx$.
Substituting this back into $I$:
$I = (x \tan \frac{x}{2} - \int \tan \frac{x}{2} \, dx) + \int \tan \frac{x}{2} \, dx + c$
$I = x \tan \frac{x}{2} + c$.

(B) Let $I = \int \sqrt{x} e^{\sqrt{x}} \, dx$.
Substitute $x = t^2$,so $dx = 2t \, dt$.
Then $I = \int t e^t (2t \, dt) = 2 \int t^2 e^t \, dt$.
Using integration by parts $\int u v \, dt = u \int v \, dt - \int (u' \int v \, dt) \, dt$:
Let $u = t^2$ and $v = e^t$.
$I = 2 [t^2 e^t - \int 2t e^t \, dt] = 2t^2 e^t - 4 \int t e^t \, dt$.
Applying integration by parts again for $\int t e^t \, dt$:
$I = 2t^2 e^t - 4 [t e^t - \int e^t \, dt] = 2t^2 e^t - 4t e^t + 4e^t + c$.
Substituting $t = \sqrt{x}$ back:
$I = 2x e^{\sqrt{x}} - 4\sqrt{x} e^{\sqrt{x}} + 4e^{\sqrt{x}} + c = (2x - 4\sqrt{x} + 4) e^{\sqrt{x}} + c$.

(A) Let $I = \int {32{x^3}{{(\log x)}^2}} dx = 32\int {{x^3}{{(\log x)}^2}dx} $.
Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = {(\log x)^2}$ and $v = {x^3}$.
$I = 32 \left[ {(\log x)^2 \cdot \frac{x^4}{4} - \int {2 \log x \cdot \frac{1}{x} \cdot \frac{x^4}{4} dx} } \right]$
$I = 32 \left[ \frac{x^4 (\log x)^2}{4} - \frac{1}{2} \int x^3 \log x dx \right]$
Now,apply integration by parts again for $\int x^3 \log x dx$ with $u = \log x$ and $v = x^3$:
$I = 32 \left[ \frac{x^4 (\log x)^2}{4} - \frac{1}{2} \left( \log x \cdot \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx \right) \right] + c$
$I = 32 \left[ \frac{x^4 (\log x)^2}{4} - \frac{x^4 \log x}{8} + \frac{1}{8} \int x^3 dx \right] + c$
$I = 32 \left[ \frac{x^4 (\log x)^2}{4} - \frac{x^4 \log x}{8} + \frac{x^4}{32} \right] + c$
$I = 8x^4 (\log x)^2 - 4x^4 \log x + x^4 + c$
$I = x^4 \{ 8(\log x)^2 - 4 \log x + 1 \} + c$.

(D) Let $x = \sin \theta$,then $dx = \cos \theta \, d\theta$.
Substituting this into the integral,we get:
$\int \sin^{-1}(3\sin \theta - 4\sin^3 \theta) \cos \theta \, d\theta$
Since $\sin(3\theta) = 3\sin \theta - 4\sin^3 \theta$,the integral becomes:
$\int \sin^{-1}(\sin 3\theta) \cos \theta \, d\theta = \int 3\theta \cos \theta \, d\theta$
Using integration by parts:
$3 \int \theta \cos \theta \, d\theta = 3 [\theta \sin \theta - \int \sin \theta \, d\theta]$
$= 3 [\theta \sin \theta + \cos \theta] + c$
Substituting back $\theta = \sin^{-1}x$ and $\cos \theta = \sqrt{1 - x^2}$:
$= 3 [x \sin^{-1}x + \sqrt{1 - x^2}] + c$.

(A) Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.
Substituting these into the integral:
$\int \cos \sqrt{x} \, dx = \int \cos(t) \cdot 2t \, dt = 2 \int t \cos(t) \, dt$.
Using integration by parts,where $u = t$ and $dv = \cos(t) \, dt$:
$\int u \, dv = uv - \int v \, du = t \sin(t) - \int \sin(t) \, dt = t \sin(t) + \cos(t)$.
Multiplying by the constant $2$:
$2(t \sin(t) + \cos(t)) + c$.
Substituting $t = \sqrt{x}$ back:
$2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + c$.

(D) Let $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
Substituting these into the integral:
$\int \tan^{-1} \left( \frac{2 \tan \theta}{1 - \tan^2 \theta} \right) \sec^2 \theta \, d\theta = \int \tan^{-1}(\tan 2\theta) \sec^2 \theta \, d\theta$
$= \int 2\theta \sec^2 \theta \, d\theta = 2 \int \theta \sec^2 \theta \, d\theta$.
Using integration by parts:
$= 2 \left[ \theta \tan \theta - \int \tan \theta \, d\theta \right] = 2 \left[ \theta \tan \theta - \log|\sec \theta| \right] + c$
$= 2 \theta \tan \theta - 2 \log|\sec \theta| + c$.
Since $\tan \theta = x$,then $\theta = \tan^{-1} x$ and $\sec \theta = \sqrt{1 + x^2}$.
$= 2x \tan^{-1} x - 2 \log(\sqrt{1 + x^2}) + c$
$= 2x \tan^{-1} x - 2 \cdot \frac{1}{2} \log(1 + x^2) + c$
$= 2x \tan^{-1} x - \log(1 + x^2) + c$.

(A) Let $I = \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx$.
Substitute $t = \sin^{-1} x$,then $x = \sin t$ and $dt = \frac{1}{\sqrt{1 - x^2}} \, dx$.
The integral becomes $I = \int t \sin t \, dt$.
Using integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = t$ and $dv = \sin t \, dt$. Then $du = dt$ and $v = -\cos t$.
$I = -t \cos t - \int (-\cos t) \, dt = -t \cos t + \sin t + c$.
Since $t = \sin^{-1} x$,we have $\sin t = x$ and $\cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2}$.
Substituting these back,we get $I = -(\sin^{-1} x) \sqrt{1 - x^2} + x + c = x - \sqrt{1 - x^2} \sin^{-1} x + c$.

(A) Let $t = \sin^{-1}x$. Then $x = \sin t$ and $dx = \cos t \, dt$.
Substituting these into the integral:
$\int \frac{t}{(1-\sin^2 t)^{3/2}} \cos t \, dt = \int \frac{t}{(\cos^2 t)^{3/2}} \cos t \, dt = \int \frac{t}{\cos^3 t} \cos t \, dt = \int t \sec^2 t \, dt$.
Using integration by parts:
$\int t \sec^2 t \, dt = t \tan t - \int \tan t \, dt = t \tan t + \log|\cos t| + c$.
Since $t = \sin^{-1}x$,then $\tan t = \frac{x}{\sqrt{1-x^2}}$ and $\cos t = \sqrt{1-x^2}$.
Substituting back:
$= \sin^{-1}x \cdot \frac{x}{\sqrt{1-x^2}} + \log(\sqrt{1-x^2}) + c = \frac{x}{\sqrt{1-x^2}} \sin^{-1}x + \frac{1}{2} \log(1-x^2) + c$.

(B) Let $I = \int \frac{x \tan^{-1} x}{(1 + x^2)^{3/2}} \, dx$.
Substitute $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
Also,$1 + x^2 = 1 + \tan^2 \theta = \sec^2 \theta$.
Substituting these into the integral:
$I = \int \frac{\tan \theta \cdot \theta \cdot \sec^2 \theta}{(\sec^2 \theta)^{3/2}} \, d\theta = \int \frac{\theta \tan \theta \sec^2 \theta}{\sec^3 \theta} \, d\theta = \int \theta \frac{\tan \theta}{\sec \theta} \, d\theta = \int \theta \sin \theta \, d\theta$.
Using integration by parts:
$I = \theta(-\cos \theta) - \int 1 \cdot (-\cos \theta) \, d\theta = -\theta \cos \theta + \sin \theta + c$.
Since $x = \tan \theta$,we have $\cos \theta = \frac{1}{\sqrt{1 + x^2}}$ and $\sin \theta = \frac{x}{\sqrt{1 + x^2}}$.
Substituting back:
$I = -(\tan^{-1} x) \frac{1}{\sqrt{1 + x^2}} + \frac{x}{\sqrt{1 + x^2}} + c = \frac{x - \tan^{-1} x}{\sqrt{1 + x^2}} + c$.

(A) Let $I = \int {{x^5}{e^{{x^2}}}} dx$.
Substitute ${x^2} = t$,then $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$.
Also,${x^4} = {t^2}$.
Substituting these into the integral:
$I = \int {{x^4} \cdot {e^{{x^2}}} \cdot x dx} = \int {{t^2}{e^t} \cdot \frac{1}{2} dt} = \frac{1}{2} \int {{t^2}{e^t} dt}$.
Using integration by parts,$\int {u v dt} = u \int v dt - \int (u' \int v dt) dt$:
$I = \frac{1}{2} [ {t^2}{e^t} - \int {2t{e^t} dt} ] = \frac{1}{2} {t^2}{e^t} - \int {t{e^t} dt}$.
Applying integration by parts again to $\int {t{e^t} dt}$:
$I = \frac{1}{2} {t^2}{e^t} - [ t{e^t} - \int {e^t dt} ] = \frac{1}{2} {t^2}{e^t} - t{e^t} + {e^t} + c$.
Substituting $t = {x^2}$ back:
$I = \frac{1}{2} {x^4}{e^{{x^2}}} - {x^2}{e^{{x^2}}} + {e^{{x^2}}} + c$.

(C) Let $I = \int e^{\sqrt{x}} \, dx$.
Substitute $\sqrt{x} = t$,so $x = t^2$ and $dx = 2t \, dt$.
Substituting these into the integral:
$I = \int e^t \cdot 2t \, dt = 2 \int t e^t \, dt$.
Using integration by parts,where $\int u \, dv = uv - \int v \, du$:
Let $u = t$ and $dv = e^t \, dt$,then $du = dt$ and $v = e^t$.
$I = 2 \left( t e^t - \int e^t \, dt \right) + A = 2(t e^t - e^t) + A = 2(t - 1) e^t + A$.
Substituting back $t = \sqrt{x}$:
$I = 2(\sqrt{x} - 1) e^{\sqrt{x}} + A$.

(C) To evaluate the integral $I = \int (\log x)^2 \, dx$,we use the method of substitution.
Let $\log x = t$,which implies $x = e^t$. Therefore,$dx = e^t \, dt$.
Substituting these into the integral,we get:
$I = \int t^2 e^t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t^2$ and $dv = e^t \, dt$:
$du = 2t \, dt$ and $v = e^t$.
$I = t^2 e^t - \int 2t e^t \, dt = t^2 e^t - 2 \int t e^t \, dt$.
Applying integration by parts again for $\int t e^t \, dt$:
$I = t^2 e^t - 2(t e^t - \int e^t \, dt) = t^2 e^t - 2t e^t + 2e^t + c$.
Substituting $t = \log x$ and $e^t = x$ back into the expression:
$I = x(\log x)^2 - 2x \log x + 2x + c$.

(A) Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = \sin^{-1}x$ and $v = x$.
$\int x \sin^{-1}x dx = \frac{x^2}{2} \sin^{-1}x - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} dx + c$
$= \frac{x^2}{2} \sin^{-1}x - \frac{1}{2} \int \frac{-(1-x^2)+1}{\sqrt{1-x^2}} dx + c$
$= \frac{x^2}{2} \sin^{-1}x + \frac{1}{2} \int \sqrt{1-x^2} dx - \frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} dx + c$
Using the standard integral $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$,we get:
$= \frac{x^2}{2} \sin^{-1}x + \frac{1}{2} \left( \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}x \right) - \frac{1}{2}\sin^{-1}x + c$
$= \frac{x^2}{2} \sin^{-1}x + \frac{x}{4}\sqrt{1-x^2} + \frac{1}{4}\sin^{-1}x - \frac{1}{2}\sin^{-1}x + c$
$= \left( \frac{x^2}{2} - \frac{1}{4} \right) \sin^{-1}x + \frac{x}{4}\sqrt{1-x^2} + c$.

(A) Let $I = \int \sec^3 x \, dx = \int \sec x \cdot \sec^2 x \, dx$.
Using integration by parts,taking $u = \sec x$ and $dv = \sec^2 x \, dx$,we have $du = \sec x \tan x \, dx$ and $v = \tan x$.
$I = \sec x \tan x - \int \tan x (\sec x \tan x) \, dx$
$I = \sec x \tan x - \int \sec x \tan^2 x \, dx$
$I = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx$
$I = \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx$
$I = \sec x \tan x - I + \log |\sec x + \tan x|$
$2I = \sec x \tan x + \log |\sec x + \tan x|$
$I = \frac{1}{2} [\sec x \tan x + \log |\sec x + \tan x|] + C$.

(C) Let $I = \int e^x \sin 2x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \sin 2x$ and $dv = e^x \, dx$. Then $du = 2 \cos 2x \, dx$ and $v = e^x$.
$I = e^x \sin 2x - \int e^x (2 \cos 2x) \, dx = e^x \sin 2x - 2 \int e^x \cos 2x \, dx$.
Now,apply integration by parts again to $\int e^x \cos 2x \, dx$. Let $u = \cos 2x$ and $dv = e^x \, dx$. Then $du = -2 \sin 2x \, dx$ and $v = e^x$.
$\int e^x \cos 2x \, dx = e^x \cos 2x - \int e^x (-2 \sin 2x) \, dx = e^x \cos 2x + 2 \int e^x \sin 2x \, dx$.
Substituting this back into the expression for $I$:
$I = e^x \sin 2x - 2(e^x \cos 2x + 2I) = e^x \sin 2x - 2e^x \cos 2x - 4I$.
$I + 4I = e^x(\sin 2x - 2 \cos 2x) + C$.
$5I = e^x(\sin 2x - 2 \cos 2x) + C$.
Comparing this with $KI = e^x(\sin 2x - 2 \cos 2x) + C$,we get $K = 5$.

(A) To solve the integral $\int x\sqrt{2x + 3} \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x$,so $du = dx$.
Let $dv = (2x + 3)^{1/2} \, dx$,so $v = \frac{(2x + 3)^{3/2}}{3/2 \times 2} = \frac{1}{3}(2x + 3)^{3/2}$.
Applying the formula:
$\int x(2x + 3)^{1/2} \, dx = x \cdot \frac{1}{3}(2x + 3)^{3/2} - \int \frac{1}{3}(2x + 3)^{3/2} \, dx$.
$= \frac{x}{3}(2x + 3)^{3/2} - \frac{1}{3} \cdot \frac{(2x + 3)^{5/2}}{5/2 \times 2} + c$.
$= \frac{x}{3}(2x + 3)^{3/2} - \frac{1}{3} \cdot \frac{(2x + 3)^{5/2}}{5} + c$.
$= \frac{x}{3}(2x + 3)^{3/2} - \frac{1}{15}(2x + 3)^{5/2} + c$.