Fundamental integration Questions in English

(A) We are given the integral $I = \int \frac{\sin \alpha}{\sqrt{1 + \cos \alpha}} d \alpha$.
Using the trigonometric identity $1 + \cos \alpha = 2 \cos^2 (\frac{\alpha}{2})$,we substitute this into the integral:
$I = \int \frac{2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})}{\sqrt{2 \cos^2 (\frac{\alpha}{2})}} d \alpha$
Assuming $\cos (\frac{\alpha}{2}) > 0$,the denominator becomes $\sqrt{2} \cos (\frac{\alpha}{2})$:
$I = \int \frac{2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})}{\sqrt{2} \cos (\frac{\alpha}{2})} d \alpha$
$I = \sqrt{2} \int \sin (\frac{\alpha}{2}) d \alpha$
Integrating $\sin (\frac{\alpha}{2})$ with respect to $\alpha$:
$I = \sqrt{2} \times (-2 \cos (\frac{\alpha}{2})) + c$
$I = -2 \sqrt{2} \cos (\frac{\alpha}{2}) + c$.

(B) We need to evaluate the integral: $\int \left( \frac{x}{a} + \frac{b}{x} + x^a + b^x + ab \right) dx$
Applying the linearity property of integration:
$= \int \frac{x}{a} dx + \int \frac{b}{x} dx + \int x^a dx + \int b^x dx + \int ab dx$
$= \frac{1}{a} \int x dx + b \int \frac{1}{x} dx + \int x^a dx + \int b^x dx + ab \int 1 dx$
Using standard integration formulas:
$= \frac{1}{a} \cdot \frac{x^2}{2} + b \log |x| + \frac{x^{a+1}}{a+1} + \frac{b^x}{\log b} + abx + C$
$= \frac{x^2}{2a} + b \log |x| + \frac{x^{a+1}}{a+1} + \frac{b^x}{\log b} + abx + C$

(B) Given $f^{\prime}(x)=\tan ^2 x+\cot ^2 x$.
Integrating both sides with respect to $x$:
$f(x)=\int(\tan ^2 x+\cot ^2 x) dx$.
Using the identity $\tan^2 x = \sec^2 x - 1$ and $\cot^2 x = \operatorname{cosec}^2 x - 1$:
$f(x)=\int(\sec^2 x - 1 + \operatorname{cosec}^2 x - 1) dx$.
$f(x)=\int(\sec^2 x + \operatorname{cosec}^2 x - 2) dx$.
$f(x)=\tan x - \cot x - 2x + C$.
Given $f\left(\frac{\pi}{4}\right)=0$:
$f\left(\frac{\pi}{4}\right)=\tan\left(\frac{\pi}{4}\right) - \cot\left(\frac{\pi}{4}\right) - 2\left(\frac{\pi}{4}\right) + C = 0$.
$1 - 1 - \frac{\pi}{2} + C = 0$.
$C = \frac{\pi}{2}$.
Therefore,$f(x)=\tan x - \cot x - 2x + \frac{\pi}{2}$.

(C) Given integral is $I = \int \frac{x^3-x^2+x-1}{x-1} dx$.
Factorize the numerator: $x^3-x^2+x-1 = x^2(x-1) + 1(x-1) = (x^2+1)(x-1)$.
Substitute this into the integral: $I = \int \frac{(x^2+1)(x-1)}{x-1} dx$.
Cancel the common term $(x-1)$: $I = \int (x^2+1) dx$.
Integrate term by term: $I = \frac{x^3}{3} + x + C$.

(B) First,simplify $f(x)$:\\ $f(x) = \tan^{-1}\left(\frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)}\right) = \tan^{-1}(\cot(x/2)) = \tan^{-1}(\tan(\pi/2 - x/2)) = \frac{\pi}{2} - \frac{x}{2}$.\\ Next,simplify $g(x)$:\\ $g(x) = \tan^{-1}\left(\frac{2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)}\right) = \tan^{-1}(\cot(x/2)) = \tan^{-1}(\tan(\pi/2 - x/2)) = \frac{\pi}{2} - \frac{x}{2}$.\\ Now,calculate the sum:\\ $f(x) + g(x) = (\frac{\pi}{2} - \frac{x}{2}) + (\frac{\pi}{2} - \frac{x}{2}) = \pi - x$.\\ Finally,integrate the sum:\\ $\int (\pi - x) \, dx = \pi x - \frac{x^2}{2} + C$.\\ Thus,option $B$ is correct.

(A) We are given the integral: $I = \int \frac{\sin^3(x) + \cos^3(x)}{\sin^2(x) \cos^2(x)} \, dx$
Divide each term in the numerator by the denominator:
$I = \int \left( \frac{\sin^3(x)}{\sin^2(x) \cos^2(x)} + \frac{\cos^3(x)}{\sin^2(x) \cos^2(x)} \right) \, dx$
$I = \int \left( \frac{\sin(x)}{\cos^2(x)} + \frac{\cos(x)}{\sin^2(x)} \right) \, dx$
Using trigonometric identities $\frac{\sin(x)}{\cos(x)} = \tan(x)$ and $\frac{1}{\cos(x)} = \sec(x)$,and $\frac{\cos(x)}{\sin(x)} = \cot(x)$ and $\frac{1}{\sin(x)} = \operatorname{cosec}(x)$:
$I = \int (\tan(x) \sec(x) + \cot(x) \operatorname{cosec}(x)) \, dx$
Integrating term by term:
$\int \sec(x) \tan(x) \, dx = \sec(x)$
$\int \operatorname{cosec}(x) \cot(x) \, dx = -\operatorname{cosec}(x)$
Thus,$I = \sec(x) - \operatorname{cosec}(x) + C$.
Hence,option $A$ is correct.

(A) We are given the integral $I = \int(1-\cos x) \operatorname{cosec}^2 x \, dx$.
Using the trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we have:
$I = \int (2 \sin^2 \frac{x}{2}) \cdot \frac{1}{\sin^2 x} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{(2 \sin \frac{x}{2} \cos \frac{x}{2})^2} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{4 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2}} \, dx$
$I = \int \frac{1}{2 \cos^2 \frac{x}{2}} \, dx$
$I = \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$
Integrating $\sec^2 \frac{x}{2}$,we get:
$I = \frac{1}{2} \cdot \frac{\tan \frac{x}{2}}{1/2} + c = \tan \frac{x}{2} + c$.
Comparing this with $f(x) + c$,we find $f(x) = \tan \frac{x}{2}$.

(B) We are given the integral $I = \int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$.
First,we rewrite the numerator as $(1+x) + \sqrt{x(1+x)}$.
This can be factored as $\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})$.
Substituting this back into the integral,we get:
$I = \int \frac{\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})}{\sqrt{x} + \sqrt{1+x}} d x$.
The term $(\sqrt{1+x} + \sqrt{x})$ cancels out from the numerator and denominator.
$I = \int \sqrt{1+x} d x$.
Using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C$,where $u = 1+x$ and $du = dx$:
$I = \frac{(1+x)^{3/2}}{3/2} + C = \frac{2}{3}(1+x)^{3/2} + C$.

(B) Let $I = \int \frac{1+x+\sqrt{x(1+x)}}{\sqrt{x}+\sqrt{1+x}} dx$.
Notice that $1+x = (\sqrt{1+x})^2$ and $x = (\sqrt{x})^2$.
So,the numerator can be written as $(\sqrt{1+x})^2 + (\sqrt{x})^2 + \sqrt{x}\sqrt{1+x}$.
This does not simplify directly,so let us rewrite the integrand:
$I = \int \frac{(\sqrt{1+x})^2 + \sqrt{x}\sqrt{1+x}}{\sqrt{x}+\sqrt{1+x}} dx$.
Factor out $\sqrt{1+x}$ from the numerator:
$I = \int \frac{\sqrt{1+x}(\sqrt{1+x} + \sqrt{x})}{\sqrt{x}+\sqrt{1+x}} dx$.
Canceling the common term $(\sqrt{x}+\sqrt{1+x})$,we get:
$I = \int \sqrt{1+x} dx$.
Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + c$:
$I = \frac{(1+x)^{3/2}}{3/2} + c = \frac{2}{3}(1+x)^{3/2} + c$.

(A) The given integral is $I = \int \frac{1}{9 \cos^2 x - 24 \sin x \cos x + 16 \sin^2 x} dx$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x}{9 - 24 \tan x + 16 \tan^2 x} dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
The integral becomes $I = \int \frac{du}{(3 - 4u)^2} = \int (3 - 4u)^{-2} du$.
Using the power rule for integration:
$I = \frac{(3 - 4u)^{-1}}{(-1) \times (-4)} + c = \frac{1}{4(3 - 4u)} + c$.
Substitute $u = \tan x$ back:
$I = \frac{1}{4(3 - 4 \tan x)} + c = \frac{1}{4(3 - 4 \frac{\sin x}{\cos x})} + c = \frac{\cos x}{4(3 \cos x - 4 \sin x)} + c$.
Thus,the correct option is $A$.

(D) We know that the Taylor series expansion for the exponential function is given by $e^u = \sum_{r=0}^{\infty} \frac{u^r}{r!}$.
Substituting $u = 2x$ into the series,we get $\sum_{r=0}^{\infty} \frac{(2x)^r}{r!} = \sum_{r=0}^{\infty} \frac{x^r 2^r}{r!} = e^{2x}$.
Now,we need to evaluate the integral $\int e^{2x} dx$.
Using the integration formula $\int e^{ax} dx = \frac{e^{ax}}{a} + c$,where $a = 2$,we get $\int e^{2x} dx = \frac{e^{2x}}{2} + c$.
Thus,the correct option is $D$.

(A) To solve the integral $I = \int \frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x$,we perform polynomial long division.
Dividing $x^4-16 x^2+2 x+8$ by $x^3-4 x^2+2$:
$x^4-16 x^2+2 x+8 = (x+4)(x^3-4 x^2+2) + (0x^2 + 0x + 0)$.
Since the remainder is $0$,the expression simplifies to:
$I = \int (x+4) d x$.
Integrating term by term:
$I = \frac{x^2}{2} + 4x + C = \frac{x^2+8x}{2} + C$.

(B) To solve the integral $\int \frac{x^4+1}{x^2+1} dx$,we perform polynomial division or algebraic manipulation.
We can write $x^4+1$ as $(x^4-1) + 2 = (x^2-1)(x^2+1) + 2$.
Thus,$\frac{x^4+1}{x^2+1} = \frac{(x^2-1)(x^2+1) + 2}{x^2+1} = x^2 - 1 + \frac{2}{x^2+1}$.
Now,integrate each term with respect to $x$:
$\int (x^2 - 1 + \frac{2}{x^2+1}) dx = \int x^2 dx - \int 1 dx + 2 \int \frac{1}{x^2+1} dx$.
$= \frac{x^3}{3} - x + 2 \tan^{-1} x + E$.
Comparing this with $Ax^3 + Bx^2 + Cx + D \tan^{-1} x + E$,we get:
$A = \frac{1}{3}$,$B = 0$,$C = -1$,$D = 2$.
Therefore,$A+B+C+D = \frac{1}{3} + 0 - 1 + 2 = \frac{1}{3} + 1 = \frac{4}{3}$.

(B) Let $I = \int \sec \left(x-\frac{\pi}{3}\right) \sec \left(x+\frac{\pi}{6}\right) d x$.
Multiply and divide by $\sin \left(\left(x+\frac{\pi}{6}\right) - \left(x-\frac{\pi}{3}\right)\right) = \sin \left(\frac{\pi}{6} + \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{2}\right) = 1$.
So,$I = \int \frac{\sin \left(\left(x+\frac{\pi}{6}\right) - \left(x-\frac{\pi}{3}\right)\right)}{\cos \left(x-\frac{\pi}{3}\right) \cos \left(x+\frac{\pi}{6}\right)} d x$.
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we get:
$I = \int \frac{\sin \left(x+\frac{\pi}{6}\right) \cos \left(x-\frac{\pi}{3}\right) - \cos \left(x+\frac{\pi}{6}\right) \sin \left(x-\frac{\pi}{3}\right)}{\cos \left(x-\frac{\pi}{3}\right) \cos \left(x+\frac{\pi}{6}\right)} d x$.
$I = \int \left( \frac{\sin \left(x+\frac{\pi}{6}\right)}{\cos \left(x+\frac{\pi}{6}\right)} - \frac{\sin \left(x-\frac{\pi}{3}\right)}{\cos \left(x-\frac{\pi}{3}\right)} \right) d x$.
$I = \int \tan \left(x+\frac{\pi}{6}\right) d x - \int \tan \left(x-\frac{\pi}{3}\right) d x$.
$I = \log \left| \sec \left(x+\frac{\pi}{6}\right) \right| - \log \left| \sec \left(x-\frac{\pi}{3}\right) \right| + c$.
$I = \log \left| \frac{\sec \left(x+\frac{\pi}{6}\right)}{\sec \left(x-\frac{\pi}{3}\right)} \right| + c$.
Note: The options provided in the question are slightly different in form. However,using $\sec \theta = 1/\cos \theta$,we can write this as $\log \left| \frac{\cos \left(x-\frac{\pi}{3}\right)}{\cos \left(x+\frac{\pi}{6}\right)} \right| + c$,which matches option $B$.

(C) To evaluate the integral $I = \int \frac{1}{x^2+x+1} \, dx$,we complete the square in the denominator:
$x^2+x+1 = (x^2+x+\frac{1}{4}) + \frac{3}{4} = (x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2$.
Substituting this into the integral:
$I = \int \frac{1}{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \, dx$.
Using the standard integral formula $\int \frac{1}{u^2+a^2} \, du = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + c$,where $u = x+\frac{1}{2}$ and $a = \frac{\sqrt{3}}{2}$:
$I = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + c$.
Simplifying the expression:
$I = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + c$.
Thus,the correct option is $C$.

(A) Let $I = \int \sin^3 x \cos^2 x \, dx$.
We can write $\sin^3 x$ as $\sin^2 x \cdot \sin x = (1 - \cos^2 x) \sin x$.
So,$I = \int (1 - \cos^2 x) \cos^2 x \sin x \, dx$.
Let $u = \cos x$,then $du = -\sin x \, dx$,or $\sin x \, dx = -du$.
Substituting these into the integral:
$I = \int (1 - u^2) u^2 (-du) = \int (u^4 - u^2) \, du$.
Integrating with respect to $u$:
$I = \frac{u^5}{5} - \frac{u^3}{3} + c$.
Substituting $u = \cos x$ back:
$I = \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + c$.
Since the provided options are in terms of $\sin x$ and $\cos x$,we check the derivative of option $A$:
$\frac{d}{dx} [\frac{\sin^4 x \cos x}{5} - \frac{\sin^2 x \cos x}{15} - \frac{2 \cos x}{15}] = \sin^3 x \cos^2 x$.
Thus,option $A$ is the correct integral.

(C) Let $I = \int \frac{x}{x \tan x + 1} \, dx = \int \frac{x \cos x}{x \sin x + \cos x} \, dx$.
Let $u = x \sin x + \cos x$.
Then $du = (\sin x + x \cos x - \sin x) \, dx = x \cos x \, dx$.
Thus,$I = \int \frac{1}{u} \, du = \log |u| + k = \log |x \sin x + \cos x| + k$.
Comparing this with $\log f(x) + k$,we get $f(x) = |x \sin x + \cos x|$.
Now,evaluate $f\left(\frac{\pi}{4}\right) = |\frac{\pi}{4} \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)|$.
$f\left(\frac{\pi}{4}\right) = |\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}| = |\frac{\pi + 4}{4 \sqrt{2}}| = \frac{\pi + 4}{4 \sqrt{2}}$.

(A) Let $I = \int (\tan^7 x + \tan x) dx$.
We can factor out $\tan x$:
$I = \int \tan x (\tan^6 x + 1) dx$.
Using the identity $\tan^2 x = \sec^2 x - 1$,we have $\tan^6 x = (\sec^2 x - 1)^3$.
Alternatively,simplify the expression as follows:
$I = \int (\tan^7 x + \tan^5 x - \tan^5 x - \tan^3 x + \tan^3 x + \tan x) dx$
$I = \int (\tan^5 x(\tan^2 x + 1) - \tan^3 x(\tan^2 x + 1) + \tan x(\tan^2 x + 1)) dx$
Since $1 + \tan^2 x = \sec^2 x$:
$I = \int (\tan^5 x \sec^2 x - \tan^3 x \sec^2 x + \tan x \sec^2 x) dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
$I = \int (u^5 - u^3 + u) du = \frac{u^6}{6} - \frac{u^4}{4} + \frac{u^2}{2} + C$.
Factor out $\frac{u^2}{12}$:
$I = \frac{u^2}{12} (2u^4 - 3u^2 + 6) + C$.
Substituting $u = \tan x$ back:
$I = \frac{\tan^2 x}{12} (2 \tan^4 x - 3 \tan^2 x + 6) + C$.

(D) We are given the integral $I = \int \frac{2 x^2 \cos \left(x^2\right)-\sin \left(x^2\right)}{x^2} d x$.
We can rewrite the integrand by splitting the fraction:
$I = \int \left( 2 \cos(x^2) - \frac{\sin(x^2)}{x^2} \right) dx$.
Alternatively,observe the derivative of the quotient $\frac{\sin(x^2)}{x}$:
$\frac{d}{dx} \left( \frac{\sin(x^2)}{x} \right) = \frac{x \cdot \frac{d}{dx}(\sin(x^2)) - \sin(x^2) \cdot \frac{d}{dx}(x)}{x^2}$
$= \frac{x \cdot (\cos(x^2) \cdot 2x) - \sin(x^2) \cdot 1}{x^2}$
$= \frac{2x^2 \cos(x^2) - \sin(x^2)}{x^2}$.
Therefore,$\int \frac{2 x^2 \cos \left(x^2\right)-\sin \left(x^2\right)}{x^2} d x = \frac{\sin \left(x^2\right)}{x} + c$.

(B) Let $I = \int \frac{\cos x + x \sin x}{x(x + \cos x)} dx$.
We can rewrite the numerator as: $\cos x + x \sin x = (x + \cos x) + (x \sin x - x) = (x + \cos x) - x(1 - \sin x)$.
Then,$I = \int \frac{(x + \cos x) - x(1 - \sin x)}{x(x + \cos x)} dx$.
Splitting the integral: $I = \int \frac{x + \cos x}{x(x + \cos x)} dx - \int \frac{x(1 - \sin x)}{x(x + \cos x)} dx$.
$I = \int \frac{1}{x} dx - \int \frac{1 - \sin x}{x + \cos x} dx$.
Let $u = x + \cos x$,then $du = (1 - \sin x) dx$.
Substituting these into the integral: $I = \ln |x| - \int \frac{1}{u} du$.
$I = \ln |x| - \ln |u| + C$.
$I = \ln |x| - \ln |x + \cos x| + C$.
$I = \ln \left| \frac{x}{x + \cos x} \right| + C$.

(B) $I = \int \frac{2}{1+x+x^2} dx = \int \frac{2}{(x+\frac{1}{2})^2 + \frac{3}{4}} dx$
Let $x + \frac{1}{2} = v$,then $dx = dv$.
Using the formula $\int \frac{1}{v^2 + a^2} dv = \frac{1}{a} \tan^{-1}(\frac{v}{a}) + c$,where $a = \frac{\sqrt{3}}{2}$:
$I = 2 \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}(\frac{v}{\frac{\sqrt{3}}{2}}) + c$
$I = \frac{4}{\sqrt{3}} \tan^{-1}(\frac{2v}{\sqrt{3}}) + c$
Substituting $v = x + \frac{1}{2}$:
$I = \frac{4}{\sqrt{3}} \tan^{-1}(\frac{2(x + \frac{1}{2})}{\sqrt{3}}) + c = \frac{4}{\sqrt{3}} \tan^{-1}(\frac{2x+1}{\sqrt{3}}) + c$

(B) Let $I = \int \frac{\sin ^{-1} \sqrt{x} - \cos ^{-1} \sqrt{x}}{\sqrt{x}(\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x})} dx$.
Since $\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x} = \frac{\pi}{2}$,we have $\cos ^{-1} \sqrt{x} = \frac{\pi}{2} - \sin ^{-1} \sqrt{x}$.
Substituting this,$I = \int \frac{\sin ^{-1} \sqrt{x} - (\frac{\pi}{2} - \sin ^{-1} \sqrt{x})}{\sqrt{x}(\frac{\pi}{2})} dx = \frac{2}{\pi} \int \frac{2 \sin ^{-1} \sqrt{x} - \frac{\pi}{2}}{\sqrt{x}} dx$.
Let $\sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,so $\frac{dx}{\sqrt{x}} = 2 dt$.
$I = \frac{2}{\pi} \int (2 \sin ^{-1} t - \frac{\pi}{2}) (2 dt) = \frac{8}{\pi} \int \sin ^{-1} t dt - 2 \int dt$.
Using $\int \sin ^{-1} t dt = t \sin ^{-1} t + \sqrt{1-t^2} + C$,we get:
$I = \frac{8}{\pi} (t \sin ^{-1} t + \sqrt{1-t^2}) - 2t + C$.
Substituting $t = \sqrt{x}$:
$I = \frac{8}{\pi} (\sqrt{x} \sin ^{-1} \sqrt{x} + \sqrt{1-x}) - 2\sqrt{x} + C$.

(C) We are given the integral $I = \int \frac{\sin ^2 \alpha-\sin ^2 x}{\cos x-\cos \alpha} d x$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we can rewrite the numerator:
$\sin^2 \alpha - \sin^2 x = (1 - \cos^2 \alpha) - (1 - \cos^2 x) = \cos^2 x - \cos^2 \alpha$.
Substituting this into the integral,we get:
$I = \int \frac{\cos^2 x - \cos^2 \alpha}{\cos x - \cos \alpha} d x$.
Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$:
$I = \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} d x$.
Canceling the common term $(\cos x - \cos \alpha)$:
$I = \int (\cos x + \cos \alpha) d x$.
Integrating with respect to $x$:
$I = \sin x + x \cos \alpha + C$.
Comparing this with $f(x) + Ax + B$,we identify $f(x) = \sin x$ and $A = \cos \alpha$.

(A) We have,$\int \frac{1}{x^2}(2 x+1)^3 d x$
Expanding the numerator using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:
$(2x+1)^3 = (2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + (1)^3 = 8x^3 + 12x^2 + 6x + 1$
Now,the integral becomes:
$\int \frac{8x^3 + 12x^2 + 6x + 1}{x^2} d x$
$= \int (8x + 12 + \frac{6}{x} + \frac{1}{x^2}) d x$
Integrating term by term:
$= 8 \int x d x + 12 \int d x + 6 \int \frac{1}{x} d x + \int x^{-2} d x$
$= 8(\frac{x^2}{2}) + 12x + 6 \log |x| + (\frac{x^{-1}}{-1}) + C$
$= 4x^2 + 12x + 6 \log |x| - \frac{1}{x} + C$

(D) Given that $\int \cos ^{-1} \sqrt{\frac{x}{a+x}} d x = f(x) + C$.
By the Fundamental Theorem of Calculus,the derivative of the integral is the integrand:
$f^{\prime}(x) = \frac{d}{dx} \int \cos ^{-1} \sqrt{\frac{x}{a+x}} d x = \cos ^{-1} \sqrt{\frac{x}{a+x}}$.
Now,we need to find $f^{\prime}(a)$.
Substitute $x = a$ into the expression for $f^{\prime}(x)$:
$f^{\prime}(a) = \cos ^{-1} \sqrt{\frac{a}{a+a}} = \cos ^{-1} \sqrt{\frac{a}{2a}}$.
Simplifying the fraction inside the square root:
$f^{\prime}(a) = \cos ^{-1} \sqrt{\frac{1}{2}} = \cos ^{-1} \left(\frac{1}{\sqrt{2}}\right)$.
Since $\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$,we have:
$f^{\prime}(a) = \frac{\pi}{4}$.

(B) Given the integral $I = \int \frac{(x - 1)^2}{(x^2 + 1)^2} dx$.
Expanding the numerator,we get $(x - 1)^2 = x^2 - 2x + 1$.
So,$I = \int \frac{x^2 + 1 - 2x}{(x^2 + 1)^2} dx$.
Splitting the integral,we have $I = \int \frac{x^2 + 1}{(x^2 + 1)^2} dx - \int \frac{2x}{(x^2 + 1)^2} dx$.
This simplifies to $I = \int \frac{1}{x^2 + 1} dx - \int 2x(x^2 + 1)^{-2} dx$.
We know that $\int \frac{1}{x^2 + 1} dx = \tan^{-1}(x)$.
For the second part,let $u = x^2 + 1$,then $du = 2x dx$.
Thus,$\int 2x(x^2 + 1)^{-2} dx = \int u^{-2} du = \frac{u^{-1}}{-1} = -\frac{1}{u} = -\frac{1}{x^2 + 1}$.
Substituting these back,$I = \tan^{-1}(x) - (-\frac{1}{x^2 + 1}) + k = \tan^{-1}(x) + \frac{1}{x^2 + 1} + k$.
Comparing this with the given form $\tan^{-1}(x) + g(x) + k$,we find $g(x) = \frac{1}{x^2 + 1}$.

(B) Given integral is $I = \int \frac{1 + x + \sqrt{x(1 + x)}}{\sqrt{x} + \sqrt{1 + x}} dx$.
We can rewrite the numerator as:
$1 + x + \sqrt{x} \cdot \sqrt{1 + x} = \sqrt{1 + x} \cdot \sqrt{1 + x} + \sqrt{x} \cdot \sqrt{1 + x}$.
Factoring out $\sqrt{1 + x}$,we get:
$\sqrt{1 + x} (\sqrt{1 + x} + \sqrt{x})$.
Substituting this back into the integral:
$I = \int \frac{\sqrt{1 + x} (\sqrt{1 + x} + \sqrt{x})}{\sqrt{x} + \sqrt{1 + x}} dx$.
Canceling the common term $(\sqrt{x} + \sqrt{1 + x})$:
$I = \int \sqrt{1 + x} dx = \int (1 + x)^{1/2} dx$.
Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + c$:
$I = \frac{(1 + x)^{3/2}}{3/2} + c = \frac{2}{3} (1 + x)^{3/2} + c$.

(A) Let $I = \int (1 - \cos x) \operatorname{cosec}^2 x \, dx$.
Using the identity $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$ and $\operatorname{cosec}^2 x = \frac{1}{\sin^2 x}$,we have:
$I = \int 2 \sin^2 \left(\frac{x}{2}\right) \cdot \frac{1}{\sin^2 x} \, dx$.
Since $\sin x = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$,then $\sin^2 x = 4 \sin^2 \left(\frac{x}{2}\right) \cos^2 \left(\frac{x}{2}\right)$.
Substituting this into the integral:
$I = \int \frac{2 \sin^2 \left(\frac{x}{2}\right)}{4 \sin^2 \left(\frac{x}{2}\right) \cos^2 \left(\frac{x}{2}\right)} \, dx$.
$I = \int \frac{1}{2 \cos^2 \left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \sec^2 \left(\frac{x}{2}\right) \, dx$.
Integrating $\sec^2 \left(\frac{x}{2}\right)$,we get $2 \tan \left(\frac{x}{2}\right)$.
$I = \frac{1}{2} \cdot 2 \tan \left(\frac{x}{2}\right) + C = \tan \left(\frac{x}{2}\right) + C$.

(C) Let $I = \int \frac{dx}{\sin x \cos x}$.
Multiply the numerator and denominator by $\sec^2 x$:
$I = \int \frac{\sec^2 x dx}{\tan x}$.
Substitute $\tan x = t$,then $\sec^2 x dx = dt$.
Therefore,$I = \int \frac{dt}{t} = \log |t| + c$.
Substituting back $t = \tan x$,we get $I = \log |\tan x| + c$.

(B) To solve the integral $I = \int \frac{x^2+1}{x^4+1} dx$,we divide the numerator and denominator by $x^2$:
$I = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx$
We can rewrite the denominator as $(x - \frac{1}{x})^2 + 2$:
$I = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + (\sqrt{2})^2} dx$
Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + c$
Substituting $t = x - \frac{1}{x} = \frac{x^2-1}{x}$ back:
$I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right) + c$
Thus,$f(x) = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)$.

(D) Let $I = \int \frac{1+\cos (4 x)}{\cot x-\tan x} d x$.
Using the identity $1+\cos(2\theta) = 2\cos^2(\theta)$,we have $1+\cos(4x) = 2\cos^2(2x)$.
Also,$\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos(2x)}{\frac{1}{2}\sin(2x)} = 2\frac{\cos(2x)}{\sin(2x)}$.
Substituting these into the integral:
$I = \int \frac{2\cos^2(2x)}{2\frac{\cos(2x)}{\sin(2x)}} dx = \int \cos(2x) \sin(2x) dx$.
Using the identity $\sin(2\theta) = 2\sin\theta \cos\theta$,we have $\sin(2x)\cos(2x) = \frac{1}{2}\sin(4x)$.
$I = \int \frac{1}{2}\sin(4x) dx = \frac{1}{2} \left( \frac{-\cos(4x)}{4} \right) + C = -\frac{1}{8}\cos(4x) + C$.
Comparing this with $A\cos(4x)+B$,we get $A = -\frac{1}{8}$.

(C) Let $I = \int \frac{dx}{\sqrt{7-6x-x^2}}$.
First,complete the square for the quadratic expression $7-6x-x^2$:
$7-6x-x^2 = 7 - (x^2 + 6x) = 7 - (x^2 + 6x + 9 - 9) = 7 - ((x+3)^2 - 9) = 16 - (x+3)^2 = 4^2 - (x+3)^2$.
Now,substitute this into the integral:
$I = \int \frac{dx}{\sqrt{4^2 - (x+3)^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$,where $a = 4$ and $x$ is replaced by $(x+3)$:
$I = \sin^{-1}\left(\frac{x+3}{4}\right) + C$.

(C) Let $I = \int (\sec^4 x + \tan^4 x) \, dx$.
Using the identity $\tan^2 x = \sec^2 x - 1$,we have $\tan^4 x = (\sec^2 x - 1)^2 = \sec^4 x - 2 \sec^2 x + 1$.
Substituting this into the integral:
$I = \int (\sec^4 x + \sec^4 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \sec^4 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \sec^2 x \cdot \sec^2 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2(1 + \tan^2 x) \sec^2 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \sec^2 x + 2 \tan^2 x \sec^2 x - 2 \sec^2 x + 1) \, dx$
$I = \int (2 \tan^2 x \sec^2 x + 1) \, dx$
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
$I = \int (2u^2 + 1) \, du = \frac{2}{3} u^3 + u + c$
$I = \frac{2}{3} \tan^3 x + \tan x + c$.

(D) Given $g\left(\frac{t+1}{2 t+1}\right)=t+1$.
Let $x = \frac{t+1}{2 t+1}$.
Then $x(2t+1) = t+1 \Rightarrow 2tx + x = t+1 \Rightarrow t(2x-1) = 1-x$.
So,$t = \frac{1-x}{2x-1}$.
Substituting $t$ into $g(x) = t+1$:
$g(x) = \frac{1-x}{2x-1} + 1 = \frac{1-x+2x-1}{2x-1} = \frac{x}{2x-1}$.
Now,$\int g(x) dx = \int \frac{x}{2x-1} dx$.
$= \frac{1}{2} \int \frac{2x}{2x-1} dx = \frac{1}{2} \int \frac{2x-1+1}{2x-1} dx$.
$= \frac{1}{2} \int \left(1 + \frac{1}{2x-1}\right) dx$.
$= \frac{1}{2} \left(x + \frac{1}{2} \log _e|2x-1|\right) + C$.
$= \frac{x}{2} + \frac{1}{4} \log _e|2x-1| + C$.

(B) Let $I = \int \frac{e^x-1}{e^x+1} dx$.
We can rewrite the integrand as: $I = \int \frac{(e^x+1)-2}{e^x+1} dx$.
This simplifies to: $I = \int (1 - \frac{2}{e^x+1}) dx = \int 1 dx - 2 \int \frac{1}{e^x+1} dx$.
To integrate $\int \frac{1}{e^x+1} dx$,multiply the numerator and denominator by $e^{-x}$:
$I = x - 2 \int \frac{e^{-x}}{1+e^{-x}} dx$.
Let $t = 1+e^{-x}$,then $dt = -e^{-x} dx$,which implies $e^{-x} dx = -dt$.
Substituting these into the integral:
$I = x - 2 \int \frac{-dt}{t} = x + 2 \int \frac{1}{t} dt$.
$I = x + 2 \log_e|t| + c = x + 2 \log_e(1+e^{-x}) + c$.
Since $1+e^{-x} = \frac{e^x+1}{e^x}$,we have:
$I = x + 2 \log_e(\frac{e^x+1}{e^x}) + c = x + 2 \log_e(e^x+1) - 2 \log_e(e^x) + c$.
$I = x + 2 \log_e(e^x+1) - 2x + c = 2 \log_e(e^x+1) - x + c$.

(A) Let $I = \int \frac{dx}{x^2+2x+2}$.
We can rewrite the denominator by completing the square:
$x^2+2x+2 = (x^2+2x+1) + 1 = (x+1)^2 + 1$.
Substituting this into the integral,we get:
$I = \int \frac{dx}{(x+1)^2 + 1}$.
Using the standard integral formula $\int \frac{du}{u^2+1} = \tan^{-1}(u) + c$,where $u = x+1$ and $du = dx$:
$I = \tan^{-1}(x+1) + c$.
Comparing this with $I = f(x) + c$,we find that $f(x) = \tan^{-1}(x+1)$.

(C) Let $I = \int(3t^2 \sin \frac{1}{t} - t \cos \frac{1}{t}) dt$.
We use the substitution $u = \frac{1}{t}$,so $du = -\frac{1}{t^2} dt$,which implies $dt = -t^2 du = -\frac{1}{u^2} du$.
The integral becomes:
$I = \int(3(\frac{1}{u})^2 \sin u - \frac{1}{u} \cos u) (- \frac{1}{u^2}) du$
$I = \int(-\frac{3}{u^4} \sin u + \frac{1}{u^3} \cos u) du$.
This does not simplify easily,so let's try to express the integrand as a derivative.
Consider $f(t) = t^3$. Then $\frac{d}{dt} (t^3 \sin \frac{1}{t}) = 3t^2 \sin \frac{1}{t} + t^3 \cos(\frac{1}{t}) \cdot (-\frac{1}{t^2}) = 3t^2 \sin \frac{1}{t} - t \cos \frac{1}{t}$.
Thus,$\int(3t^2 \sin \frac{1}{t} - t \cos \frac{1}{t}) dt = t^3 \sin \frac{1}{t} + c$.
Comparing this with $f(t) \sin(\frac{1}{t}) + c$,we get $f(t) = t^3$.
Therefore,$f(2) = 2^3 = 8$.

(C) Let $I = \int [(\log_{2} x)^2 + 2 \log_{2} x] dx$.
We know that $\frac{d}{dx} [x(\log_{2} x)^2] = 1 \cdot (\log_{2} x)^2 + x \cdot 2(\log_{2} x) \cdot \frac{d}{dx}(\log_{2} x)$.
Since $\frac{d}{dx}(\log_{2} x) = \frac{1}{x \ln 2}$,this does not simplify directly to the integrand.
Let's use integration by parts on the first term: $\int (\log_{2} x)^2 dx$.
Let $u = (\log_{2} x)^2$ and $dv = dx$. Then $du = 2(\log_{2} x) \cdot \frac{1}{x \ln 2} dx$ and $v = x$.
$\int (\log_{2} x)^2 dx = x(\log_{2} x)^2 - \int x \cdot \frac{2 \log_{2} x}{x \ln 2} dx = x(\log_{2} x)^2 - \frac{2}{\ln 2} \int \log_{2} x dx$.
This suggests the original problem likely assumes $\log x$ as the natural logarithm $\ln x$.
Assuming $\log x = \ln x$:
$I = \int [(\ln x)^2 + 2 \ln x] dx$.
Using the derivative rule: $\frac{d}{dx} [x(\ln x)^2] = (\ln x)^2 + x \cdot 2 \ln x \cdot \frac{1}{x} = (\ln x)^2 + 2 \ln x$.
Therefore,$\int [(\ln x)^2 + 2 \ln x] dx = x(\ln x)^2 + c$.

(A) We are given the integral $I = \int [ \cos(x) \cdot \frac{d}{dx}(\csc(x)) ] dx$.
First,we know that $\frac{d}{dx}(\csc(x)) = -\csc(x) \cot(x)$.
Substituting this into the integral,we get:
$I = \int [ \cos(x) \cdot (-\csc(x) \cot(x)) ] dx$
$I = - \int [ \cos(x) \cdot \frac{1}{\sin(x)} \cdot \frac{\cos(x)}{\sin(x)} ] dx$
$I = - \int [ \frac{\cos^2(x)}{\sin^2(x)} ] dx$
$I = - \int \cot^2(x) dx$
Using the identity $\cot^2(x) = \csc^2(x) - 1$,we have:
$I = - \int (\csc^2(x) - 1) dx$
$I = - \int \csc^2(x) dx + \int 1 dx$
Since $\int \csc^2(x) dx = -\cot(x)$,we get:
$I = -(-\cot(x)) + x + c$
$I = \cot(x) + x + c$
Comparing this with $f(x) + g(x) + c$,we have $f(x) = \cot(x)$ and $g(x) = x$ (or vice versa).
Therefore,$f(x) \cdot g(x) = x \cot(x)$.

(A) We need to evaluate the integral $I = \int x \tan^2 x dx$.
Using the identity $\tan^2 x = \sec^2 x - 1$,we get:
$I = \int x(\sec^2 x - 1) dx = \int x \sec^2 x dx - \int x dx$.
Applying integration by parts to $\int x \sec^2 x dx$ (taking $u = x$ and $dv = \sec^2 x dx$):
$I = [x \tan x - \int 1 \cdot \tan x dx] - \frac{x^2}{2} + C$.
Since $\int \tan x dx = \log_e(\sec x)$:
$I = x \tan x - \log_e(\sec x) - \frac{x^2}{2} + C$.
Thus,the correct option is $A$.

(B) We know that for all $x \in \mathbb{R}$,$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.
Substituting this into the integral,we get:
$I = \int x^{2020} \left(\frac{\pi}{2}\right) dx$
$I = \frac{\pi}{2} \int x^{2020} dx$
Using the power rule for integration,$\int x^n dx = \frac{x^{n+1}}{n+1} + C$:
$I = \frac{\pi}{2} \cdot \frac{x^{2021}}{2021} + C$
Since $\frac{\pi}{2} = \tan^{-1} x + \cot^{-1} x$,we can write:
$I = \frac{x^{2021}}{2021} (\tan^{-1} x + \cot^{-1} x) + C$
Thus,option $B$ is correct.

(A) Let $I = \int \frac{x+\sin x}{1+\cos x} d x$.
Using the identities $1+\cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get:
$I = \int \frac{x}{2 \cos^2 \frac{x}{2}} d x + \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} d x$
$I = \frac{1}{2} \int x \sec^2 \frac{x}{2} d x + \int \tan \frac{x}{2} d x$.
Applying integration by parts to the first term:
$\int x \sec^2 \frac{x}{2} d x = x \cdot \frac{\tan \frac{x}{2}}{1/2} - \int 1 \cdot \frac{\tan \frac{x}{2}}{1/2} d x = 2x \tan \frac{x}{2} - 2 \int \tan \frac{x}{2} d x$.
Substituting this back into the expression for $I$:
$I = \frac{1}{2} (2x \tan \frac{x}{2} - 2 \int \tan \frac{x}{2} d x) + \int \tan \frac{x}{2} d x$
$I = x \tan \frac{x}{2} - \int \tan \frac{x}{2} d x + \int \tan \frac{x}{2} d x$
$I = x \tan \frac{x}{2} + C$.

(B) We are given the integral $\int \frac{x^8+4}{x^4-2 x^2+2} d x$.
First,we rewrite the numerator by adding and subtracting $4x^4$:
$\int \frac{x^8+4x^4+4-4x^4}{x^4-2 x^2+2} d x = \int \frac{(x^4+2)^2 - (2x^2)^2}{x^4-2 x^2+2} d x$.
Using the difference of squares formula $a^2-b^2 = (a-b)(a+b)$,we have:
$\int \frac{(x^4+2-2x^2)(x^4+2+2x^2)}{x^4-2 x^2+2} d x = \int (x^4+2x^2+2) d x$.
Integrating term by term,we get:
$\frac{x^5}{5} + \frac{2x^3}{3} + 2x + k$.
Comparing this with $Ax^5+Bx^3+Cx+k$,we find $A = \frac{1}{5}$,$B = \frac{2}{3}$,and $C = 2$.
Now,calculate $5A+3B+C$:
$5(\frac{1}{5}) + 3(\frac{2}{3}) + 2 = 1 + 2 + 2 = 5$.

(A) We have,$I = \int \frac{x^4+x^2+1}{x^2-x+1} dx$.
Since $x^4+x^2+1 = (x^2+1)^2 - x^2 = (x^2+x+1)(x^2-x+1)$,
we can simplify the integrand:
$I = \int \frac{(x^2+x+1)(x^2-x+1)}{x^2-x+1} dx = \int (x^2+x+1) dx$.
Integrating term by term,we get:
$I = \frac{x^3}{3} + \frac{x^2}{2} + x + c$.

(C) Let $I = \int \frac{(x+1)^2}{x(x^2+1)} dx$.
Expanding the numerator,we get $(x+1)^2 = x^2 + 2x + 1$.
So,$I = \int \frac{x^2 + 2x + 1}{x(x^2+1)} dx$.
We can rewrite the numerator as $(x^2+1) + 2x$.
Thus,$I = \int \frac{(x^2+1) + 2x}{x(x^2+1)} dx$.
Splitting the integral,we get $I = \int \frac{x^2+1}{x(x^2+1)} dx + \int \frac{2x}{x(x^2+1)} dx$.
$I = \int \frac{1}{x} dx + 2 \int \frac{1}{x^2+1} dx$.
Integrating,we get $I = \log |x| + 2 \tan^{-1}(x) + C$.
Hence,option $(C)$ is correct.

(B) Given the integral $I = \int \left( \frac{2x^3 - 3x + 5}{2x^2} \right) dx$.
Dividing each term in the numerator by $2x^2$,we get:
$I = \int \left( \frac{2x^3}{2x^2} - \frac{3x}{2x^2} + \frac{5}{2x^2} \right) dx = \int \left( x - \frac{3}{2x} + \frac{5}{2x^2} \right) dx$.
Integrating term by term:
$I = \frac{x^2}{2} - \frac{3}{2} \ln|x| - \frac{5}{2} \left( -\frac{1}{x} \right) + C = \frac{x^2}{2} - \frac{3}{2} \ln|x| + \frac{5}{2x} + C$.
However,the expression $\ln(x)$ is only defined for $x > 0$. Therefore,the integral is valid for $x > 0$.
Hence,option $B$ is correct.

(A) Given,$\int \frac{x^2}{2 x^2+6 \sqrt{2} x+9} d x$.
Note that $2 x^2+6 \sqrt{2} x+9 = (\sqrt{2} x+3)^2$.
We can write the numerator as:
$x^2 = \frac{1}{2} (2 x^2 + 6 \sqrt{2} x + 9) - 3 \sqrt{2} x - \frac{9}{2}$.
So,the integral becomes:
$\int \left( \frac{1}{2} - \frac{3 \sqrt{2} x + \frac{9}{2}}{(\sqrt{2} x + 3)^2} \right) d x = \frac{x}{2} - \int \frac{3 \sqrt{2} x + \frac{9}{2}}{(\sqrt{2} x + 3)^2} d x$.
Let $u = \sqrt{2} x + 3$,then $du = \sqrt{2} dx$,so $dx = \frac{du}{\sqrt{2}}$ and $x = \frac{u-3}{\sqrt{2}}$.
The integral part is $\int \frac{3 \sqrt{2} (\frac{u-3}{\sqrt{2}}) + \frac{9}{2}}{u^2} \frac{du}{\sqrt{2}} = \frac{1}{\sqrt{2}} \int \frac{3u - 9 + \frac{9}{2}}{u^2} du = \frac{1}{\sqrt{2}} \int \frac{3u - \frac{9}{2}}{u^2} du = \frac{1}{\sqrt{2}} \int \left( \frac{3}{u} - \frac{9}{2u^2} \right) du$.
$= \frac{1}{\sqrt{2}} [3 \log |u| + \frac{9}{2u}] + c = \frac{3}{\sqrt{2}} \log |\sqrt{2} x + 3| + \frac{9}{2 \sqrt{2} (\sqrt{2} x + 3)} + c$.
Substituting back into the expression:
$\frac{x}{2} - \left( \frac{3}{\sqrt{2}} \log |\sqrt{2} x + 3| + \frac{9}{2 \sqrt{2} (\sqrt{2} x + 3)} \right) + c$.
To match the form in the options,we factor out $\frac{1}{2 \sqrt{2}}$:
$= \frac{1}{2 \sqrt{2}} [\sqrt{2} x - 6 \log |\sqrt{2} x + 3| - \frac{9}{\sqrt{2} x + 3}] + c$.
Adding and subtracting $3$ inside the bracket:
$= \frac{1}{2 \sqrt{2}} [(\sqrt{2} x + 3) - 3 - 6 \log |\sqrt{2} x + 3| - \frac{9}{\sqrt{2} x + 3}] + c$.
Since the constant term can be absorbed into $c$,this matches option $(a)$.