Integration by Parts Questions in English

(A) Let $I = \int (\log x)^2 x^3 dx$.
Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = (\log x)^2$ and $v = x^3$. Then $u' = \frac{2 \log x}{x}$ and $\int v dx = \frac{x^4}{4}$.
$I = (\log x)^2 \frac{x^4}{4} - \int \frac{2 \log x}{x} \cdot \frac{x^4}{4} dx$
$I = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \int x^3 \log x dx$.
Now,evaluate $\int x^3 \log x dx$ using integration by parts again with $u = \log x$ and $v = x^3$.
$\int x^3 \log x dx = (\log x) \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx$
$= \frac{x^4 \log x}{4} - \frac{1}{4} \int x^3 dx = \frac{x^4 \log x}{4} - \frac{x^4}{16}$.
Substituting this back into the expression for $I$:
$I = \frac{x^4}{4} (\log x)^2 - \frac{1}{2} \left( \frac{x^4 \log x}{4} - \frac{x^4}{16} \right) + C$
$I = \frac{x^4}{4} (\log x)^2 - \frac{x^4 \log x}{8} + \frac{x^4}{32} + C$
Factor out $\frac{x^4}{32}$:
$I = \frac{x^4}{32} \left( 8(\log x)^2 - 4 \log x + 1 \right) + C$.
Comparing this with $\frac{x^4}{32} f(x) + C$,we get $f(x) = 8(\log x)^2 - 4 \log x + 1$.

(B) The primitive means the indefinite integral. Let $I = \int \cos(\log x) dx$.
Using integration by parts,let $u = \cos(\log x)$ and $dv = dx$. Then $du = -\sin(\log x) \cdot \frac{1}{x} dx$ and $v = x$.
$I = x \cos(\log x) - \int x \cdot (-\sin(\log x)) \cdot \frac{1}{x} dx = x \cos(\log x) + \int \sin(\log x) dx$.
Now,integrate $\int \sin(\log x) dx$ by parts: $u = \sin(\log x)$,$dv = dx$.
$I = x \cos(\log x) + [x \sin(\log x) - \int x \cdot \cos(\log x) \cdot \frac{1}{x} dx]$.
$I = x \cos(\log x) + x \sin(\log x) - I$.
$2I = x [\cos(\log x) + \sin(\log x)] + C$.
$I = \frac{x}{2} [\cos(\log x) + \sin(\log x)] + C$.
Comparing this with $f(x)\{\cos(g(x)) + \sin(h(x))\}$,we get $f(x) = \frac{x}{2}$,$g(x) = \log x$,and $h(x) = \log x$.
Thus,$f^{\prime}(x) = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.
Therefore,option $B$ is correct.

(C) We are given the integral $I(x) = \int x^2(\log x)^2 dx$. Using integration by parts,where $u = (\log x)^2$ and $dv = x^2 dx$,we have $du = \frac{2 \log x}{x} dx$ and $v = \frac{x^3}{3}$.
$I(x) = \frac{x^3}{3}(\log x)^2 - \int \frac{x^3}{3} \cdot \frac{2 \log x}{x} dx$
$I(x) = \frac{x^3}{3}(\log x)^2 - \frac{2}{3} \int x^2 \log x dx$
Applying integration by parts again for $\int x^2 \log x dx$ with $u = \log x$ and $dv = x^2 dx$:
$\int x^2 \log x dx = \frac{x^3}{3} \log x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3} \log x - \frac{x^3}{9} + C_1$
Substituting this back into the expression for $I(x)$:
$I(x) = \frac{x^3}{3}(\log x)^2 - \frac{2}{3} [\frac{x^3}{3} \log x - \frac{x^3}{9}] + C$
$I(x) = \frac{x^3}{3}(\log x)^2 - \frac{2x^3}{9} \log x + \frac{2x^3}{27} + C$
$I(x) = \frac{x^3}{27} [9(\log x)^2 - 6 \log x + 2] + C$
Given $I(1) = 0$,we substitute $x = 1$:
$I(1) = \frac{1}{27} [9(0)^2 - 6(0) + 2] + C = 0$
$\frac{2}{27} + C = 0 \Rightarrow C = -\frac{2}{27}$
Thus,$I(x) = \frac{x^3}{27} [9(\log x)^2 - 6 \log x + 2] - \frac{2}{27}$.

(A) Given,$\int x^3 e^{2 x} d x = \frac{e^{2 x}}{8} f(x) + c$.
Applying integration by parts repeatedly:
$\int x^3 e^{2 x} d x = x^3 \frac{e^{2 x}}{2} - \int 3x^2 \frac{e^{2 x}}{2} d x$
$= \frac{x^3 e^{2 x}}{2} - \frac{3}{2} \left( x^2 \frac{e^{2 x}}{2} - \int 2x \frac{e^{2 x}}{2} d x \right)$
$= \frac{x^3 e^{2 x}}{2} - \frac{3x^2 e^{2 x}}{4} + \frac{3}{2} \int x e^{2 x} d x$
$= \frac{x^3 e^{2 x}}{2} - \frac{3x^2 e^{2 x}}{4} + \frac{3}{2} \left( x \frac{e^{2 x}}{2} - \int \frac{e^{2 x}}{2} d x \right)$
$= \frac{x^3 e^{2 x}}{2} - \frac{3x^2 e^{2 x}}{4} + \frac{3x e^{2 x}}{4} - \frac{3}{4} \frac{e^{2 x}}{2} + c$
$= \frac{e^{2 x}}{8} (4x^3 - 6x^2 + 6x - 3) + c$.
Comparing with the given form,$f(x) = 4x^3 - 6x^2 + 6x - 3$.
For $f(x) = 1$,we have $4x^3 - 6x^2 + 6x - 4 = 0$,which simplifies to $2x^3 - 3x^2 + 3x - 2 = 0$.
Factoring,$2(x^3 - 1) - 3x(x - 1) = 0 \Rightarrow (x - 1)(2(x^2 + x + 1) - 3x) = 0$.
$(x - 1)(2x^2 - x + 2) = 0$.
The roots are $x = 1$ (real) and the roots of $2x^2 - x + 2 = 0$ (complex).
The sum of the complex roots is $-\frac{b}{a} = -(\frac{-1}{2}) = \frac{1}{2}$.

(A) Let $I = \int (\log x)^3 x^5 dx$. Using integration by parts $\int u v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = (\log x)^3$ and $v = x^5$.
$I = (\log x)^3 \frac{x^6}{6} - \int 3(\log x)^2 \frac{1}{x} \frac{x^6}{6} dx = \frac{x^6}{6}(\log x)^3 - \frac{1}{2} \int x^5 (\log x)^2 dx$.
Applying parts again for $\int x^5 (\log x)^2 dx$:
$I = \frac{x^6}{6}(\log x)^3 - \frac{1}{2} \left[ (\log x)^2 \frac{x^6}{6} - \int 2(\log x) \frac{1}{x} \frac{x^6}{6} dx \right] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} \int x^5 \log x dx$.
Applying parts again for $\int x^5 \log x dx$:
$I = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} \left[ \log x \frac{x^6}{6} - \int \frac{1}{x} \frac{x^6}{6} dx \right] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36} \log x - \frac{1}{36} \int x^5 dx$.
$I = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36} \log x - \frac{1}{36} \frac{x^6}{6} + c = x^6 \left[ \frac{(\log x)^3}{6} - \frac{(\log x)^2}{12} + \frac{\log x}{36} - \frac{1}{216} \right] + c$.

(B) To evaluate $\int x^5 e^{-2 x} d x$,we use the formula for integration by parts repeatedly or the tabular method (Bernoulli's formula): $\int u v' dx = u v - u' v_1 + u'' v_2 - u''' v_3 + \dots$
Here,$u = x^5$ and $v' = e^{-2 x}$.
$u = x^5, u' = 5x^4, u'' = 20x^3, u''' = 60x^2, u^{(4)} = 120x, u^{(5)} = 120, u^{(6)} = 0$.
$v = e^{-2 x}, v_1 = \frac{e^{-2 x}}{-2}, v_2 = \frac{e^{-2 x}}{4}, v_3 = \frac{e^{-2 x}}{-8}, v_4 = \frac{e^{-2 x}}{16}, v_5 = \frac{e^{-2 x}}{-32}, v_6 = \frac{e^{-2 x}}{64}$.
Applying the formula:
$\int x^5 e^{-2 x} d x = x^5 \left(\frac{e^{-2 x}}{-2}\right) - 5x^4 \left(\frac{e^{-2 x}}{4}\right) + 20x^3 \left(\frac{e^{-2 x}}{-8}\right) - 60x^2 \left(\frac{e^{-2 x}}{16}\right) + 120x \left(\frac{e^{-2 x}}{-32}\right) - 120 \left(\frac{e^{-2 x}}{64}\right) + c$
$= -e^{-2 x} \left[ \frac{x^5}{2} + \frac{5x^4}{4} + \frac{20x^3}{8} + \frac{60x^2}{16} + \frac{120x}{32} + \frac{120}{64} \right] + c$
$= -e^{-2 x} \left[ \frac{x^5}{2} + \frac{5x^4}{4} + \frac{5x^3}{2} + \frac{15x^2}{4} + \frac{15x}{4} + \frac{15}{8} \right] + c$.

(C) We know that $\sin x \cos x = \frac{1}{2} \sin 2x$.
Substituting this into the integral,we get:
$I = \int x^2 \left(\frac{1}{2} \sin 2x\right) dx = \frac{1}{2} \int x^2 \sin 2x \, dx$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,let $u = x^2$ and $dv = \sin 2x \, dx$.
Then $du = 2x \, dx$ and $v = -\frac{\cos 2x}{2}$.
$I = \frac{1}{2} \left[ -\frac{x^2 \cos 2x}{2} - \int \left(-\frac{\cos 2x}{2}\right) (2x) \, dx \right]$
$I = -\frac{x^2 \cos 2x}{4} + \frac{1}{2} \int x \cos 2x \, dx$.
Applying integration by parts again for $\int x \cos 2x \, dx$:
Let $u = x$ and $dv = \cos 2x \, dx$. Then $du = dx$ and $v = \frac{\sin 2x}{2}$.
$\int x \cos 2x \, dx = \frac{x \sin 2x}{2} - \int \frac{\sin 2x}{2} \, dx = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4}$.
Substituting this back:
$I = -\frac{x^2 \cos 2x}{4} + \frac{1}{2} \left( \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \right) + c$
$I = -\frac{x^2 \cos 2x}{4} + \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + c$.
Comparing this with the options,the expression can be rewritten as $\frac{1-2x^2}{8} \cos 2x + \frac{x}{4} \sin 2x + c$.

(A) Using the formula for integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = x^3$ and $v = e^{5x}$.
$\int x^3 e^{5x} dx = x^3 \frac{e^{5x}}{5} - \int 3x^2 \frac{e^{5x}}{5} dx = \frac{x^3 e^{5x}}{5} - \frac{3}{5} \int x^2 e^{5x} dx$.
Now,$\int x^2 e^{5x} dx = x^2 \frac{e^{5x}}{5} - \int 2x \frac{e^{5x}}{5} dx = \frac{x^2 e^{5x}}{5} - \frac{2}{5} \int x e^{5x} dx$.
And $\int x e^{5x} dx = x \frac{e^{5x}}{5} - \int 1 \frac{e^{5x}}{5} dx = \frac{x e^{5x}}{5} - \frac{e^{5x}}{25}$.
Substituting these back:
$\int x^3 e^{5x} dx = \frac{x^3 e^{5x}}{5} - \frac{3}{5} [\frac{x^2 e^{5x}}{5} - \frac{2}{5} (\frac{x e^{5x}}{5} - \frac{e^{5x}}{25})] + C$.
$= \frac{x^3 e^{5x}}{5} - \frac{3 x^2 e^{5x}}{25} + \frac{6 x e^{5x}}{125} - \frac{6 e^{5x}}{625} + C$.
$= \frac{e^{5x}}{625} [125 x^3 - 75 x^2 + 30 x - 6] + C$.
Since $5^4 = 625$,we have $f(x) = 125 x^3 - 75 x^2 + 30 x - 6$.

(C) Let $I = \int \log \left(a^2+x^2\right) d x$.
Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Let $u = \log(a^2+x^2)$ and $v = 1$.
Then $u' = \frac{2x}{a^2+x^2}$ and $\int v dx = x$.
$I = x \log(a^2+x^2) - \int \frac{2x^2}{a^2+x^2} dx + C$.
$I = x \log(a^2+x^2) - 2 \int \frac{x^2+a^2-a^2}{a^2+x^2} dx + C$.
$I = x \log(a^2+x^2) - 2 \int (1 - \frac{a^2}{a^2+x^2}) dx + C$.
$I = x \log(a^2+x^2) - 2x + 2a^2 \int \frac{1}{a^2+x^2} dx + C$.
Since $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a})$,we get:
$I = x \log(a^2+x^2) - 2x + 2a^2 (\frac{1}{a} \tan^{-1}(\frac{x}{a})) + C$.
$I = x \log(a^2+x^2) - 2x + 2a \tan^{-1}(\frac{x}{a}) + C$.
Comparing with $h(x)+C$,we have $h(x) = x \log(a^2+x^2) - 2x + 2a \tan^{-1}(\frac{x}{a})$.

(D) We use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = (x+1)^2$ and $dv = e^x \, dx$.
Then $du = 2(x+1) \, dx$ and $v = e^x$.
$\int (x+1)^2 e^x \, dx = (x+1)^2 e^x - \int 2(x+1) e^x \, dx$.
Now,apply integration by parts again for $\int (x+1) e^x \, dx$:
Let $u = (x+1)$ and $dv = e^x \, dx$.
Then $du = dx$ and $v = e^x$.
$\int (x+1) e^x \, dx = (x+1) e^x - \int e^x \, dx = (x+1) e^x - e^x = x e^x$.
Substituting this back:
$\int (x+1)^2 e^x \, dx = (x+1)^2 e^x - 2(x e^x) + C$
$= (x^2 + 2x + 1) e^x - 2x e^x + C$
$= (x^2 + 1) e^x + C$.

(A) 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 $\int u \, dv = uv - \int v \, du$:
$I_n = x(\log x)^n - \int x \cdot n(\log x)^{n-1} \cdot \frac{1}{x} \, dx$
$I_n = x(\log x)^n - n \int (\log x)^{n-1} \, dx$
Since $I_{n-1} = \int (\log x)^{n-1} \, dx$,we have:
$I_n = x(\log x)^n - n I_{n-1}$
Rearranging the terms,we get:
$I_n + n I_{n-1} = x(\log x)^n$.

(A) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$.
Substitute $x = \tan \theta$,then $d x = \sec ^2 \theta d \theta$.
Since $\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) = \sin ^{-1}(\sin 2 \theta) = 2 \theta$ (assuming $|x| \le 1$),we have:
$I = \int 2 \theta \sec ^2 \theta d \theta$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u = \theta$ and $dv = \sec ^2 \theta d \theta$:
$I = 2 \left[ \theta \tan \theta - \int \tan \theta d \theta \right]$.
$I = 2 [\theta \tan \theta + \log |\cos \theta|] + C$.
Since $\tan \theta = x$,we have $\theta = \tan ^{-1} x$ and $\cos \theta = \frac{1}{\sqrt{1+x^2}}$.
$I = 2 [x \tan ^{-1} x + \log |\frac{1}{\sqrt{1+x^2}}|] + C$.
$I = 2 x \tan ^{-1} x + 2 \log (1+x^2)^{-1/2} + C$.
$I = 2 x \tan ^{-1} x - \log (1+x^2) + C$.
Comparing this with $f(x) - \log (1+x^2)$,we get $f(x) = 2 x \tan ^{-1} x$.

(B) Let $I = \int (\log x)^3 dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int t^3 e^t dt$.
Using the integration by parts formula $\int e^t f(t) dt = e^t [f(t) - f'(t) + f''(t) - f'''(t) + \dots]$,where $f(t) = t^3$:
$f'(t) = 3t^2$,$f''(t) = 6t$,and $f'''(t) = 6$.
Therefore,$I = e^t [t^3 - 3t^2 + 6t - 6] + C$.
Substituting back $t = \log x$ and $e^t = x$:
$I = x [(\log x)^3 - 3(\log x)^2 + 6 \log x - 6] + C$.

(A) Let $I = \int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$.
Substitute $x = \cos 2\theta$,which implies $dx = -2\sin 2\theta d\theta$.
Then $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} = \sqrt{\frac{2\sin^2 \theta}{2\cos^2 \theta}} = \tan \theta$.
So,$I = \int \theta (-2\sin 2\theta) d\theta = -2 \int \theta \sin 2\theta d\theta$.
Using integration by parts: $I = -2 \left[ \theta \left(-\frac{\cos 2\theta}{2}\right) - \int 1 \cdot \left(-\frac{\cos 2\theta}{2}\right) d\theta \right] = \theta \cos 2\theta - \int \cos 2\theta d\theta = \theta \cos 2\theta - \frac{\sin 2\theta}{2} + c$.
Since $x = \cos 2\theta$,$\theta = \frac{1}{2} \cos^{-1} x$ and $\sin 2\theta = \sqrt{1-x^2}$.
Substituting these back: $I = \frac{1}{2} \cos^{-1} x \cdot x - \frac{1}{2} \sqrt{1-x^2} + c = \frac{1}{2} (x \cos^{-1} x - \sqrt{1-x^2}) + c$.

(A) Let $I = \int \frac{x^2 \operatorname{Tan}^{-1} x}{(1+x^2)^2} dx$.
Using integration by parts,let $u = \operatorname{Tan}^{-1} x$ and $dv = \frac{x^2}{(1+x^2)^2} dx$.
Then $du = \frac{1}{1+x^2} dx$.
To find $v$,we write $\int \frac{x^2}{(1+x^2)^2} dx = \int \frac{x^2+1-1}{(1+x^2)^2} dx = \int \frac{1}{1+x^2} dx - \int \frac{1}{(1+x^2)^2} dx$.
Using the substitution $x = \tan \theta$,$dx = \sec^2 \theta d\theta$,we get $\int \frac{1}{(1+x^2)^2} dx = \int \frac{\sec^2 \theta}{\sec^4 \theta} d\theta = \int \cos^2 \theta d\theta = \frac{1}{2} \int (1 + \cos 2\theta) d\theta = \frac{1}{2} (\theta + \frac{\sin 2\theta}{2}) = \frac{1}{2} \operatorname{Tan}^{-1} x + \frac{x}{2(1+x^2)}$.
Thus,$v = \operatorname{Tan}^{-1} x - (\frac{1}{2} \operatorname{Tan}^{-1} x + \frac{x}{2(1+x^2)}) = \frac{1}{2} \operatorname{Tan}^{-1} x - \frac{x}{2(1+x^2)}$.
Now,$I = uv - \int v du = \operatorname{Tan}^{-1} x (\frac{1}{2} \operatorname{Tan}^{-1} x - \frac{x}{2(1+x^2)}) - \int (\frac{1}{2} \operatorname{Tan}^{-1} x - \frac{x}{2(1+x^2)}) \frac{1}{1+x^2} dx$.
$I = \frac{(\operatorname{Tan}^{-1} x)^2}{2} - \frac{x \operatorname{Tan}^{-1} x}{2(1+x^2)} - \frac{1}{2} \int \frac{\operatorname{Tan}^{-1} x}{1+x^2} dx + \frac{1}{2} \int \frac{x}{(1+x^2)^2} dx$.
$I = \frac{(\operatorname{Tan}^{-1} x)^2}{2} - \frac{x \operatorname{Tan}^{-1} x}{2(1+x^2)} - \frac{(\operatorname{Tan}^{-1} x)^2}{4} - \frac{1}{4(1+x^2)} + c$.
$I = \frac{(\operatorname{Tan}^{-1} x)^2}{4} - \frac{x \operatorname{Tan}^{-1} x}{2(1+x^2)} - \frac{1}{4(1+x^2)} + c$.
Comparing with the options,option $A$ is equivalent to $\frac{(\operatorname{Tan}^{-1} x)^2}{4} - \frac{2x \operatorname{Tan}^{-1} x + 1 - x^2}{4(1+x^2)} + c$.

(D) Let $I = \int \frac{\log x}{(1+x)^3} dx$.
Using integration by parts,let $u = \log x$ and $dv = (1+x)^{-3} dx$.
Then $du = \frac{1}{x} dx$ and $v = \int (1+x)^{-3} dx = \frac{(1+x)^{-2}}{-2} = -\frac{1}{2(1+x)^2}$.
Applying the formula $\int u dv = uv - \int v du$:
$I = -\frac{\log x}{2(1+x)^2} - \int \left(-\frac{1}{2(1+x)^2}\right) \frac{1}{x} dx$
$I = -\frac{\log x}{2(1+x)^2} + \frac{1}{2} \int \frac{1}{x(1+x)^2} dx$.
Using partial fractions for $\frac{1}{x(1+x)^2} = \frac{A}{x} + \frac{B}{1+x} + \frac{C}{(1+x)^2}$:
$1 = A(1+x)^2 + Bx(1+x) + Cx$.
For $x=0$,$A=1$.
For $x=-1$,$C=-1$.
Comparing coefficients of $x^2$: $A+B=0 \implies B=-1$.
So,$\int \frac{1}{x(1+x)^2} dx = \int \left(\frac{1}{x} - \frac{1}{1+x} - \frac{1}{(1+x)^2}\right) dx = \log|x| - \log|1+x| + \frac{1}{1+x} = \log\left|\frac{x}{1+x}\right| + \frac{1}{1+x}$.
Substituting back: $I = -\frac{\log x}{2(1+x)^2} + \frac{1}{2} \left(\log\left|\frac{x}{1+x}\right| + \frac{1}{1+x}\right) + c = \frac{1}{2} \left[ \frac{1}{1+x} - \frac{\log x}{(1+x)^2} + \log\left|\frac{x}{1+x}\right| \right] + c$.

(C) We use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $I = \int x^3 \sin 3x \, dx$.
Using the tabular method for integration by parts:
$u = x^3$,$dv = \sin 3x \, dx$
$u' = 3x^2$,$v = -\frac{1}{3} \cos 3x$
$u'' = 6x$,$v_1 = -\frac{1}{9} \sin 3x$
$u''' = 6$,$v_2 = \frac{1}{27} \cos 3x$
$u'''' = 0$,$v_3 = \frac{1}{81} \sin 3x$
$I = (x^3)(-\frac{1}{3} \cos 3x) - (3x^2)(-\frac{1}{9} \sin 3x) + (6x)(\frac{1}{27} \cos 3x) - (6)(\frac{1}{81} \sin 3x) + c$
$I = -\frac{x^3}{3} \cos 3x + \frac{x^2}{3} \sin 3x + \frac{2x}{9} \cos 3x - \frac{2}{27} \sin 3x + c$
Multiply by $\frac{27}{27}$ to match the form $\frac{1}{27}[f(x) \cos 3x + g(x) \sin 3x]$:
$I = \frac{1}{27}[-9x^3 \cos 3x + 9x^2 \sin 3x + 6x \cos 3x - 2 \sin 3x] + c$
$I = \frac{1}{27}[(-9x^3 + 6x) \cos 3x + (9x^2 - 2) \sin 3x] + c$
Thus,$f(x) = -9x^3 + 6x$ and $g(x) = 9x^2 - 2$.
$f(1) = -9(1)^3 + 6(1) = -9 + 6 = -3$.
$g(1) = 9(1)^2 - 2 = 9 - 2 = 7$.
$f(1) + g(1) = -3 + 7 = 4$.

(A) Let $I = \int \frac{x^2(x \sec^2 x + \tan x)}{(x \tan x + 1)^2} dx$.
Using integration by parts,let $u = x^2$ and $dv = \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx$.
First,find $v = \int \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx$.
Let $t = x \tan x + 1$,then $dt = (x \sec^2 x + \tan x) dx$.
So,$v = \int \frac{1}{t^2} dt = -\frac{1}{t} = -\frac{1}{x \tan x + 1}$.
Now,applying the integration by parts formula $\int u dv = uv - \int v du$:
$I = x^2 \left( -\frac{1}{x \tan x + 1} \right) - \int \left( -\frac{1}{x \tan x + 1} \right) (2x) dx$.
$I = -\frac{x^2}{x \tan x + 1} + \int \frac{2x}{x \tan x + 1} dx$.
$I = -\frac{x^2}{x \tan x + 1} + \int \frac{2x \cos x}{x \sin x + \cos x} dx$.
Comparing this with the given expression,$f(x) = \int \frac{2x \cos x}{x \sin x + \cos x} dx$.
Let $u = x \sin x + \cos x$,then $du = (x \cos x + \sin x - \sin x) dx = x \cos x dx$.
Thus,$f(x) = \int \frac{2}{u} du = 2 \log |u| + c = 2 \log |x \sin x + \cos x| + c$.

(C) We use integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = (\log x)^3$ and $v = x^5$.
$\int (\log x)^3 x^5 dx = (\log x)^3 \frac{x^6}{6} - \int 3(\log x)^2 \frac{1}{x} \frac{x^6}{6} dx = \frac{x^6}{6}(\log x)^3 - \frac{1}{2} \int x^5 (\log x)^2 dx$.
Applying integration by parts again:
$= \frac{x^6}{6}(\log x)^3 - \frac{1}{2} [(\log x)^2 \frac{x^6}{6} - \int 2(\log x) \frac{1}{x} \frac{x^6}{6} dx] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} \int x^5 \log x dx$.
Applying integration by parts once more:
$= \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{1}{6} [(\log x) \frac{x^6}{6} - \int \frac{1}{x} \frac{x^6}{6} dx] = \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36}\log x - \frac{1}{36} \int x^5 dx$.
$= \frac{x^6}{6}(\log x)^3 - \frac{x^6}{12}(\log x)^2 + \frac{x^6}{36}\log x - \frac{x^6}{216} + k$.
Factor out $\frac{x^6}{216}$:
$= \frac{x^6}{216} [36(\log x)^3 - 18(\log x)^2 + 6(\log x) - 1] + k$.
Comparing with the given form,$A = 216, B = 36, C = -18, D = 6$.
Thus,$A - (B + C + D) = 216 - (36 - 18 + 6) = 216 - 24 = 192$.

(A) Using integration by parts,$I_n = \int x^n \sin x \, dx = -x^n \cos x + n \int x^{n-1} \cos x \, dx = -x^n \cos x + n(x^{n-1} \sin x - (n-1) \int x^{n-2} \sin x \, dx)$.
So,$I_n = -x^n \cos x + n x^{n-1} \sin x - n(n-1) I_{n-2}$.
This gives the recurrence relation: $I_n + n(n-1) I_{n-2} = -x^n \cos x + n x^{n-1} \sin x$.
For $n=6$: $I_6 + 30 I_4 = -x^6 \cos x + 6 x^5 \sin x$.
For $n=4$: $I_4 + 12 I_2 = -x^4 \cos x + 4 x^3 \sin x$.
Multiplying the second equation by $-30$: $-30 I_4 - 360 I_2 = 30 x^4 \cos x - 120 x^3 \sin x$.
Adding the two equations: $I_6 - 360 I_2 = (30 x^4 - x^6) \cos x + (6 x^5 - 120 x^3) \sin x$.
Comparing with $f(x) \cos x + g(x) \sin x$,we have $f(x) = 30 x^4 - x^6$ and $g(x) = 6 x^5 - 120 x^3$.
Then $f(1) = 30(1)^4 - (1)^6 = 29$ and $g(1) = 6(1)^5 - 120(1)^3 = -114$.
Thus,$f(1) + g(1) = 29 - 114 = -85$.

(C) Let $I = \int \frac{\tan ^{-1} x}{x^3} d x$.
Using integration by parts,let $u = \tan ^{-1} x$ and $dv = x^{-3} dx$.
Then $du = \frac{1}{1+x^2} dx$ and $v = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$.
$I = u v - \int v du = \tan ^{-1} x \left(-\frac{1}{2x^2}\right) - \int \left(-\frac{1}{2x^2}\right) \frac{1}{1+x^2} dx$.
$I = -\frac{\tan ^{-1} x}{2x^2} + \frac{1}{2} \int \frac{1}{x^2(1+x^2)} dx$.
Using partial fractions,$\frac{1}{x^2(1+x^2)} = \frac{1}{x^2} - \frac{1}{1+x^2}$.
$I = -\frac{\tan ^{-1} x}{2x^2} + \frac{1}{2} \int \left(\frac{1}{x^2} - \frac{1}{1+x^2}\right) dx$.
$I = -\frac{\tan ^{-1} x}{2x^2} + \frac{1}{2} \left(-\frac{1}{x} - \tan ^{-1} x\right) + C$.
$I = -\frac{1}{2x} - \left(\frac{1}{2x^2} + \frac{1}{2}\right) \tan ^{-1} x + C$.

(D) We use the method of integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = (\log x)^2$ and $v = x^3$.
Then $u' = 2 \log x \cdot \frac{1}{x}$ and $\int v dx = \frac{x^4}{4}$.
$\int x^3(\log x)^2 dx = \frac{x^4}{4}(\log x)^2 - \int (2 \log x \cdot \frac{1}{x} \cdot \frac{x^4}{4}) dx = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} \int x^3 \log x dx$.
Now,integrate $\int x^3 \log x dx$ by parts again:
$\int x^3 \log x dx = \log x \cdot \frac{x^4}{4} - \int \frac{1}{x} \cdot \frac{x^4}{4} dx = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 dx = \frac{x^4}{4} \log x - \frac{x^4}{16}$.
Substituting this back:
$\int x^3(\log x)^2 dx = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} [\frac{x^4}{4} \log x - \frac{x^4}{16}] + K = x^4 [\frac{1}{4}(\log x)^2 - \frac{1}{8} \log x + \frac{1}{32}] + K$.
Comparing with $x^4[A(\log x)^2 + B(\log x) + C]$,we get $A = \frac{1}{4}$,$B = -\frac{1}{8}$,and $C = \frac{1}{32}$.
Thus,$A + B + C = \frac{1}{4} - \frac{1}{8} + \frac{1}{32} = \frac{8 - 4 + 1}{32} = \frac{5}{32}$.

(A) Let $I = \int x^2 \cdot \frac{x \sec^2 x+\tan x}{(x \tan x+1)^2} dx$.
Using integration by parts,let $u = x^2$ and $dv = \frac{x \sec^2 x+\tan x}{(x \tan x+1)^2} dx$.
Then $du = 2x dx$ and $v = \int \frac{d(x \tan x)}{(x \tan x+1)^2} = -\frac{1}{x \tan x+1}$.
So,$I = x^2 \left(-\frac{1}{x \tan x+1}\right) - \int 2x \left(-\frac{1}{x \tan x+1}\right) dx$.
$I = -\frac{x^2}{x \tan x+1} + 2 \int \frac{x}{x \tan x+1} dx$.
Multiply numerator and denominator by $\cos x$:
$I = -\frac{x^2}{x \tan x+1} + 2 \int \frac{x \cos x}{x \sin x + \cos x} dx$.
Let $t = x \sin x + \cos x$,then $dt = (x \cos x + \sin x - \sin x) dx = x \cos x dx$.
$I = -\frac{x^2}{x \tan x+1} + 2 \int \frac{dt}{t} = -\frac{x^2}{x \tan x+1} + 2 \log|x \sin x + \cos x| + C$.
Comparing with the given form,$A = 2$,$B = -1$,and $f(x) = x^2$.
Thus,$f(A+B) = f(2-1) = f(1) = (1)^2 = 1$.

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

(C) To evaluate the integral $I = \int (\log x)^2 dx$,we use the method of integration by parts: $\int u dv = uv - \int v du$.
Let $u = (\log x)^2$ and $dv = dx$.
Then $du = 2 \log x \cdot \frac{1}{x} dx$ and $v = x$.
Applying the formula:
$I = x(\log x)^2 - \int x \cdot \frac{2 \log x}{x} dx + c$
$I = x(\log x)^2 - 2 \int \log x dx + c$
Now,integrate $\int \log x dx$ using integration by parts again (let $u = \log x, dv = dx$):
$\int \log x dx = x \log x - \int x \cdot \frac{1}{x} dx = x \log x - x$.
Substituting this back into the expression for $I$:
$I = x(\log x)^2 - 2(x \log x - x) + c$
$I = x(\log x)^2 - 2x \log x + 2x + c$
$I = x(\log x)^2 - 2x(\log x - 1) + c$.

(C) Let $I = \int (x+x^2) \log(1+x^2) dx$. Using integration by parts,taking $\log(1+x^2)$ as the first function:
$I = \log(1+x^2) \int (x+x^2) dx - \int \left( \frac{d}{dx} \log(1+x^2) \int (x+x^2) dx \right) dx$
$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} \right) - \int \frac{2x}{1+x^2} \left( \frac{x^2}{2} + \frac{x^3}{3} \right) dx$
$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} \right) - \int \left( \frac{x^3}{1+x^2} + \frac{2x^4}{3(1+x^2)} \right) dx$
Evaluating the integrals:
$\int \frac{x^3}{1+x^2} dx = \int \frac{x(x^2+1-1)}{1+x^2} dx = \int x dx - \int \frac{x}{1+x^2} dx = \frac{x^2}{2} - \frac{1}{2} \log(1+x^2)$
$\int \frac{2x^4}{3(1+x^2)} dx = \frac{2}{3} \int \frac{x^4-1+1}{1+x^2} dx = \frac{2}{3} \int (x^2-1) dx + \frac{2}{3} \int \frac{1}{1+x^2} dx = \frac{2x^3}{9} - \frac{2x}{3} + \frac{2}{3} \tan^{-1} x$
Substituting back:
$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} \right) - \left( \frac{x^2}{2} - \frac{1}{2} \log(1+x^2) + \frac{2x^3}{9} - \frac{2x}{3} + \frac{2}{3} \tan^{-1} x \right) + c$
$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} + \frac{1}{2} \right) - \frac{2}{3} \tan^{-1} x - \frac{2x^3}{9} - \frac{x^2}{2} + \frac{2x}{3} + c$
Comparing with the given expression,$F(x) = \frac{x^2}{2} + \frac{x^3}{3} + \frac{1}{2}$.

(C) Let $I_1 = \int \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) d x$.
For $|x| < 1$,we know that $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right) = 2 \tan ^{-1} x$.
Thus,$I_1 = \int 2 \tan ^{-1} x d x = 2 \int \tan ^{-1} x d x$.
Using integration by parts,let $u = \tan ^{-1} x$ and $dv = dx$. Then $du = \frac{1}{1+x^2} dx$ and $v = x$.
$I_1 = 2 \left( x \tan ^{-1} x - \int \frac{x}{1+x^2} dx \right)$.
$I_1 = 2 x \tan ^{-1} x - \int \frac{2x}{1+x^2} dx$.
$I_1 = 2 x \tan ^{-1} x - \log(1+x^2) + C$.
Since $2 \tan ^{-1} x = \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$,we substitute back:
$I_1 = x \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) - \log(1+x^2) + C$.

(A) Let $I = \int (\log x)^3 x^4 \, dx$. Using 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 $dv = x^4 \, dx$:
$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} [\frac{x^5}{5}(\log x)^2 - \int \frac{x^5}{5} \cdot 2(\log x) \cdot \frac{1}{x} \, dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5(\log x)^2 + \frac{6}{25} \int x^4 \log x \, dx$.
Applying integration by parts for $\int x^4 \log x \, dx$ with $u = \log x$ and $dv = x^4 \, dx$:
$I = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5(\log x)^2 + \frac{6}{25} [\frac{x^5}{5} \log x - \int \frac{x^5}{5} \cdot \frac{1}{x} \, dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25} x^5(\log x)^2 + \frac{6}{125} x^5 \log x - \frac{6}{625} x^5 + c$.
Factoring out $\frac{x^5}{625}$:
$I = \frac{x^5}{625} [125(\log x)^3 - 75(\log x)^2 + 30(\log x) - 6] + c$.
Substituting $p = \log x$,we get $\frac{x^5}{625} [125 p^3 - 75 p^2 + 30 p - 6] + c$.

(A) Let $I = \int x^4 e^{2 x} d x$.
Using integration by parts,$\int u v d x = u \int v d x - \int (u' \int v d x) d x$.
Taking $u = x^4$ and $v = e^{2 x}$:
$I = x^4 \frac{e^{2 x}}{2} - \int 4 x^3 \frac{e^{2 x}}{2} d x = \frac{x^4 e^{2 x}}{2} - 2 \int x^3 e^{2 x} d x$.
Applying integration by parts again:
$\int x^3 e^{2 x} d x = x^3 \frac{e^{2 x}}{2} - \int 3 x^2 \frac{e^{2 x}}{2} d x = \frac{x^3 e^{2 x}}{2} - \frac{3}{2} \int x^2 e^{2 x} d x$.
Continuing the process:
$\int x^2 e^{2 x} d x = \frac{x^2 e^{2 x}}{2} - \int x e^{2 x} d x = \frac{x^2 e^{2 x}}{2} - (\frac{x e^{2 x}}{2} - \frac{e^{2 x}}{4})$.
Substituting back into the expression for $I$:
$I = \frac{x^4 e^{2 x}}{2} - 2 [\frac{x^3 e^{2 x}}{2} - \frac{3}{2} (\frac{x^2 e^{2 x}}{2} - \frac{x e^{2 x}}{2} + \frac{e^{2 x}}{4})] + C$.
$I = \frac{x^4 e^{2 x}}{2} - x^3 e^{2 x} + \frac{3}{2} x^2 e^{2 x} - \frac{3}{2} x e^{2 x} + \frac{3}{4} e^{2 x} + C$.
$I = \frac{e^{2 x}}{4} (2 x^4 - 4 x^3 + 6 x^2 - 6 x + 3) + C$.

(C) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} dx$.
We can rewrite the integral as $I = \int x^2 \cdot \frac{x}{\sqrt{1+x^2}} dx$.
Using integration by parts,let $u = x^2$ and $dv = \frac{x}{\sqrt{1+x^2}} dx$.
Then $du = 2x dx$ and $v = \int (1+x^2)^{-1/2} x dx = \sqrt{1+x^2}$.
Applying the formula $\int u dv = uv - \int v du$:
$I = x^2 \sqrt{1+x^2} - \int 2x \sqrt{1+x^2} dx$.
For the remaining integral,let $t = 1+x^2$,so $dt = 2x dx$.
$I = x^2 \sqrt{1+x^2} - \int t^{1/2} dt = x^2 \sqrt{1+x^2} - \frac{t^{3/2}}{3/2} + C$.
$I = x^2 \sqrt{1+x^2} - \frac{2}{3}(1+x^2)^{3/2} + C$.

(A) Given that,$I_n = \int x^n \cdot e^{cx} \, dx$.
Using the method of integration by parts,where $\int u \, dv = uv - \int v \, du$:
Let $u = x^n$ and $dv = e^{cx} \, dx$.
Then $du = n x^{n-1} \, dx$ and $v = \frac{e^{cx}}{c}$.
Substituting these into the formula:
$I_n = x^n \cdot \frac{e^{cx}}{c} - \int \frac{e^{cx}}{c} \cdot n x^{n-1} \, dx$.
$I_n = \frac{x^n e^{cx}}{c} - \frac{n}{c} \int x^{n-1} e^{cx} \, dx$.
Since $I_{n-1} = \int x^{n-1} e^{cx} \, dx$,we have:
$I_n = \frac{x^n e^{cx}}{c} - \frac{n}{c} I_{n-1}$.
Multiplying both sides by $c$:
$c I_n = x^n e^{cx} - n I_{n-1}$.
Rearranging the terms:
$c I_n + n I_{n-1} = x^n e^{cx}$.

(A) We need to evaluate $I = \int e^{-3x}(x^2 + \sin 4x) dx = \int x^2 e^{-3x} dx + \int e^{-3x} \sin 4x dx$.
For the first part,using integration by parts $\int u dv = uv - \int v du$ with $u = x^2$ and $dv = e^{-3x} dx$:
$\int x^2 e^{-3x} dx = x^2 \left(\frac{e^{-3x}}{-3}\right) - \int 2x \left(\frac{e^{-3x}}{-3}\right) dx = -\frac{x^2}{3} e^{-3x} + \frac{2}{3} \int x e^{-3x} dx$.
Applying integration by parts again:
$\int x e^{-3x} dx = x \left(\frac{e^{-3x}}{-3}\right) - \int 1 \left(\frac{e^{-3x}}{-3}\right) dx = -\frac{x}{3} e^{-3x} - \frac{1}{9} e^{-3x}$.
So,$\int x^2 e^{-3x} dx = -\frac{x^2}{3} e^{-3x} - \frac{2}{9} x e^{-3x} - \frac{2}{27} e^{-3x}$.
For the second part,using the formula $\int e^{ax} \sin bx dx = \frac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx)$:
$\int e^{-3x} \sin 4x dx = \frac{e^{-3x}}{(-3)^2 + 4^2} (-3 \sin 4x - 4 \cos 4x) = -\frac{e^{-3x}}{25} (3 \sin 4x + 4 \cos 4x)$.
Combining both parts:
$I = -e^{-3x} \left( \frac{x^2}{3} + \frac{2x}{9} + \frac{2}{27} + \frac{3}{25} \sin 4x + \frac{4}{25} \cos 4x \right) + C$.

(C) Using integration by parts $\int u \, dv = uv - \int v \, du$ repeatedly:
$\int x^3 \sin 3x \, dx = x^3 \left( \frac{-\cos 3x}{3} \right) - \int 3x^2 \left( \frac{-\cos 3x}{3} \right) dx = -\frac{x^3 \cos 3x}{3} + \int x^2 \cos 3x \, dx$
$= -\frac{x^3 \cos 3x}{3} + \left( x^2 \frac{\sin 3x}{3} - \int 2x \frac{\sin 3x}{3} \, dx \right)$
$= -\frac{x^3 \cos 3x}{3} + \frac{x^2 \sin 3x}{3} - \frac{2}{3} \int x \sin 3x \, dx$
Now,$\int x \sin 3x \, dx = x \left( \frac{-\cos 3x}{3} \right) - \int 1 \left( \frac{-\cos 3x}{3} \right) dx = -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9}$
Substituting back:
$\int x^3 \sin 3x \, dx = -\frac{x^3 \cos 3x}{3} + \frac{x^2 \sin 3x}{3} - \frac{2}{3} \left( -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} \right)$
$= -\frac{x^3 \cos 3x}{3} + \frac{x^2 \sin 3x}{3} + \frac{2x \cos 3x}{9} - \frac{2 \sin 3x}{27}$
$= \cos 3x \left( -\frac{x^3}{3} + \frac{2x}{9} \right) + \sin 3x \left( \frac{x^2}{3} - \frac{2}{27} \right) + c$
Comparing with $f(x) \cos 3x + g(x) \sin 3x + c$,we get $f(x) = -\frac{x^3}{3} + \frac{2x}{9}$ and $g(x) = \frac{x^2}{3} - \frac{2}{27}$
Then $27(f(x) + x g(x)) = 27 \left( -\frac{x^3}{3} + \frac{2x}{9} + x \left( \frac{x^2}{3} - \frac{2}{27} \right) \right)$
$= 27 \left( -\frac{x^3}{3} + \frac{2x}{9} + \frac{x^3}{3} - \frac{2x}{27} \right) = 27 \left( \frac{6x - 2x}{27} \right) = 4x$

(D) We are given $I_m = \int x^m \cos n x \, dx$. Using integration by parts,let $u = x^m$ and $dv = \cos n x \, dx$. Then $du = m x^{m-1} \, dx$ and $v = \frac{\sin n x}{n}$.
$I_m = \frac{x^m \sin n x}{n} - \int \frac{m x^{m-1} \sin n x}{n} \, dx = \frac{x^m \sin n x}{n} - \frac{m}{n} \int x^{m-1} \sin n x \, dx$.
Now,integrate $\int x^{m-1} \sin n x \, dx$ by parts again. Let $u = x^{m-1}$ and $dv = \sin n x \, dx$. Then $du = (m-1) x^{m-2} \, dx$ and $v = -\frac{\cos n x}{n}$.
$\int x^{m-1} \sin n x \, dx = -\frac{x^{m-1} \cos n x}{n} + \int \frac{(m-1) x^{m-2} \cos n x}{n} \, dx$.
Substituting this back into the expression for $I_m$:
$I_m = \frac{x^m \sin n x}{n} - \frac{m}{n} \left( -\frac{x^{m-1} \cos n x}{n} + \frac{m-1}{n} \int x^{m-2} \cos n x \, dx \right)$.
$I_m = \frac{x^m \sin n x}{n} + \frac{m x^{m-1} \cos n x}{n^2} - \frac{m(m-1)}{n^2} I_{m-2}$.
Comparing this with the given equation $I_m = g(x) - \frac{m(m-1)}{n^2} I_{m-2}$,we get $g(x) = \frac{x^m \sin n x}{n} + \frac{m x^{m-1} \cos n x}{n^2}$.

(B) Given $\int \phi(x) dx = \psi(x)$. We need to evaluate $I = \int \phi(h(x)) \cdot h(x) h'(x) dx$.
Let $h(x) = t$,then $h'(x) dx = dt$.
The integral becomes $I = \int \phi(t) \cdot t dt$.
Using integration by parts,let $u = t$ and $dv = \phi(t) dt$. Then $du = dt$ and $v = \int \phi(t) dt = \psi(t)$.
Applying the formula $\int u dv = uv - \int v du$:
$I = t \psi(t) - \int \psi(t) dt$.
Substituting $t = h(x)$ back into the expression:
$I = h(x) \psi(h(x)) - \int \psi(h(x)) h'(x) dx + c$.
This can be written as $(\psi \circ h)(x) h(x) - \int (\psi \circ h)(x) h'(x) dx + c$.

(A) Using integration by parts $\int u v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = (\log x)^3$ and $v = x^4$:
$\int x^4(\log x)^3 dx = \frac{x^5}{5}(\log x)^3 - \int 3(\log x)^2 \cdot \frac{1}{x} \cdot \frac{x^5}{5} dx = \frac{x^5}{5}(\log x)^3 - \frac{3}{5} \int x^4(\log x)^2 dx$
Applying parts again:
$= \frac{x^5}{5}(\log x)^3 - \frac{3}{5} [\frac{x^5}{5}(\log x)^2 - \int 2(\log x) \cdot \frac{1}{x} \cdot \frac{x^5}{5} dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{25} \int x^4 \log x dx$
Applying parts again:
$= \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{25} [\frac{x^5}{5} \log x - \int \frac{1}{x} \cdot \frac{x^5}{5} dx] = \frac{x^5}{5}(\log x)^3 - \frac{3}{25}x^5(\log x)^2 + \frac{6}{125}x^5 \log x - \frac{6}{125} \int x^4 dx$
$= x^5 [\frac{1}{5}(\log x)^3 - \frac{3}{25}(\log x)^2 + \frac{6}{125} \log x - \frac{6}{625}] + k$
Comparing coefficients: $A = \frac{1}{5}, B = -\frac{3}{25}, C = \frac{6}{125}, D = -\frac{6}{625}$
$A + B + C + 5D = \frac{1}{5} - \frac{3}{25} + \frac{6}{125} + 5(-\frac{6}{625}) = \frac{25 - 15 + 6 - 6}{125} = \frac{10}{125} = \frac{2}{25}$

(C) Let $I_n = \int_0^1 e^x(x-1)^n dx$. Using integration by parts,we have $\int u dv = uv - \int v du$. Let $u = (x-1)^n$ and $dv = e^x dx$. Then $du = n(x-1)^{n-1} dx$ and $v = e^x$.
$I_n = [e^x(x-1)^n]_0^1 - n \int_0^1 e^x(x-1)^{n-1} dx = (0 - (-1)^n) - n I_{n-1} = (-1)^{n+1} - n I_{n-1}$.
For $n=1$: $I_1 = (-1)^2 - 1 I_0 = 1 - \int_0^1 e^x dx = 1 - (e-1) = 2-e$.
For $n=2$: $I_2 = (-1)^3 - 2 I_1 = -1 - 2(2-e) = -1 - 4 + 2e = 2e-5$.
For $n=3$: $I_3 = (-1)^4 - 3 I_2 = 1 - 3(2e-5) = 1 - 6e + 15 = 16-6e$.
Comparing this with the given value $16-6e$,we find $n=3$.

(A) Let $I = \int \cos(\ln x) \, dx$.
Using integration by parts,let $u = \cos(\ln x)$ and $dv = dx$. Then $du = -\sin(\ln x) \cdot \frac{1}{x} \, dx$ and $v = x$.
$I = x \cos(\ln x) - \int x \cdot (-\sin(\ln x)) \cdot \frac{1}{x} \, dx$
$I = x \cos(\ln x) + \int \sin(\ln x) \, dx$.
Now,integrate $\int \sin(\ln x) \, dx$ by parts again,with $u = \sin(\ln x)$ and $dv = dx$. Then $du = \cos(\ln x) \cdot \frac{1}{x} \, dx$ and $v = x$.
$\int \sin(\ln x) \, dx = x \sin(\ln x) - \int x \cdot \cos(\ln x) \cdot \frac{1}{x} \, dx = x \sin(\ln x) - I$.
Substituting this back into the equation for $I$:
$I = x \cos(\ln x) + x \sin(\ln x) - I$
$2I = x \{\cos(\ln x) + \sin(\ln x)\}$
$I = \frac{x}{2} \{\cos(\ln x) + \sin(\ln x)\} + c$.

(B) Let $I = \int \cos x \log \left(\tan \frac{x}{2}\right) dx$.
Using integration by parts,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = \log \left(\tan \frac{x}{2}\right)$ and $v = \cos x$.
Then $u' = \frac{1}{\tan(x/2)} \cdot \sec^2(x/2) \cdot \frac{1}{2} = \frac{1}{2 \sin(x/2) \cos(x/2)} = \frac{1}{\sin x}$ and $\int v dx = \sin x$.
$I = \log \left(\tan \frac{x}{2}\right) \cdot \sin x - \int \left( \frac{1}{\sin x} \cdot \sin x \right) dx$.
$I = \sin x \log \left(\tan \frac{x}{2}\right) - \int 1 dx$.
$I = \sin x \log \left(\tan \frac{x}{2}\right) - x + c$.
Comparing this with the given expression $\sin x \log \left(\tan \frac{x}{2}\right) + f(x)$,we get $f(x) = c - x$.

(C) We are given the integral $I = \int \frac{x^2 \, dx}{(x \sin x + \cos x)^2}$.
We can rewrite the integrand as $I = \int \frac{x}{(x \sin x + \cos x)^2} \cdot \frac{x}{\cos x} \, dx$.
Using integration by parts,let $u = \frac{x}{\cos x}$ and $dv = \frac{x \, dx}{(x \sin x + \cos x)^2}$.
Then $du = \frac{\cos x - x(-\sin x)}{\cos^2 x} \, dx = \frac{\cos x + x \sin x}{\cos^2 x} \, dx$.
To find $v$,let $t = x \sin x + \cos x$,then $dt = (\sin x + x \cos x - \sin x) \, dx = x \cos x \, dx$. This does not simplify directly,so we use the derivative of $\frac{-1}{x \sin x + \cos x}$ which is $\frac{-(x \cos x)}{(x \sin x + \cos x)^2}$.
Thus,$v = \frac{-1}{x \sin x + \cos x} \cdot \frac{1}{\cos x}$ is not quite right. Let's use $I = \int \frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x + \cos x)^2} \, dx$.
Let $u = \frac{x}{\cos x}$ and $dv = \frac{x \cos x}{(x \sin x + \cos x)^2} \, dx$. Then $v = \frac{-1}{x \sin x + \cos x}$.
$I = \frac{-x}{\cos x(x \sin x + \cos x)} - \int \left( \frac{-1}{x \sin x + \cos x} \right) \left( \frac{\cos x + x \sin x}{\cos^2 x} \right) \, dx$.
$I = \frac{-x}{\cos x(x \sin x + \cos x)} + \int \sec^2 x \, dx$.
$I = \frac{-x}{\cos x(x \sin x + \cos x)} + \tan x + c$.
Comparing this with $f(x) + \tan x + c$,we get $f(x) = \frac{-x}{\cos x(x \sin x + \cos x)}$.

(B) Using 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 $v = x$.
The integral becomes $x \tan^{-1} x - \int \frac{x}{1+x^2} dx$.
To solve $\int \frac{x}{1+x^2} dx$,let $t = 1+x^2$,so $dt = 2x \, dx$,which means $x \, dx = \frac{1}{2} dt$.
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,we get $x \tan^{-1} x - \frac{1}{2} \log(1+x^2) + C$.
Comparing this with $Ax \tan^{-1} x + B \log(1+x^2) + C$,we get $A = 1$ and $B = -1/2$.
Therefore,$A + B = 1 - 1/2 = 1/2$.