Integration by Parts Questions in English

(B) Let $I = \int \frac{\log \left(x^2+a^2\right)}{x^2} \,d x$.
Using integration by parts,$\int u v \,d x = u \int v \,d x - \int \left( \frac{du}{dx} \int v \,d x \right) d x$.
Let $u = \log \left(x^2+a^2\right)$ and $v = x^{-2}$.
Then $\frac{du}{dx} = \frac{2x}{x^2+a^2}$ and $\int v \,d x = -\frac{1}{x}$.
$I = \log \left(x^2+a^2\right) \cdot \left(-\frac{1}{x}\right) - \int \left( \frac{2x}{x^2+a^2} \cdot \left(-\frac{1}{x}\right) \right) d x$.
$I = -\frac{\log \left(x^2+a^2\right)}{x} + 2 \int \frac{1}{x^2+a^2} \,d x$.
Using the standard integral $\int \frac{1}{x^2+a^2} \,d x = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + c$.
$I = -\frac{\log \left(x^2+a^2\right)}{x} + \frac{2}{a} \tan^{-1} \left(\frac{x}{a}\right) + c$.

(A) Given $\int e^{x^2} \cdot x^3 \, dx = e^{x^2} f(x) + C$.
Let $I = \int e^{x^2} \cdot x^2 \cdot x \, dx$.
Substitute $t = x^2$,then $dt = 2x \, dx$,so $x \, dx = \frac{1}{2} dt$.
$I = \frac{1}{2} \int t e^t \, dt$.
Using integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = t$ and $dv = e^t \, dt$,then $du = dt$ and $v = e^t$.
$I = \frac{1}{2} [t e^t - \int e^t \, dt] = \frac{1}{2} [t e^t - e^t] + C = \frac{1}{2} e^t (t - 1) + C$.
Substituting $t = x^2$ back: $I = \frac{1}{2} e^{x^2} (x^2 - 1) + C$.
Comparing with $e^{x^2} f(x) + C$,we get $f(x) = \frac{1}{2} (x^2 - 1)$.
Check $f(1) = \frac{1}{2} (1^2 - 1) = 0$,which is consistent.
Therefore,$f(2) = \frac{1}{2} (2^2 - 1) = \frac{1}{2} (4 - 1) = \frac{3}{2}$.

(D) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$.
Substitute $x = \tan \theta$,then $dx = \sec ^2 \theta d \theta$.
Since $|x| < 1$,we have $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) = \sin ^{-1}(\sin 2 \theta) = 2 \theta$.
Thus,$I = \int 2 \theta \sec ^2 \theta d \theta = 2 \int \theta \sec ^2 \theta d \theta$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u = \theta$ and $dv = \sec ^2 \theta d \theta$:
$I = 2 [\theta \tan \theta - \int \tan \theta d \theta] = 2 [\theta \tan \theta - \log |\sec \theta|] + c$.
Since $\tan \theta = x$ and $\sec \theta = \sqrt{1+x^2}$,we have $\theta = \tan ^{-1} x$.
$I = 2 [x \tan ^{-1} x - \log |\sqrt{1+x^2}|] + c = 2 x \tan ^{-1} x - 2 \log (1+x^2)^{1/2} + c$.
$I = 2 x \tan ^{-1} x - \log |1+x^2| + c$.

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

(A) Let $I = \int \log (2+x)^{2+x} \, dx$.
Using the property $\log(a^b) = b \log a$,we get:
$I = \int (2+x) \log(2+x) \, dx$.
Let $u = 2+x$,then $du = dx$.
The integral becomes $I = \int u \log u \, du$.
Using integration by parts $\int f(u)g'(u) \, du = f(u)g(u) - \int f'(u)g(u) \, du$,let $f(u) = \log u$ and $g'(u) = u$.
Then $f'(u) = \frac{1}{u}$ and $g(u) = \frac{u^2}{2}$.
$I = \frac{u^2}{2} \log u - \int \frac{1}{u} \cdot \frac{u^2}{2} \, du$
$I = \frac{u^2}{2} \log u - \frac{1}{2} \int u \, du$
$I = \frac{u^2}{2} \log u - \frac{1}{2} \cdot \frac{u^2}{2} + c$
$I = \frac{u^2}{2} \left( \log u - \frac{1}{2} \right) + c$
Since $\frac{1}{2} = \log \sqrt{e}$,we have:
$I = \frac{u^2}{2} (\log u - \log \sqrt{e}) + c = \frac{u^2}{2} \log \left( \frac{u}{\sqrt{e}} \right) + c$.
Substituting $u = 2+x$ back:
$I = \frac{(2+x)^2}{2} \log \left( \frac{2+x}{\sqrt{e}} \right) + c$.

(A) To evaluate $\int x^2 \cos x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x^2$ and $dv = \cos x \, dx$.
Then $du = 2x \, dx$ and $v = \sin x$.
Applying the formula:
$\int x^2 \cos x \, dx = x^2 \sin x - \int (\sin x)(2x) \, dx = x^2 \sin x - 2 \int x \sin x \, dx$.
Now,apply integration by parts again for $\int x \sin x \, dx$:
Let $u = x$ and $dv = \sin x \, dx$.
Then $du = dx$ and $v = -\cos x$.
$\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x$.
Substituting this back into the original expression:
$\int x^2 \cos x \, dx = x^2 \sin x - 2(-x \cos x + \sin x) + c = x^2 \sin x + 2x \cos x - 2 \sin x + c$.

(A) Let $I = \int e^x \cos x \, dx$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = \cos x$ and $dv = e^x \, dx$.
Then $du = -\sin x \, dx$ and $v = e^x$.
$I = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx$.
Now,evaluate $\int e^x \sin x \, dx$ using integration by parts again.
Let $u = \sin x$ and $dv = e^x \, dx$.
Then $du = \cos x \, dx$ and $v = e^x$.
$\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx = e^x \sin x - I$.
Substituting this back into the equation for $I$:
$I = e^x \cos x + e^x \sin x - I$.
$2I = e^x (\cos x + \sin x)$.
$I = \frac{e^x (\cos x + \sin x)}{2} + c$.

(B) We know that $\tan^{-1} \left( \frac{2x}{1-x^2} \right) = 2 \tan^{-1} x$ for $|x| < 1$.
Substituting this into the integral,we get $I = \int 2 \tan^{-1} x \, dx$.
Using integration by parts,let $u = \tan^{-1} x$ and $dv = 2 \, dx$.
Then $du = \frac{1}{1+x^2} \, dx$ and $v = 2x$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
$I = 2x \tan^{-1} x - \int \frac{2x}{1+x^2} \, dx$.
Let $t = 1+x^2$,then $dt = 2x \, dx$.
So,$\int \frac{2x}{1+x^2} \, dx = \int \frac{1}{t} \, dt = \log|t| + c = \log(1+x^2) + c$.
Thus,$I = 2x \tan^{-1} x - \log(1+x^2) + c$.
Rearranging the terms,we get $I - 2x \tan^{-1} x = -\log(1+x^2) + c$.

(C) Let $I = \int \log (1+x)^{1+x} \,dx$.
Using the property $\log(a^b) = b \log a$,we get:
$I = \int (1+x) \log (1+x) \,dx$.
Substitute $t = 1+x$,then $dt = dx$.
$I = \int t \log t \,dt$.
Using integration by parts $\int u v \,dt = u \int v \,dt - \int (u' \int v \,dt) \,dt$,where $u = \log t$ and $v = t$:
$I = \log t \cdot \frac{t^2}{2} - \int (\frac{1}{t} \cdot \frac{t^2}{2}) \,dt$.
$I = \frac{t^2}{2} \log t - \frac{1}{2} \int t \,dt$.
$I = \frac{t^2}{2} \log t - \frac{1}{2} \cdot \frac{t^2}{2} + c$.
$I = \frac{t^2}{2} [\log t - \frac{1}{2}] + c$.
Substituting $t = 1+x$ back:
$I = \frac{(1+x)^2}{2} [\log (1+x) - \frac{1}{2}] + c$.

(C) We use the method of integration by parts for the given integral: $\int (f(x) g^{\prime \prime}(x) - f^{\prime \prime}(x) g(x)) \, dx$.
First,consider the integral $\int f(x) g^{\prime \prime}(x) \, dx$. Using integration by parts,let $u = f(x)$ and $dv = g^{\prime \prime}(x) \, dx$. Then $du = f^{\prime}(x) \, dx$ and $v = g^{\prime}(x)$.
So,$\int f(x) g^{\prime \prime}(x) \, dx = f(x) g^{\prime}(x) - \int f^{\prime}(x) g^{\prime}(x) \, dx$.
Next,consider the integral $\int f^{\prime \prime}(x) g(x) \, dx$. Using integration by parts,let $u = g(x)$ and $dv = f^{\prime \prime}(x) \, dx$. Then $du = g^{\prime}(x) \, dx$ and $v = f^{\prime}(x)$.
So,$\int f^{\prime \prime}(x) g(x) \, dx = g(x) f^{\prime}(x) - \int g^{\prime}(x) f^{\prime}(x) \, dx$.
Subtracting the two results:
$\int f(x) g^{\prime \prime}(x) \, dx - \int f^{\prime \prime}(x) g(x) \, dx = [f(x) g^{\prime}(x) - \int f^{\prime}(x) g^{\prime}(x) \, dx] - [g(x) f^{\prime}(x) - \int g^{\prime}(x) f^{\prime}(x) \, dx]$.
The integral terms $\int f^{\prime}(x) g^{\prime}(x) \, dx$ cancel out.
Thus,the result is $f(x) g^{\prime}(x) - f^{\prime}(x) g(x) + C$.

(A) We are given $f(x) = 1+x$ and $g(x) = \log x$.
Therefore,$g(f(x)) = \log(1+x)$.
We need to evaluate the integral $\int \log(1+x) \, dx$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = \log(1+x)$ and $dv = dx$.
Then $du = \frac{1}{1+x} \, dx$ and $v = x$.
$\int \log(1+x) \, dx = x \log(1+x) - \int \frac{x}{1+x} \, dx + c$.
We can rewrite the integrand as $\frac{x}{1+x} = \frac{1+x-1}{1+x} = 1 - \frac{1}{1+x}$.
So,$\int \log(1+x) \, dx = x \log(1+x) - \int (1 - \frac{1}{1+x}) \, dx + c$.
$= x \log(1+x) - (x - \log(1+x)) + c$.
$= x \log(1+x) - x + \log(1+x) + c$.
$= (1+x) \log(1+x) - x + c$.

(B) Let $\log x = t$,then $x = e^t$. Differentiating with respect to $t$,we get $dx = e^t \, dt$.
Substituting these into the integral,we have:
$I = \int \sin(t) e^t \, dt$.
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 for $\int e^t \cos t \, dt$:
$I = e^t \sin t - [e^t \cos t - \int e^t (-\sin t) \, dt]$.
$I = e^t \sin t - e^t \cos t - \int e^t \sin t \, dt$.
$I = e^t \sin t - e^t \cos t - I$.
$2I = e^t(\sin t - \cos t) + c$.
Substituting $t = \log x$ and $e^t = x$:
$2I = x(\sin(\log x) - \cos(\log x)) + c$.
$I = \frac{x}{2}(\sin(\log x) - \cos(\log x)) + c$.

(D) To evaluate the integral $I = \int \cos^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \cos^{-1} x$ and $dv = dx$.
Then $du = -\frac{1}{\sqrt{1-x^2}} \, dx$ and $v = x$.
Applying the formula: $I = x \cos^{-1} x - \int x \left( -\frac{1}{\sqrt{1-x^2}} \right) \, dx$.
$I = x \cos^{-1} x + \int \frac{x}{\sqrt{1-x^2}} \, dx$.
To solve $\int \frac{x}{\sqrt{1-x^2}} \, dx$,let $t = 1-x^2$,so $dt = -2x \, dx$ or $x \, dx = -\frac{1}{2} \, dt$.
The integral becomes $\int \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \left( \frac{t^{1/2}}{1/2} \right) = -\sqrt{t} = -\sqrt{1-x^2}$.
Therefore,$I = x \cos^{-1} x - \sqrt{1-x^2} + c$.

(B) To evaluate the integral $I = \int \sec^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \sec^{-1} x$ and $dv = dx$.
Then $du = \frac{1}{|x| \sqrt{x^2 - 1}} \, dx$ and $v = x$.
Applying the formula:
$I = x \sec^{-1} x - \int x \cdot \frac{1}{|x| \sqrt{x^2 - 1}} \, dx$.
Assuming $x > 1$,we have $|x| = x$,so:
$I = x \sec^{-1} x - \int \frac{1}{\sqrt{x^2 - 1}} \, dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \log \left| x + \sqrt{x^2 - a^2} \right| + C$,we get:
$I = x \sec^{-1} x - \log \left| x + \sqrt{x^2 - 1} \right| + C$.

(B) Let $I = \int x^{3} e^{x^{2}} dx$.
Substitute $x^{2} = t$,then $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$.
Substituting these into the integral,we get $I = \frac{1}{2} \int t e^{t} dt$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = e^{t} dt$.
Then $du = dt$ and $v = e^{t}$.
So,$I = \frac{1}{2} [t e^{t} - \int e^{t} dt] = \frac{1}{2} [t e^{t} - e^{t}] + c$.
Factoring out $e^{t}$,we get $I = \frac{1}{2} e^{t}(t - 1) + c$.
Substituting back $t = x^{2}$,we obtain $I = \frac{1}{2} e^{x^{2}}(x^{2} - 1) + c$.

(C) Let $I = \int \cot x \cdot \log [\log (\sin x)] d x$.
Substitute $t = \log (\sin x)$.
Then $dt = \frac{1}{\sin x} \cdot \cos x \, dx = \cot x \, dx$.
Thus,the integral becomes $I = \int \log t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \log t$ and $dv = dt$:
$I = t \log t - \int t \cdot \frac{1}{t} \, dt$.
$I = t \log t - \int 1 \, dt$.
$I = t \log t - t + c = t(\log t - 1) + c$.
Substituting back $t = \log (\sin x)$:
$I = \log (\sin x) [\log (\log (\sin x)) - 1] + c$.

(A) Let $I = \int \sin^{-1} x \cdot 1 \, dx$.
Using integration by parts,$\int u \cdot v \, dx = u \int v \, dx - \int (u' \cdot \int v \, dx) \, dx$.
Here,$u = \sin^{-1} x$ and $v = 1$.
$I = \sin^{-1} x \cdot x - \int \frac{1}{\sqrt{1 - x^2}} \cdot x \, dx$.
$I = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} \, dx$.
To solve $\int \frac{x}{\sqrt{1 - x^2}} \, dx$,let $1 - x^2 = t$,then $-2x \, dx = dt$,so $x \, dx = -\frac{1}{2} dt$.
$\int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \cdot \frac{t^{1/2}}{1/2} = -\sqrt{t} = -\sqrt{1 - x^2}$.
Substituting this back into the expression for $I$:
$I = x \sin^{-1} x - (-\sqrt{1 - x^2}) + c = x \sin^{-1} x + \sqrt{1 - x^2} + c$.

(B) Let $I = \int (1+x) \log x \, dx$.
Using integration by parts,let $u = \log x$ and $dv = (1+x) \, dx$.
Then $du = \frac{1}{x} \, dx$ and $v = x + \frac{x^2}{2}$.
Using the formula $\int u \, dv = uv - \int v \, du$:
$I = \log x \left(x + \frac{x^2}{2}\right) - \int \left(x + \frac{x^2}{2}\right) \cdot \frac{1}{x} \, dx$.
$I = \left(x + \frac{x^2}{2}\right) \log x - \int \left(1 + \frac{x}{2}\right) \, dx$.
$I = \left(x + \frac{x^2}{2}\right) \log x - \left(x + \frac{x^2}{4}\right) + C$.

(B) Let $I = \int \cos (\log x) \cdot 1 \, dx \dots (i)$
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = \cos (\log x)$ and $dv = 1 \, dx$.
Then $du = -\sin (\log x) \cdot \frac{1}{x} \, dx$ and $v = x$.
$I = x \cos (\log x) - \int x \cdot (-\sin (\log x)) \cdot \frac{1}{x} \, dx$
$I = x \cos (\log x) + \int \sin (\log x) \, dx$
Now,apply integration by parts again to $\int \sin (\log x) \cdot 1 \, dx$:
$I = x \cos (\log x) + [x \sin (\log x) - \int x \cdot \cos (\log x) \cdot \frac{1}{x} \, dx] + C$
$I = x \cos (\log x) + x \sin (\log x) - \int \cos (\log x) \, dx + C$
$I = x [\cos (\log x) + \sin (\log x)] - I + C$
$2I = x [\sin (\log x) + \cos (\log x)] + C$
$I = \frac{x}{2} [\sin (\log x) + \cos (\log x)] + C$

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

(D) Let $I = \int e^{x^2} \cdot x^3 \, dx$.
Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} dt$.
Substituting these into the integral,we get:
$I = \frac{1}{2} \int e^t \cdot t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = e^t \, dt$:
$I = \frac{1}{2} (t e^t - \int e^t \, dt) = \frac{1}{2} (t e^t - e^t) + c$.
$I = \frac{1}{2} e^t (t - 1) + c$.
Substituting $t = x^2$ back,we get:
$I = \frac{1}{2} e^{x^2} (x^2 - 1) + c$.
Comparing this with the given form $e^{x^2} f(x) + c$,we find $f(x) = \frac{x^2 - 1}{2}$.
Since $f(1) = \frac{1^2 - 1}{2} = 0$,the condition is satisfied.

(C) We need to evaluate the integral $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$.
Applying the linearity property of integrals:
$I = \int \frac{1}{\log x} dx - \int \frac{1}{(\log x)^2} dx$.
Using integration by parts on the first term $\int \frac{1}{\log x} dx$,let $u = \frac{1}{\log x}$ and $dv = dx$.
Then $du = -\frac{1}{(\log x)^2} \cdot \frac{1}{x} dx$ and $v = x$.
Using the formula $\int u dv = uv - \int v du$:
$\int \frac{1}{\log x} dx = \frac{x}{\log x} - \int x \left( -\frac{1}{(\log x)^2} \cdot \frac{1}{x} \right) dx$
$= \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx$.
Substituting this back into the original expression for $I$:
$I = \left( \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx \right) - \int \frac{1}{(\log x)^2} dx$
$I = \frac{x}{\log x} + C$.

(A) Let $I = \int e^{\sin x} \sin 2x \, dx$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int e^{\sin x} (2 \sin x \cos x) \, dx$.
Let $t = \sin x$,then $dt = \cos x \, dx$.
Substituting these into the integral:
$I = \int e^t (2t) \, dt = 2 \int t e^t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = e^t \, dt$:
$I = 2 [t e^t - \int e^t \, dt] = 2 [t e^t - e^t] + c$.
$I = 2 e^t (t - 1) + c$.
Substituting $t = \sin x$ back:
$I = 2 e^{\sin x} (\sin x - 1) + c$.
Thus,the correct option is $A$.

(D) We need to evaluate the integral $I = \int \log x^2 \, dx$.
Using the property of logarithms,$\log x^2 = 2 \log x$.
So,$I = \int 2 \log x \, dx = 2 \int \log x \, dx$.
Using integration by parts,$\int \log x \, dx = x \log x - x + C_1$.
Therefore,$I = 2(x \log x - x) + C = 2x \log x - 2x + C$.
We can rewrite this as $2x(\log x - 1) = 2x(\log x - \log e) = 2x \log \left(\frac{x}{e}\right) + C$.
Thus,the correct option is $D$.

(D) Let $I = \int e^x \cos 2x \, dx$.
Using the integration by parts formula $\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$,let $u = \cos 2x$ and $v = e^x$.
$I = \cos 2x \cdot e^x - \int (-2 \sin 2x) \cdot e^x \, dx = e^x \cos 2x + 2 \int e^x \sin 2x \, dx$.
Applying integration by parts again to $\int e^x \sin 2x \, dx$ with $u = \sin 2x$ and $v = e^x$:
$\int e^x \sin 2x \, dx = \sin 2x \cdot e^x - \int (2 \cos 2x) \cdot e^x \, dx = e^x \sin 2x - 2I$.
Substituting this back into the equation for $I$:
$I = e^x \cos 2x + 2(e^x \sin 2x - 2I) = e^x \cos 2x + 2e^x \sin 2x - 4I$.
$5I = e^x(\cos 2x + 2 \sin 2x)$.
$I = \frac{e^x}{5}(\cos 2x + 2 \sin 2x) + C$.
Thus,the correct option is $D$.

(C) To evaluate the integral $I = \int (x^2 + 3x + 2) e^x dx$,we use the integration by parts formula: $\int u dv = uv - \int v du$.
Let $u = x^2 + 3x + 2$ and $dv = e^x dx$.
Then $du = (2x + 3) dx$ and $v = e^x$.
$I = (x^2 + 3x + 2) e^x - \int (2x + 3) e^x dx$.
Now,apply integration by parts again for $\int (2x + 3) e^x dx$:
Let $u_1 = 2x + 3$ and $dv_1 = e^x dx$.
Then $du_1 = 2 dx$ and $v_1 = e^x$.
$\int (2x + 3) e^x dx = (2x + 3) e^x - \int 2 e^x dx = (2x + 3) e^x - 2e^x = (2x + 1) e^x$.
Substituting this back into the expression for $I$:
$I = (x^2 + 3x + 2) e^x - (2x + 1) e^x + C$.
$I = (x^2 + 3x + 2 - 2x - 1) e^x + C$.
$I = (x^2 + x + 1) e^x + C$.
Thus,the correct option is $C$.

(D) Let $I = \int[(3x-1) \cos x + (1-2x) \sin x] dx$.
Using integration by parts,$\int u dv = uv - \int v du$:
For $\int (3x-1) \cos x dx$: Let $u = 3x-1, dv = \cos x dx \Rightarrow du = 3 dx, v = \sin x$.
$\int (3x-1) \cos x dx = (3x-1) \sin x - \int 3 \sin x dx = (3x-1) \sin x + 3 \cos x$.
For $\int (1-2x) \sin x dx$: Let $u = 1-2x, dv = \sin x dx \Rightarrow du = -2 dx, v = -\cos x$.
$\int (1-2x) \sin x dx = (1-2x)(-\cos x) - \int (-\cos x)(-2) dx = (2x-1) \cos x - 2 \sin x$.
Adding these: $I = (3x-1) \sin x + 3 \cos x + (2x-1) \cos x - 2 \sin x + C$.
$I = (3x-1-2) \sin x + (3+2x-1) \cos x + C = (3x-3) \sin x + (2x+2) \cos x + C$.
Comparing with $f(x) \cos x + g(x) \sin x + C$,we get $f(x) = 2x+2$ and $g(x) = 3x-3 = 3(x-1)$.

(A) We have,$I = \int e^{\sin x} \sin 2x \, dx$
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int e^{\sin x} (2 \sin x \cos x) \, dx = 2 \int e^{\sin x} \sin x \cos x \, dx$
Let $\sin x = t$,then $\cos x \, dx = dt$.
Substituting these into the integral:
$I = 2 \int e^t \cdot t \, dt$
Using integration by parts $\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = e^t \, dt$:
$I = 2 [t e^t - \int e^t \, dt] = 2 [t e^t - e^t] + C$
$I = 2 e^t (t - 1) + C$
Substituting $t = \sin x$ back:
$I = 2 e^{\sin x} (\sin x - 1) + C$

(D) Given,for $x > 0$,we need to evaluate $I = \int \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) dx$.
Using the trigonometric substitution $x = \tan \theta$,we know that for $x > 0$,$\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) = 2 \tan^{-1} x$.
Substituting this into the integral,we get $I = \int 2 \tan^{-1} x dx = 2 \int \tan^{-1} x dx$.
Using integration by parts,let $u = \tan^{-1} x$ and $dv = dx$. Then $du = \frac{1}{1+x^{2}} dx$ and $v = x$.
$I = 2 \left( x \tan^{-1} x - \int \frac{x}{1+x^{2}} dx \right)$.
To solve $\int \frac{x}{1+x^{2}} dx$,let $t = 1+x^{2}$,then $dt = 2x dx$,so $x dx = \frac{1}{2} dt$.
Thus,$\int \frac{x}{1+x^{2}} dx = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log(1+x^{2})$.
Substituting this back,$I = 2x \tan^{-1} x - 2 \left( \frac{1}{2} \log(1+x^{2}) \right) + C = 2x \tan^{-1} x - \log(1+x^{2}) + C$.

(D) To evaluate the integral $ I = \int x^{3} \sin 3 x \, dx $,we use the method of integration by parts or the tabular method (Bernoulli's formula).
Using the formula $ \int u v \, dx = u v_1 - u' v_2 + u'' v_3 - u''' v_4 + \dots $,where $ u = x^3 $ and $ v = \sin 3x $:
$ u = x^3, u' = 3x^2, u'' = 6x, u''' = 6, u'''' = 0 $
$ v_1 = \int \sin 3x \, dx = -\frac{\cos 3x}{3} $
$ v_2 = \int -\frac{\cos 3x}{3} \, dx = -\frac{\sin 3x}{9} $
$ v_3 = \int -\frac{\sin 3x}{9} \, dx = \frac{\cos 3x}{27} $
$ v_4 = \int \frac{\cos 3x}{27} \, dx = \frac{\sin 3x}{81} $
Substituting these into the formula:
$ I = x^3 \left( -\frac{\cos 3x}{3} \right) - (3x^2) \left( -\frac{\sin 3x}{9} \right) + (6x) \left( \frac{\cos 3x}{27} \right) - (6) \left( \frac{\sin 3x}{81} \right) + C $
$ I = -\frac{x^3 \cos 3x}{3} + \frac{x^2 \sin 3x}{3} + \frac{2x \cos 3x}{9} - \frac{2 \sin 3x}{27} + C $

(C) We use the formula for integration by parts: $\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$.
Let $I = \int e^{x} \cdot x^{5} \, dx$.
Using the generalized integration by parts formula $\int e^{x} f(x) \, dx = e^{x} [f(x) - f'(x) + f''(x) - f'''(x) + f^{(4)}(x) - f^{(5)}(x)] + C$.
Here,$f(x) = x^{5}$.
Then,$f'(x) = 5x^{4}$,$f''(x) = 20x^{3}$,$f'''(x) = 60x^{2}$,$f^{(4)}(x) = 120x$,and $f^{(5)}(x) = 120$.
Substituting these into the formula:
$I = e^{x} [x^{5} - 5x^{4} + 20x^{3} - 60x^{2} + 120x - 120] + C$.

(B) Given $I_{n}=\int(\log x)^{n} dx$.
Using integration by parts,let $u = (\log x)^{n}$ and $dv = dx$.
Then $du = n(\log x)^{n-1} \cdot \frac{1}{x} dx$ and $v = x$.
Applying the formula $\int u dv = uv - \int v du$:
$I_{n} = x(\log x)^{n} - \int x \cdot n(\log x)^{n-1} \cdot \frac{1}{x} dx$
$I_{n} = x(\log x)^{n} - n \int (\log x)^{n-1} dx$
$I_{n} = x(\log x)^{n} - n I_{n-1}$
Rearranging the terms,we get:
$I_{n} + n I_{n-1} = x(\log x)^{n} + C$.

(C) Let $m = \frac{1}{n}$. As $n \rightarrow \infty$,$m \rightarrow 0$.
$f(x) = \lim _{m \rightarrow 0} \frac{1}{m^2} \left(x^m - x^{\frac{m}{m+1}}\right)$
$f(x) = \lim _{m \rightarrow 0} \frac{x^{\frac{m}{m+1}}}{m^2} \left(x^{m - \frac{m}{m+1}} - 1\right) = \lim _{m \rightarrow 0} \frac{x^{\frac{m}{m+1}}}{m^2} \left(x^{\frac{m^2}{m+1}} - 1\right)$
$f(x) = \lim _{m \rightarrow 0} x^{\frac{m}{m+1}} \cdot \left(\frac{x^{\frac{m^2}{m+1}} - 1}{\frac{m^2}{m+1}}\right) \cdot \frac{1}{m+1}$
Using the standard limit $\lim _{t \rightarrow 0} \frac{a^t - 1}{t} = \log a$,we get:
$f(x) = x^0 \cdot \log x \cdot \frac{1}{0+1} = \log x$
Now,$\int x f(x) d x = \int x \log x d x$.
Using integration by parts: $\int u v d x = u \int v d x - \int (u' \int v d x) d x$.
Let $u = \log x$ and $v = x$.
$\int x \log x d x = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} d x$
$= \frac{x^2}{2} \log x - \int \frac{x}{2} d x = \frac{x^2}{2} \log x - \frac{x^2}{4} + C$.

(A) Given $\int f(x) dx = \psi(x)$.
Let $I = \int x^5 f(x^3) dx = \int x^3 f(x^3) x^2 dx$.
Substitute $t = x^3$,then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{dt}{3}$.
Substituting these into the integral,we get $I = \int t f(t) \frac{dt}{3} = \frac{1}{3} \int t f(t) dt$.
Using integration by parts,where $u = t$ and $dv = f(t) dt$ (so $du = dt$ and $v = \psi(t)$):
$I = \frac{1}{3} [t \psi(t) - \int \psi(t) dt]$.
Now,substitute $t = x^3$ back into the expression:
$I = \frac{1}{3} [x^3 \psi(x^3) - \int \psi(x^3) (3x^2 dx)]$.
$I = \frac{1}{3} x^3 \psi(x^3) - \int x^2 \psi(x^3) dx$.

(B) Given $I_n = \int (\ln x)^n dx$.
Using integration by parts,let $u = (\ln x)^n$ and $dv = dx$.
Then $du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx$ and $v = x$.
$I_n = x(\ln x)^n - \int x \cdot \frac{n(\ln x)^{n-1}}{x} dx + k$.
$I_n = x(\ln x)^n - n \int (\ln x)^{n-1} dx + k$.
Since $I_{n-1} = \int (\ln x)^{n-1} dx$,we have:
$I_n = x(\ln x)^n - n I_{n-1} + k$.
Rearranging the terms,we get:
$I_n + n I_{n-1} = x(\ln x)^n + k$.

(C) Given $I_{m, n} = \int e^{mx} \cdot x^n \, dx$.
Using integration by parts,let $u = x^n$ and $dv = e^{mx} \, dx$.
Then $du = nx^{n-1} \, dx$ and $v = \frac{e^{mx}}{m}$.
$I_{m, n} = \int u \, dv = uv - \int v \, du$.
$I_{m, n} = x^n \left( \frac{e^{mx}}{m} \right) - \int \left( \frac{e^{mx}}{m} \right) (nx^{n-1}) \, dx$.
$I_{m, n} = \frac{x^n e^{mx}}{m} - \frac{n}{m} \int e^{mx} x^{n-1} \, dx$.
Since $I_{m, n-1} = \int e^{mx} x^{n-1} \, dx$,we have:
$I_{m, n} = \frac{x^n e^{mx}}{m} - \frac{n}{m} I_{m, n-1} + c$.
Rearranging the terms,we get:
$I_{m, n} + \frac{n}{m} I_{m, n-1} = \frac{x^n e^{mx}}{m} + c$.
Hence,option $C$ is correct.

(A) Let $x = \tan \theta$,then $dx = \sec^2 \theta d\theta$.
Substituting this into the integral,we get:
$\int \operatorname{Cos}^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) \sec^2 \theta d\theta = \int \operatorname{Cos}^{-1}(\cos 2\theta) \sec^2 \theta d\theta = \int 2\theta \sec^2 \theta d\theta$.
Using integration by parts,let $u = 2\theta$ and $dv = \sec^2 \theta d\theta$.
Then $du = 2 d\theta$ and $v = \tan \theta$.
$\int 2\theta \sec^2 \theta d\theta = 2\theta \tan \theta - \int 2 \tan \theta d\theta$.
$= 2\theta \tan \theta - 2 \ln|\sec \theta| + c$.
Since $x = \tan \theta$,then $\theta = \tan^{-1} x$ and $\sec \theta = \sqrt{1+\tan^2 \theta} = \sqrt{1+x^2}$.
Substituting back,we get:
$2x \tan^{-1} x - 2 \ln(\sqrt{1+x^2}) + c = 2x \tan^{-1} x - \ln(1+x^2) + c$.
This can be written as $2[x \tan^{-1} x - \frac{1}{2} \ln(1+x^2)] + c = 2[x \tan^{-1} x - \ln \sqrt{1+x^2}] + c$.

(A) Let $I = \int x \operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right) dx$.
Substitute $x = \operatorname{Tan} \theta$,so $dx = \operatorname{Sec}^2 \theta d\theta$.
Since $x > 0$,$\theta \in (0, \pi/2)$.
The expression becomes $\operatorname{Cos}^{-1}\left(\frac{1-\operatorname{Tan}^2 \theta}{1+\operatorname{Tan}^2 \theta}\right) = \operatorname{Cos}^{-1}(\operatorname{Cos} 2\theta) = 2\theta$.
Thus,$I = \int \operatorname{Tan} \theta (2\theta) \operatorname{Sec}^2 \theta d\theta = 2 \int \theta \operatorname{Tan} \theta \operatorname{Sec}^2 \theta d\theta$.
Using integration by parts,let $u = \theta$ and $dv = \operatorname{Tan} \theta \operatorname{Sec}^2 \theta d\theta$.
Then $du = d\theta$ and $v = \frac{\operatorname{Tan}^2 \theta}{2}$.
$I = 2 \left[ \theta \cdot \frac{\operatorname{Tan}^2 \theta}{2} - \int \frac{\operatorname{Tan}^2 \theta}{2} d\theta \right] = \theta \operatorname{Tan}^2 \theta - \int (\operatorname{Sec}^2 \theta - 1) d\theta$.
$I = \theta \operatorname{Tan}^2 \theta - \operatorname{Tan} \theta + \theta + c = \theta (1 + \operatorname{Tan}^2 \theta) - \operatorname{Tan} \theta + c$.
Substituting back $\theta = \operatorname{Tan}^{-1} x$ and $\operatorname{Tan} \theta = x$:
$I = (1+x^2) \operatorname{Tan}^{-1} x - x + c$.

(D) Let $I = \int \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) dx$.
Put $x = a \tan^2 \theta$,then $dx = 2a \tan \theta \sec^2 \theta d\theta$.
Then $\sqrt{\frac{x}{a+x}} = \sqrt{\frac{a \tan^2 \theta}{a(1 + \tan^2 \theta)}} = \sqrt{\sin^2 \theta} = \sin \theta$.
So,$I = \int \theta (2a \tan \theta \sec^2 \theta) d\theta = 2a \int \theta \tan \theta \sec^2 \theta d\theta$.
Using integration by parts,let $u = \theta$ and $dv = \tan \theta \sec^2 \theta d\theta$.
Then $du = d\theta$ and $v = \frac{\tan^2 \theta}{2}$.
$I = 2a \left[ \theta \frac{\tan^2 \theta}{2} - \int \frac{\tan^2 \theta}{2} d\theta \right] = a \theta \tan^2 \theta - a \int (\sec^2 \theta - 1) d\theta$.
$I = a \theta \tan^2 \theta - a(\tan \theta - \theta) + c = a \theta (\tan^2 \theta + 1) - a \tan \theta + c$.
Since $\tan^2 \theta = \frac{x}{a}$,we have $\theta = \tan^{-1} \sqrt{\frac{x}{a}}$.
$I = a \tan^{-1} \sqrt{\frac{x}{a}} (\frac{x}{a} + 1) - a \sqrt{\frac{x}{a}} + c = (x+a) \tan^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + c$.

(B) Let $I = \int \frac{e^{\sqrt{x}}}{\sqrt{x}} (x + \sqrt{x}) dx$.
Substitute $\sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{dx}{\sqrt{x}} = 2 dt$.
Substituting these into the integral,we get:
$I = \int e^t (t^2 + t) (2 dt) = 2 \int (t^2 + t) e^t dt$.
Using the integration by parts formula $\int u v' dt = uv - \int u' v dt$,let $u = t^2 + t$ and $v' = e^t$. Then $u' = 2t + 1$ and $v = e^t$.
$I = 2 [(t^2 + t) e^t - \int (2t + 1) e^t dt]$.
Now,evaluate $\int (2t + 1) e^t dt$ using integration by parts again:
Let $u = 2t + 1$ and $v' = e^t$. Then $u' = 2$ and $v = e^t$.
$\int (2t + 1) e^t dt = (2t + 1) e^t - \int 2 e^t dt = (2t + 1) e^t - 2 e^t = (2t - 1) e^t$.
Substituting this back into the expression for $I$:
$I = 2 [ (t^2 + t) e^t - (2t - 1) e^t ] + K = 2 e^t [ t^2 + t - 2t + 1 ] + K = 2 e^t [ t^2 - t + 1 ] + K$.
Since $t = \sqrt{x}$,we have $I = e^{\sqrt{x}} [ 2x - 2\sqrt{x} + 2 ] + K$.
Comparing this with $e^{\sqrt{x}} [ Ax + B\sqrt{x} + C ] + K$,we get $A = 2$,$B = -2$,and $C = 2$.
Therefore,$A + B + C = 2 - 2 + 2 = 2$.

(C) Using integration by parts,let $u = \log \cot (\frac{x}{2})$ and $dv = \cos x dx$.
Then $du = \frac{1}{\cot (\frac{x}{2})} \cdot (-\csc^2 (\frac{x}{2})) \cdot \frac{1}{2} dx = -\frac{1}{2 \sin (\frac{x}{2}) \cos (\frac{x}{2})} dx = -\frac{1}{\sin x} dx$.
And $v = \int \cos x dx = \sin x$.
Using the formula $\int u dv = uv - \int v du$:
$\int \cos x \log \cot (\frac{x}{2}) dx = \sin x \log \cot (\frac{x}{2}) - \int \sin x \cdot (-\frac{1}{\sin x}) dx$
$= \sin x \log \cot (\frac{x}{2}) + \int 1 dx$
$= \sin x \log \cot (\frac{x}{2}) + x + c$.

(B) Let $I = \int \frac{x \log x}{(x^2-1)^{3/2}} dx$.
Using integration by parts,let $u = \log x$ and $dv = \frac{x}{(x^2-1)^{3/2}} dx$.
Then $du = \frac{1}{x} dx$ and $v = \int (x^2-1)^{-3/2} x dx = -(x^2-1)^{-1/2} = -\frac{1}{\sqrt{x^2-1}}$.
Using the formula $\int u dv = uv - \int v du$:
$I = -\frac{\log x}{\sqrt{x^2-1}} - \int \left(-\frac{1}{\sqrt{x^2-1}}\right) \frac{1}{x} dx$
$I = -\frac{\log x}{\sqrt{x^2-1}} + \int \frac{1}{x \sqrt{x^2-1}} dx$
Since $\int \frac{1}{x \sqrt{x^2-1}} dx = \sec^{-1} x + C$,
$I = \sec^{-1} x - \frac{\log x}{\sqrt{x^2-1}} + C$.

(A) Let $I = \int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x$.
We can rewrite the integrand as:
$I = \int \frac{x\sqrt{1+\frac{1}{x^2}} \left[\log \left(x^2(1+\frac{1}{x^2})\right)-2 \log x\right]}{x^4} d x$
$I = \int \frac{\sqrt{1+\frac{1}{x^2}} \left[\log x^2 + \log(1+\frac{1}{x^2}) - 2 \log x\right]}{x^3} d x$
Since $\log x^2 = 2 \log x$,the expression simplifies to:
$I = \int \frac{\sqrt{1+\frac{1}{x^2}} \log(1+\frac{1}{x^2})}{x^3} d x$.
Let $1+\frac{1}{x^2} = t^2$. Then $-\frac{2}{x^3} dx = 2t dt$,which implies $-\frac{dx}{x^3} = t dt$.
Substituting these into the integral:
$I = -\int t \cdot \log(t^2) \cdot t dt = -\int t^2 \cdot 2 \log t dt = -2 \int t^2 \log t dt$.
Using integration by parts:
$I = -2 \left[ \frac{t^3}{3} \log t - \int \frac{t^3}{3} \cdot \frac{1}{t} dt \right] = -2 \left[ \frac{t^3}{3} \log t - \frac{t^3}{9} \right] = \frac{2}{9} t^3 - \frac{2}{3} t^3 \log t$.
$I = \frac{1}{9} t^3 [2 - 6 \log t] = \frac{1}{9} t^3 [2 - 3 \log t^2]$.
Substituting $t^2 = 1+\frac{1}{x^2}$ and $t = (1+\frac{1}{x^2})^{1/2}$:
$I = \frac{1}{9} (1+\frac{1}{x^2})^{3/2} [2 - 3 \log (1+\frac{1}{x^2})] + c$.

(D) Let $I = \int \operatorname{Tan}^{-1}\left(x^{\frac{1}{3}}\right) d x$.
Substitute $x = t^3$,then $dx = 3t^2 dt$.
$I = \int \operatorname{Tan}^{-1}(t) \cdot 3t^2 dt = 3 \int t^2 \operatorname{Tan}^{-1}(t) dt$.
Using integration by parts,let $u = \operatorname{Tan}^{-1}(t)$ and $dv = t^2 dt$.
Then $du = \frac{1}{1+t^2} dt$ and $v = \frac{t^3}{3}$.
$I = 3 \left[ \frac{t^3}{3} \operatorname{Tan}^{-1}(t) - \int \frac{t^3}{3(1+t^2)} dt \right] = t^3 \operatorname{Tan}^{-1}(t) - \int \frac{t^3}{1+t^2} dt$.
Since $\frac{t^3}{1+t^2} = \frac{t(t^2+1)-t}{1+t^2} = t - \frac{t}{1+t^2}$,
$I = t^3 \operatorname{Tan}^{-1}(t) - \int \left( t - \frac{t}{1+t^2} \right) dt = t^3 \operatorname{Tan}^{-1}(t) - \frac{t^2}{2} + \frac{1}{2} \log(1+t^2) + c$.
Substituting $t = x^{\frac{1}{3}}$,we get $I = x \operatorname{Tan}^{-1}(x^{\frac{1}{3}}) - \frac{1}{2} x^{\frac{2}{3}} + \frac{1}{2} \log(1+x^{\frac{2}{3}}) + c$.

(C) Let $I = \int \cos^{-1}(2x^2-1) \, dx$.
Substitute $x = \cos \theta$,then $dx = -\sin \theta \, d\theta$.
The integral becomes $I = \int \cos^{-1}(2 \cos^2 \theta - 1) \cdot (-\sin \theta) \, d\theta$.
Using the identity $\cos 2\theta = 2 \cos^2 \theta - 1$,we get $I = \int \cos^{-1}(\cos 2\theta) \cdot (-\sin \theta) \, d\theta = \int 2\theta \cdot (-\sin \theta) \, d\theta = -2 \int \theta \sin \theta \, d\theta$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \theta$ and $dv = \sin \theta \, d\theta$:
$I = -2 [\theta(-\cos \theta) - \int (-\cos \theta) \, d\theta] = -2 [-\theta \cos \theta + \sin \theta] + c = 2\theta \cos \theta - 2 \sin \theta + c$.
Since $x = \cos \theta$,then $\theta = \cos^{-1} x$ and $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1-x^2}$.
Substituting these back,$I = 2x \cos^{-1} x - 2\sqrt{1-x^2} + c = 2(x \cos^{-1} x - \sqrt{1-x^2}) + c$.

(A) To evaluate $I = \int (\log x)^3 x^4 \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = (\log x)^3$ and $dv = x^4 \, dx$.
Then $du = 3(\log x)^2 \cdot \frac{1}{x} \, dx$ and $v = \frac{x^5}{5}$.
$I = \frac{x^5}{5}(\log x)^3 - \int \frac{x^5}{5} \cdot 3(\log x)^2 \cdot \frac{1}{x} \, dx = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} \int x^4 (\log x)^2 \, dx$.
Applying integration by parts again for $\int x^4 (\log x)^2 \, dx$ with $u = (\log x)^2$ and $dv = x^4 \, dx$:
$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} \left[ \frac{x^5}{5}(\log x)^2 - \int \frac{x^5}{5} \cdot 2 \log x \cdot \frac{1}{x} \, dx \right] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5 (\log x)^2 + \frac{6}{25} \int x^4 \log x \, dx$.
Applying integration by parts one last time for $\int x^4 \log x \, dx$ with $u = \log x$ and $dv = x^4 \, dx$:
$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5 (\log x)^2 + \frac{6}{25} \left[ \frac{x^5}{5} \log x - \int \frac{x^5}{5} \cdot \frac{1}{x} \, dx \right] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5 (\log x)^2 + \frac{6}{125} x^5 \log x - \frac{6}{125} \int x^4 \, dx$.
$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5 (\log x)^2 + \frac{6}{125} x^5 \log x - \frac{6}{625} x^5 + c$.
Factoring out $x^5$,we get $x^5 \left[ \frac{1}{5}(\log x)^3 - \frac{3}{25}(\log x)^2 + \frac{6}{125} \log x - \frac{6}{625} \right] + c$.

(B) We know that $\cos^2 x = \frac{1 + \cos 2x}{2}$.
Substituting this into the integral,we get $\int x^2 \left(\frac{1 + \cos 2x}{2}\right) dx = \frac{1}{2} \int x^2 dx + \frac{1}{2} \int x^2 \cos 2x dx$.
$= \frac{1}{2} \cdot \frac{x^3}{3} + \frac{1}{2} \left[ x^2 \cdot \frac{\sin 2x}{2} - \int 2x \cdot \frac{\sin 2x}{2} dx \right]$
$= \frac{x^3}{6} + \frac{x^2 \sin 2x}{4} - \frac{1}{2} \int x \sin 2x dx$.
Using integration by parts for $\int x \sin 2x dx = x \left(-\frac{\cos 2x}{2}\right) - \int 1 \cdot \left(-\frac{\cos 2x}{2}\right) dx = -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4}$.
Substituting back: $\frac{x^3}{6} + \frac{x^2 \sin 2x}{4} - \frac{1}{2} \left( -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} \right) + c = \frac{x^3}{6} + \left( \frac{x^2}{4} - \frac{1}{8} \right) \sin 2x + \left( \frac{x}{4} \right) \cos 2x + c$.
Comparing with $\frac{1}{6} f(x) + g(x) \sin 2x + h(x) \cos 2x + c$,we get $f(x) = x^3$,$g(x) = \frac{x^2}{4} - \frac{1}{8}$,and $h(x) = \frac{x}{4}$.
Then $f(1) = 1^3 = 1$,$g(2) = \frac{2^2}{4} - \frac{1}{8} = 1 - \frac{1}{8} = \frac{7}{8}$,and $h(\frac{1}{2}) = \frac{1/2}{4} = \frac{1}{8}$.
Finally,$f(1) + g(2) + h(\frac{1}{2}) = 1 + \frac{7}{8} + \frac{1}{8} = 1 + 1 = 2$.

(A) Let $I = \int (\log_{e} 2x)^3 dx$.
Substitute $t = \log_{e} 2x$,then $2x = e^t$,so $x = \frac{1}{2} e^t$ and $dx = \frac{1}{2} e^t dt$.
$I = \int t^3 \cdot \frac{1}{2} e^t dt = \frac{1}{2} \int t^3 e^t dt$.
Using integration by parts $\int u dv = uv - \int v du$,where $u = t^3$ and $dv = e^t dt$:
$I = \frac{1}{2} [t^3 e^t - \int 3t^2 e^t dt] = \frac{1}{2} t^3 e^t - \frac{3}{2} \int t^2 e^t dt$.
Applying integration by parts again:
$\int t^2 e^t dt = t^2 e^t - 2 \int t e^t dt = t^2 e^t - 2(t e^t - e^t) = t^2 e^t - 2t e^t + 2e^t$.
Substituting back:
$I = \frac{1}{2} t^3 e^t - \frac{3}{2} [t^2 e^t - 2t e^t + 2e^t] + c = \frac{1}{2} e^t [t^3 - 3t^2 + 6t - 6] + c$.
Since $e^t = 2x$ and $t = \log_{e} 2x$:
$I = \frac{1}{2} (2x) [(\log_{e} 2x)^3 - 3(\log_{e} 2x)^2 + 6(\log_{e} 2x) - 6] + c = x[(\log_{e} 2x)^3 - 3(\log_{e} 2x)^2 + 6(\log_{e} 2x) - 6] + c$.

(A) Let $I = \int e^{2x+3} \sin 6x \, dx$.
Using integration by parts $\int u \, dv = uv - \int v \, du$,let $u = \sin 6x$ and $dv = e^{2x+3} \, dx$.
Then $du = 6 \cos 6x \, dx$ and $v = \frac{e^{2x+3}}{2}$.
$I = \frac{e^{2x+3}}{2} \sin 6x - \int \frac{e^{2x+3}}{2} \cdot 6 \cos 6x \, dx = \frac{e^{2x+3}}{2} \sin 6x - 3 \int e^{2x+3} \cos 6x \, dx$.
Applying integration by parts again for $\int e^{2x+3} \cos 6x \, dx$:
$u = \cos 6x, dv = e^{2x+3} \, dx \implies du = -6 \sin 6x \, dx, v = \frac{e^{2x+3}}{2}$.
$I = \frac{e^{2x+3}}{2} \sin 6x - 3 \left[ \frac{e^{2x+3}}{2} \cos 6x - \int \frac{e^{2x+3}}{2} (-6 \sin 6x) \, dx \right]$.
$I = \frac{e^{2x+3}}{2} \sin 6x - \frac{3}{2} e^{2x+3} \cos 6x - 9 \int e^{2x+3} \sin 6x \, dx$.
$I = \frac{e^{2x+3}}{2} \sin 6x - \frac{3}{2} e^{2x+3} \cos 6x - 9I$.
$10I = \frac{e^{2x+3}}{2} (\sin 6x - 3 \cos 6x) + C$.
$I = \frac{e^{2x+3}}{20} (\sin 6x - 3 \cos 6x) + C = \frac{e^{2x+3}}{40} (2 \sin 6x - 6 \cos 6x) + C$.

(C) Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = (\log x)^2$ and $dv = x^3 \, dx$.
Then $du = 2(\log x) \cdot \frac{1}{x} \, dx$ and $v = \frac{x^4}{4}$.
$I = \int x^3(\log x)^2 \, dx = (\log x)^2 \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \cdot 2(\log x) \cdot \frac{1}{x} \, dx$
$I = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} \int x^3 \log x \, dx$
Now,apply integration by parts again for $\int x^3 \log x \, dx$:
Let $u = \log x$ and $dv = x^3 \, dx$.
Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^4}{4}$.
$\int x^3 \log x \, dx = \frac{x^4}{4} \log x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{x^4}{4} \log x - \frac{x^4}{16}$.
Substituting this back:
$I = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} \left[ \frac{x^4}{4} \log x - \frac{x^4}{16} \right] + C$.