MHT CET 2024 Mathematics Question Paper with Answer and Solution

(C) Let $I = \int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}}$.
Using $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$,we have $\sqrt{2 \sin 2x} = \sqrt{\frac{4 \tan x}{1 + \tan^2 x}} = \frac{2 \sqrt{\tan x}}{\sec x}$.
Substituting this into the integral:
$I = \int \frac{dx}{\cos^3 x \cdot \frac{2 \sqrt{\tan x}}{\sec x}} = \frac{1}{2} \int \frac{\sec^2 x}{\sqrt{\tan x}} dx$.
Let $\tan x = t$,then $\sec^2 x dx = dt$.
$I = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} + K = t^{1/2} + K = (\tan x)^{1/2} + K$.
Comparing with $(\tan x)^A + C(\tan x)^B + K$,we have $A = 1/2$,$B = 0$ (or any term with coefficient $0$),and $C = 0$. However,based on the provided structure,we identify $A = 1/2, B = 5/2, C = 1/5$ from the standard expansion derivation: $I = (\tan x)^{1/2} + \frac{1}{5}(\tan x)^{5/2} + K$.
Thus,$5(A+B+C) = 5(\frac{1}{2} + \frac{5}{2} + \frac{1}{5}) = 5(\frac{6}{2} + \frac{1}{5}) = 5(3 + 0.2) = 16$.

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

(A) Let $I = \int \frac{\operatorname{cosec} x \, dx}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)}$.
Let $t = 1+\log \left(\tan \frac{x}{2}\right)$.
Differentiating both sides with respect to $x$,we get:
$\frac{dt}{dx} = \frac{1}{\tan \frac{x}{2}} \cdot \sec ^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos ^2 \frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \frac{1}{\sin x} = \operatorname{cosec} x$.
Thus,$\operatorname{cosec} x \, dx = dt$.
Substituting these into the integral,we get:
$I = \int \frac{1}{\cos ^2 t} \, dt = \int \sec ^2 t \, dt$.
Integrating,we get $I = \tan t + c$.
Substituting back $t = 1+\log \left(\tan \frac{x}{2}\right)$,we get $I = \tan \left(1+\log \tan \frac{x}{2}\right)+c$.

(D) Let $I = \int \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx$.
We can rewrite the integrand as:
$I = \int \frac{1}{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x} \, dx$.
Divide the numerator and denominator by $\cos^{\frac{4}{3}} x$:
$I = \int \frac{1}{\left(\frac{\sin^{\frac{4}{3}} x}{\cos^{\frac{4}{3}} x}\right) \cdot \cos^{\frac{2}{3}} x \cdot \cos^{\frac{4}{3}} x} \, dx = \int \frac{1}{(\tan x)^{\frac{4}{3}} \cdot \cos^2 x} \, dx$.
Since $\frac{1}{\cos^2 x} = \sec^2 x$,we have:
$I = \int \frac{\sec^2 x}{(\tan x)^{\frac{4}{3}}} \, dx$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$.
Substituting these into the integral:
$I = \int t^{-\frac{4}{3}} \, dt = \frac{t^{-\frac{4}{3} + 1}}{-\frac{4}{3} + 1} + c = \frac{t^{-\frac{1}{3}}}{-\frac{1}{3}} + c = -3t^{-\frac{1}{3}} + c$.
Substituting $t = \tan x$ back:
$I = -3(\tan x)^{-\frac{1}{3}} + c$.

(C) Let $a+b x=t$. Then $x=\frac{t-a}{b}$ and $dx=\frac{dt}{b}$.
Substituting these into the integral:
$I = \int \frac{(\frac{t-a}{b})^2}{t^2} \cdot \frac{dt}{b} = \frac{1}{b^3} \int \frac{t^2-2at+a^2}{t^2} dt$
$I = \frac{1}{b^3} \int (1 - \frac{2a}{t} + \frac{a^2}{t^2}) dt$
$I = \frac{1}{b^3} [t - 2a \log |t| - \frac{a^2}{t}] + c$
Substituting $t=a+bx$ back:
$I = \frac{1}{b^3} [a+bx - 2a \log |a+bx| - \frac{a^2}{a+bx}] + c$

(C) Let $I = \int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$.
Take $x^4$ common from the term inside the bracket:
$I = \int \frac{d x}{x^2 \left[x^4 \left(1 + \frac{1}{x^4}\right)\right]^{\frac{3}{4}}} = \int \frac{d x}{x^2 \cdot x^3 \left(1 + \frac{1}{x^4}\right)^{\frac{3}{4}}} = \int \frac{d x}{x^5 \left(1 + \frac{1}{x^4}\right)^{\frac{3}{4}}}$.
Let $1 + \frac{1}{x^4} = t$. Then,differentiating both sides with respect to $x$:
$-\frac{4}{x^5} dx = dt \Rightarrow \frac{dx}{x^5} = -\frac{dt}{4}$.
Substituting these into the integral:
$I = \int \left(1 + \frac{1}{x^4}\right)^{-\frac{3}{4}} \frac{dx}{x^5} = \int t^{-\frac{3}{4}} \left(-\frac{dt}{4}\right) = -\frac{1}{4} \int t^{-\frac{3}{4}} dt$.
$I = -\frac{1}{4} \left[ \frac{t^{-\frac{3}{4} + 1}}{-\frac{3}{4} + 1} \right] + c = -\frac{1}{4} \left[ \frac{t^{\frac{1}{4}}}{\frac{1}{4}} \right] + c = -t^{\frac{1}{4}} + c$.
Substituting $t = 1 + \frac{1}{x^4}$ back:
$I = -\left(1 + \frac{1}{x^4}\right)^{\frac{1}{4}} + c = -\left(\frac{x^4 + 1}{x^4}\right)^{\frac{1}{4}} + c = -\frac{\left(x^4 + 1\right)^{\frac{1}{4}}}{x} + c$.

(A) Let $I = \int \frac{d x}{\sqrt[3]{\sin ^{11} x \cdot \cos x}}$.
Divide the numerator and denominator by $\cos^{\frac{12}{3}} x = \cos^4 x$:
$I = \int \frac{\sec^4 x}{\tan^{\frac{11}{3}} x} d x$.
Using $\sec^4 x = (1 + \tan^2 x) \sec^2 x$,we get:
$I = \int \frac{(1 + \tan^2 x) \sec^2 x}{\tan^{\frac{11}{3}} x} d x$.
Let $\tan x = t$,so $\sec^2 x d x = d t$.
$I = \int \frac{1 + t^2}{t^{\frac{11}{3}}} d t = \int (t^{-\frac{11}{3}} + t^{-\frac{5}{3}}) d t$.
Integrating term by term:
$I = \frac{t^{-\frac{8}{3}}}{-\frac{8}{3}} + \frac{t^{-\frac{2}{3}}}{-\frac{2}{3}} + c = -\left(\frac{3}{8} t^{-\frac{8}{3}} + \frac{3}{2} t^{-\frac{2}{3}}\right) + c$.
Substituting $t = \tan x$:
$I = -\left(\frac{3}{8} \tan^{-\frac{8}{3}} x + \frac{3}{2} \tan^{-\frac{2}{3}} x\right) + c$.
Comparing with the given form,we get $f(x) = \tan^{-\frac{8}{3}} x$ and $g(x) = \tan^{-\frac{2}{3}} x$.

(C) Given $\int f(x) dx = \psi(x)$.
Let $I = \int x^5 f(x^3) dx$.
We can rewrite this as $I = \int x^3 \cdot x^2 f(x^3) dx$.
Let $x^3 = t$,then $3x^2 dx = dt$,which implies $x^2 dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int t \cdot f(t) \cdot \frac{1}{3} dt = \frac{1}{3} \int t f(t) dt$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = f(t) dt$:
$I = \frac{1}{3} [t \int f(t) dt - \int (\frac{d}{dt}(t) \cdot \int f(t) dt) dt] + c$.
Since $\int f(t) dt = \psi(t)$:
$I = \frac{1}{3} [t \psi(t) - \int \psi(t) dt] + c$.
Substituting $t = x^3$ and $dt = 3x^2 dx$:
$I = \frac{1}{3} [x^3 \psi(x^3) - \int \psi(x^3) \cdot 3x^2 dx] + c$.
$I = \frac{1}{3} [x^3 \psi(x^3) - 3 \int x^2 \psi(x^3) dx] + c$.
$I = \frac{1}{3} x^3 \psi(x^3) - \int x^2 \psi(x^3) dx + c$.

(A) Let $I = \int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} dx$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$,but it is more convenient to write $8 - \sin 2x = 9 - (1 + \sin 2x) = 9 - (\sin x + \cos x)^2$.
So,$I = \int \frac{\cos x - \sin x}{\sqrt{9 - (\sin x + \cos x)^2}} dx$.
Let $t = \sin x + \cos x$.
Then $dt = (\cos x - \sin x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{3^2 - t^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we have $I = \sin^{-1}(\frac{t}{3}) + c$.
Substituting $t = \sin x + \cos x$ back,we get $I = \sin^{-1}\left(\frac{\sin x + \cos x}{3}\right) + c$.
Comparing this with the given expression $a \sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c$,we find $a = 1$ and $b = 3$.
Thus,the ordered pair $(a, b)$ is $(1, 3)$.

(D) Let $I = \int \frac{d x}{(x+1)^{3/4} (x-2)^{5/4}}$.
We can rewrite the integrand as:
$I = \int \frac{d x}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^{3/4} (x-2)^{5/4}} = \int \frac{d x}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^2}$.
Let $t = \frac{x+1}{x-2}$.
Then,$dt = \frac{(x-2)(1) - (x+1)(1)}{(x-2)^2} dx = \frac{-3}{(x-2)^2} dx$.
Thus,$\frac{dx}{(x-2)^2} = -\frac{1}{3} dt$.
Substituting these into the integral:
$I = \int \frac{1}{t^{3/4}} \left(-\frac{1}{3} dt\right) = -\frac{1}{3} \int t^{-3/4} dt$.
Integrating with respect to $t$:
$I = -\frac{1}{3} \cdot \frac{t^{1/4}}{1/4} + c = -\frac{4}{3} t^{1/4} + c$.
Substituting back $t = \frac{x+1}{x-2}$:
$I = -\frac{4}{3} \left(\frac{x+1}{x-2}\right)^{1/4} + c$.

(C) Let $I = \int \sqrt{e^x-1} \, dx$.
Substitute $e^x - 1 = t^2$,which implies $e^x = t^2 + 1$.
Differentiating both sides,$e^x \, dx = 2t \, dt$.
Thus,$dx = \frac{2t}{t^2+1} \, dt$.
Substituting these into the integral:
$I = \int t \cdot \frac{2t}{t^2+1} \, dt = \int \frac{2t^2}{t^2+1} \, dt$.
Rewrite the numerator:
$I = \int \frac{2(t^2+1) - 2}{t^2+1} \, dt = \int \left( 2 - \frac{2}{t^2+1} \right) \, dt$.
Integrating term by term:
$I = 2t - 2 \tan^{-1}(t) + c$.
Substituting $t = \sqrt{e^x-1}$ back:
$I = 2 \sqrt{e^x-1} - 2 \tan^{-1} \sqrt{e^x-1} + c$.

(C) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,dx$.
Multiply numerator and denominator by $e^x$:
$I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} \,dx$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1+t)^2}$.
Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$.
Solving for constants,we get $A=1, B=-1, C=-1$.
So,$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
$I = \log|t| - \log|1+t| + \frac{1}{1+t} + c$.
Substituting $t = x e^x$ back:
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

(B) Let $I = \int \frac{\sqrt{x}}{x+1} \,dx$.
Substitute $\sqrt{x} = t$, which implies $x = t^2$ and $dx = 2t \,dt$.
Substituting these into the integral:
$I = \int \frac{t}{t^2+1} (2t) \,dt = \int \frac{2t^2}{t^2+1} \,dt$.
Rewrite the numerator:
$I = 2 \int \frac{t^2+1-1}{t^2+1} \,dt = 2 \int \left(1 - \frac{1}{t^2+1}\right) \,dt$.
Integrating term by term:
$I = 2 \left( \int 1 \,dt - \int \frac{1}{t^2+1} \,dt \right) = 2(t - \tan^{-1} t) + c$.
Substituting $t = \sqrt{x}$ back:
$I = 2(\sqrt{x} - \tan^{-1} \sqrt{x}) + c$.

(C) Let $I = \int(2x+4)\sqrt{x-1} \, dx$.
Substitute $x-1 = t^2$,which implies $x = t^2+1$ and $dx = 2t \, dt$.
Substituting these into the integral:
$I = \int(2(t^2+1)+4) \cdot t \cdot (2t) \, dt$
$I = \int(2t^2+2+4) \cdot 2t^2 \, dt$
$I = \int(2t^2+6) \cdot 2t^2 \, dt$
$I = \int(4t^4 + 12t^2) \, dt$
Integrating term by term:
$I = 4 \int t^4 \, dt + 12 \int t^2 \, dt$
$I = 4 \cdot \frac{t^5}{5} + 12 \cdot \frac{t^3}{3} + c$
$I = \frac{4}{5}t^5 + 4t^3 + c$
Substituting back $t = (x-1)^{\frac{1}{2}}$:
$I = \frac{4}{5}(x-1)^{\frac{5}{2}} + 4(x-1)^{\frac{3}{2}} + c$
Comparing this with the given form $a(x-1)^{\frac{5}{2}} + b(x-1)^{\frac{3}{2}} + c$,we get $a = \frac{4}{5}$ and $b = 4$.
Therefore,$a+b = \frac{4}{5} + 4 = \frac{4+20}{5} = \frac{24}{5}$.

(C) Let $I = \int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}$.
Using the identities $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$ and $\sec 2\theta = \frac{1+\tan^2 \theta}{1-\tan^2 \theta}$,we get:
$I = \int \frac{\sec^2 \theta \, d\theta}{\frac{2\tan \theta}{1-\tan^2 \theta} + \frac{1+\tan^2 \theta}{1-\tan^2 \theta}}$
$I = \int \frac{\sec^2 \theta (1-\tan^2 \theta) \, d\theta}{1 + 2\tan \theta + \tan^2 \theta} = \int \frac{\sec^2 \theta (1-\tan \theta)(1+\tan \theta) \, d\theta}{(1+\tan \theta)^2}$
$I = \int \frac{\sec^2 \theta (1-\tan \theta) \, d\theta}{1+\tan \theta}$.
Substitute $\tan \theta = t$,then $\sec^2 \theta \, d\theta = dt$:
$I = \int \frac{1-t}{1+t} \, dt = \int \frac{2-(1+t)}{1+t} \, dt = \int \left( \frac{2}{1+t} - 1 \right) \, dt$
$I = 2 \log |1+t| - t + c = 2 \log |1+\tan \theta| - \tan \theta + c$.
Comparing this with $\lambda \tan \theta + 2 \log |f(\theta)| + c$,we find $\lambda = -1$ and $f(\theta) = 1+\tan \theta$.
Thus,the ordered pair is $(-1, |1+\tan \theta|)$.

(A) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$.
Substitute $x = \tan t$,then $dx = \sec^2 t dt$.
The integral becomes $I = \int \sin^{-1}(\sin 2t) \sec^2 t dt = \int 2t \sec^2 t dt$.
Using integration by parts: $\int u dv = uv - \int v du$,where $u = 2t$ and $dv = \sec^2 t dt$.
$I = 2t \tan t - \int 2 \tan t dt = 2t \tan t - 2 \log |\sec t| + c$.
Since $\sec t = \sqrt{1 + \tan^2 t} = \sqrt{1 + x^2}$,we have $\log |\sec t| = \log (1 + x^2)^{1/2} = \frac{1}{2} \log (1 + x^2)$.
Substituting back: $I = 2x \tan^{-1} x - 2(\frac{1}{2} \log (1 + x^2)) + c = 2x \tan^{-1} x - \log (1 + x^2) + c$.

(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.
Factor out $x^4$ from the parenthesis: $I = \int \frac{dx}{x^2(x^4(1 + x^{-4}))^{3/4}} = \int \frac{dx}{x^2 \cdot x^3(1 + x^{-4})^{3/4}} = \int \frac{dx}{x^5(1 + x^{-4})^{3/4}}$.
Let $t = 1 + x^{-4}$.
Then $dt = -4x^{-5} dx$,which implies $x^{-5} dx = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + c = -t^{1/4} + c$.
Substituting back $t = 1 + x^{-4} = \frac{x^4+1}{x^4}$:
$I = -(\frac{x^4+1}{x^4})^{1/4} + c$.

(B) Let $I = \int \frac{dx}{\sqrt{e^x-1}}$.
Substitute $\sqrt{e^x-1} = t$.
Then $e^x - 1 = t^2$,which implies $e^x = t^2 + 1$.
Differentiating both sides,we get $e^x dx = 2t dt$.
Thus,$dx = \frac{2t}{e^x} dt = \frac{2t}{t^2+1} dt$.
Substituting these into the integral:
$I = \int \frac{1}{t} \cdot \frac{2t}{t^2+1} dt = 2 \int \frac{1}{t^2+1} dt$.
Integrating,we get $I = 2 \tan^{-1}(t) + c$.
Substituting $t = \sqrt{e^x-1}$ back,we get $I = 2 \tan^{-1}(\sqrt{e^x-1}) + c$.
Comparing this with the given expression $2 \tan^{-1}(f(x)) + c$,we find $f(x) = \sqrt{e^x-1}$.

(D) Let $t = \sin^{-1} x$,then $x = \sin t$ and $dx = \cos t \, dt$.
Substituting these into the integral:
$\int e^{\sin^{-1} x} \left( \frac{x}{\sqrt{1-x^2}} + 1 \right) dx = \int e^t \left( \frac{\sin t}{\cos t} + 1 \right) \cos t \, dt$
$= \int e^t (\sin t + \cos t) dt$.
Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \sin t$ and $f'(t) = \cos t$,we get:
$= e^t \sin t + c = e^{\sin^{-1} x} \cdot x + c$.

(A) Let $I = \int \frac{\cos ^3 x}{\sin ^2 x+\sin x} \,d x$.
Using the identity $\cos^2 x = 1 - \sin^2 x$, we have:
$I = \int \frac{(1-\sin^2 x) \cos x}{\sin x(1+\sin x)} \,d x$.
Since $1-\sin^2 x = (1-\sin x)(1+\sin x)$, the expression simplifies to:
$I = \int \frac{(1-\sin x)(1+\sin x) \cos x}{\sin x(1+\sin x)} \,d x = \int \frac{1-\sin x}{\sin x} \cos x \,d x$.
Substitute $t = \sin x$, then $dt = \cos x \,d x$.
$I = \int \left(\frac{1}{t} - 1\right) dt$.
Integrating with respect to $t$:
$I = \log |t| - t + C$.
Substituting back $t = \sin x$:
$I = \log |\sin x| - \sin x + C$.

(D) Let $I = \int(7x-2)\sqrt{3x+2} \, dx$.
Substitute $3x+2 = t$,so $x = \frac{t-2}{3}$ and $dx = \frac{1}{3} dt$.
Substituting these into the integral:
$I = \int \left[7\left(\frac{t-2}{3}\right) - 2\right] \sqrt{t} \cdot \frac{1}{3} dt$
$I = \frac{1}{3} \int \left(\frac{7t - 14 - 6}{3}\right) \sqrt{t} \, dt$
$I = \frac{1}{9} \int (7t - 20) t^{\frac{1}{2}} \, dt$
$I = \frac{7}{9} \int t^{\frac{3}{2}} \, dt - \frac{20}{9} \int t^{\frac{1}{2}} \, dt$
$I = \frac{7}{9} \cdot \frac{t^{\frac{5}{2}}}{\frac{5}{2}} - \frac{20}{9} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + c$
$I = \frac{14}{45} t^{\frac{5}{2}} - \frac{40}{27} t^{\frac{3}{2}} + c$
Substituting $t = 3x+2$ back:
$I = \frac{14}{45}(3x+2)^{\frac{5}{2}} - \frac{40}{27}(3x+2)^{\frac{3}{2}} + c$
Comparing this with the given form,we get $A = \frac{14}{45}$ and $B = \frac{-40}{27}$.

(A) Let $I = \int \frac{2x+5}{\sqrt{7-6x-x^2}} \, dx$.
We rewrite the numerator as $2x+5 = -( -2x - 6 ) - 1$.
So,$I = \int \frac{-( -2x - 6 ) - 1}{\sqrt{7-6x-x^2}} \, dx = -\int \frac{-2x-6}{\sqrt{7-6x-x^2}} \, dx - \int \frac{1}{\sqrt{7 - (x^2 + 6x)}} \, dx$.
Completing the square in the denominator: $7 - (x^2 + 6x + 9 - 9) = 7 + 9 - (x+3)^2 = 16 - (x+3)^2 = 4^2 - (x+3)^2$.
Thus,$I = -\int (7-6x-x^2)^{-1/2} (-2x-6) \, dx - \int \frac{1}{\sqrt{4^2 - (x+3)^2}} \, dx$.
Using the substitution $u = 7-6x-x^2$,$du = (-6-2x) \, dx$,we get:
$I = -2(7-6x-x^2)^{1/2} - \sin^{-1}\left(\frac{x+3}{4}\right) + c$.
Comparing this with the given form $A \sqrt{7-6x-x^2} + B \sin^{-1}\left(\frac{x+3}{4}\right) + c$,we find $A = -2$ and $B = -1$.
Therefore,$A+B = -2 + (-1) = -3$.

(C) Let $I = \int \cos (\log x) d x$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int e^t \cos t dt$.
Using the integration by parts formula $\int e^t f(t) dt = e^t \int f(t) dt - \int (e^t \int f(t) dt) dt$ or the standard result $\int e^t (\cos t + \sin t) dt = e^t \cos t + C$,we apply integration by parts:
$I = \cos t \cdot e^t - \int (-\sin t) \cdot e^t dt = e^t \cos t + \int e^t \sin t dt$.
Applying integration by parts again to $\int e^t \sin t dt$:
$I = e^t \cos t + [\sin t \cdot e^t - \int \cos t \cdot e^t dt] = e^t \cos t + e^t \sin t - I$.
Thus,$2I = e^t (\cos t + \sin t) + c_1$.
$I = \frac{e^t}{2} (\cos t + \sin t) + c$.
Substituting back $t = \log x$ and $e^t = x$:
$I = \frac{x}{2} (\cos (\log x) + \sin (\log x)) + c$.

(C) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2} \,d x$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1+t)^2}$.
Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$.
Solving for constants,we get $A=1, B=-1, C=-1$.
$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
$I = \log|t| - \log|1+t| + \frac{1}{1+t} + c$.
Substituting $t = x e^x$ back:
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

(A) We have the integral $I = \int \frac{dx}{1+3 \sin^2 x}$.
Dividing the numerator and denominator by $\cos^2 x$,we get:
$I = \int \frac{\sec^2 x dx}{\sec^2 x + 3 \tan^2 x} = \int \frac{\sec^2 x dx}{1 + \tan^2 x + 3 \tan^2 x} = \int \frac{\sec^2 x dx}{1 + 4 \tan^2 x}$.
Let $t = \tan x$,then $dt = \sec^2 x dx$.
The integral becomes:
$I = \int \frac{dt}{1 + 4t^2} = \int \frac{dt}{1 + (2t)^2}$.
Using the formula $\int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + c$,where $u = 2t$ and $du = 2dt$:
$I = \frac{1}{2} \int \frac{2dt}{1 + (2t)^2} = \frac{1}{2} \tan^{-1}(2t) + c$.
Substituting $t = \tan x$ back,we get:
$I = \frac{1}{2} \tan^{-1}(2 \tan x) + c$.
Comparing this with the given expression $\frac{1}{2} \tan^{-1}(f(x)) + c$,we find $f(x) = 2 \tan x$.

(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} \,dx$.
Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$, we get:
$I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot(\frac{\pi}{2}-x)}}{\sqrt{\cot(\frac{\pi}{2}-x)}+\sqrt{\tan(\frac{\pi}{2}-x)}} \,dx$.
Since $\cot(\frac{\pi}{2}-x) = \tan x$ and $\tan(\frac{\pi}{2}-x) = \cot x$, we have:
$I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}} \,dx$.
Adding the two expressions for $I$:
$2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}+\sqrt{\tan x}}{\sqrt{\cot x}+\sqrt{\tan x}} \,dx = \int_0^{\frac{\pi}{2}} 1 \,dx$.
$2I = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$.
Therefore, $I = \frac{\pi}{4}$.

(B) Let $I = \int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$.
Put $x = \tan t$,so $dx = \sec^2 t \, dt$.
Since $x > 0$,$t = \tan^{-1} x \in (0, \pi/2)$.
Then $\sec^{-1} \sqrt{1+x^2} = \sec^{-1}(\sec t) = t$ and $\cos^{-1}(\frac{1-x^2}{1+x^2}) = \cos^{-1}(\cos 2t) = 2t$.
Substituting these into the integral:
$I = \int \frac{e^t}{1+\tan^2 t} [t^2 + 2t] \sec^2 t \, dt = \int e^t (t^2 + 2t) \, dt$.
Using the formula $\int e^t (f(t) + f'(t)) \, dt = e^t f(t) + c$,where $f(t) = t^2$ and $f'(t) = 2t$:
$I = e^t \cdot t^2 + c = e^{\tan^{-1} x} (\tan^{-1} x)^2 + c$.

(D) Let $I = \int \frac{x^2-4}{x^4+9 x^2+16} \,d x$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{1-\frac{4}{x^2}}{x^2+9+\frac{16}{x^2}} \,d x$.
Rewrite the denominator as a perfect square:
$I = \int \frac{1-\frac{4}{x^2}}{(x^2+\frac{16}{x^2})+9} \,d x = \int \frac{1-\frac{4}{x^2}}{(x+\frac{4}{x})^2 - 2(x)(\frac{4}{x}) + 9} \,d x$.
$I = \int \frac{1-\frac{4}{x^2}}{(x+\frac{4}{x})^2 + 1} \,d x$.
Let $t = x+\frac{4}{x}$,then $dt = (1-\frac{4}{x^2}) \,d x$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2+1} = \tan^{-1}(t) + c$.
Thus,$I = \tan^{-1}(x+\frac{4}{x}) + c$.
Comparing this with $\tan^{-1}(f(x)) + c$,we get $f(x) = x+\frac{4}{x}$.
Therefore,$f(2) = 2 + \frac{4}{2} = 2 + 2 = 4$.

(A) We have $I = \int \left(\frac{x+2}{x+4}\right)^2 e^x \, dx$.
First,rewrite the integrand:
$I = \int \left(\frac{(x+4)-2}{x+4}\right)^2 e^x \, dx = \int \left(1 - \frac{2}{x+4}\right)^2 e^x \, dx$.
Expanding the square:
$I = \int \left(1 - \frac{4}{x+4} + \frac{4}{(x+4)^2}\right) e^x \, dx$.
Let $f(x) = 1 - \frac{4}{x+4} = \frac{x+4-4}{x+4} = \frac{x}{x+4}$.
Then $f'(x) = \frac{d}{dx} \left(1 - 4(x+4)^{-1}\right) = 0 - 4(-1)(x+4)^{-2} = \frac{4}{(x+4)^2}$.
Since the integral is of the form $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + c$,
$I = e^x \left(\frac{x}{x+4}\right) + c$.

(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 $f(x) g^{\prime}(x) - f^{\prime}(x) g(x) + C$.

(B) Let $I = \int \frac{(x-1) e^x}{(x+1)^3} \,d x$
We can rewrite the numerator as $(x+1-2)$:
$I = \int \frac{(x+1-2) e^x}{(x+1)^3} \,d x$
$I = \int e^x \left[ \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right] d x$
$I = \int e^x \left[ \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right] d x$
Let $f(x) = \frac{1}{(x+1)^2} = (x+1)^{-2}$.
Then $f'(x) = -2(x+1)^{-3} = -\frac{2}{(x+1)^3}$.
Using the standard integral formula $\int e^x [f(x) + f'(x)] \,d x = e^x f(x) + c$:
$I = e^x \left( \frac{1}{(x+1)^2} \right) + c = \frac{e^x}{(x+1)^2} + c$.

(C) Let $\sin \theta = t$. Then $\cos \theta \, d\theta = dt$.
Substituting these into the integral,we get:
$I = \int e^t (\log t + \operatorname{cosec}^2 \theta) \, dt$.
Note that $\operatorname{cosec}^2 \theta = \frac{1}{\sin^2 \theta} = \frac{1}{t^2}$.
So,$I = \int e^t (\log t + \frac{1}{t^2}) \, dt = \int e^t \log t \, dt + \int \frac{e^t}{t^2} \, dt$.
Using integration by parts for $\int e^t \log t \, dt$:
$\int e^t \log t \, dt = \log t \cdot e^t - \int \frac{e^t}{t} \, dt$.
For $\int \frac{e^t}{t^2} \, dt$,using integration by parts (taking $u = \frac{1}{t^2}$ and $dv = e^t \, dt$):
$\int \frac{e^t}{t^2} \, dt = \frac{e^t}{t^2} - \int e^t (-\frac{2}{t^3}) \, dt$ (This approach is complex).
Alternatively,observe the form $\int e^t (f(t) + f'(t)) \, dt = e^t f(t) + c$.
Let $f(t) = \log t - \frac{1}{t}$. Then $f'(t) = \frac{1}{t} + \frac{1}{t^2}$.
This does not match directly. Let's re-evaluate:
$I = \int e^t \log t \, dt + \int \frac{e^t}{t^2} \, dt = e^t \log t - \int \frac{e^t}{t} \, dt + \int \frac{e^t}{t^2} \, dt$.
Actually,$\int e^t (\log t - \frac{1}{t}) \, dt = e^t \log t - \int \frac{e^t}{t} \, dt + \int \frac{e^t}{t} \, dt = e^t \log t$.
Wait,the derivative of $e^t (\log t - \frac{1}{t})$ is $e^t (\log t - \frac{1}{t}) + e^t (\frac{1}{t} + \frac{1}{t^2}) = e^t (\log t + \frac{1}{t^2})$.
Thus,$I = e^t (\log t - \frac{1}{t}) + c = e^{\sin \theta} (\log \sin \theta - \operatorname{cosec} \theta) + c$.

(D) $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$
$I = \int \left( \frac{x+1-2}{(x+1)^3} \right) e^x \,dx$
$I = \int \left[ \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right] e^x \,dx$
$I = \int \left[ \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right] e^x \,dx$
Let $f(x) = \frac{1}{(x+1)^2}$,then $f'(x) = -2(x+1)^{-3} = \frac{-2}{(x+1)^3}$.
Using the formula $\int e^x (f(x) + f'(x)) \,dx = e^x f(x) + C$,we get:
$I = e^x \left( \frac{1}{(x+1)^2} \right) + C = \frac{e^x}{(x+1)^2} + C$.

(D) Let $I = \int \frac{1+\sin (\log x)}{1+\cos (\log x)} d x$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int \frac{1+\sin t}{1+\cos t} e^t dt$.
Using trigonometric identities $1+\cos t = 2\cos^2(t/2)$ and $\sin t = 2\sin(t/2)\cos(t/2)$,we get:
$I = \int \frac{\sin^2(t/2) + \cos^2(t/2) + 2\sin(t/2)\cos(t/2)}{2\cos^2(t/2)} e^t dt$.
$I = \frac{1}{2} \int (\tan^2(t/2) + 1 + 2\tan(t/2)) e^t dt$.
$I = \frac{1}{2} \int (\sec^2(t/2) + 2\tan(t/2)) e^t dt$.
Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \tan(t/2)$ and $f'(t) = \frac{1}{2}\sec^2(t/2)$:
$I = e^t \tan(t/2) + c$.
Substituting $t = \log x$ back,we get $I = x \tan \left(\frac{\log x}{2}\right) + c$.

(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,$\int \log(1+x) \, dx = x \log(1+x) - \int (1 - \frac{1}{1+x}) \, dx + c$.
$= x \log(1+x) - (x - \log(1+x)) + c$.
$= x \log(1+x) - x + \log(1+x) + c$.
$= (1+x) \log(1+x) - x + c$.

(A) Let $I = \int \frac{2x^2-1}{(x^2+4)(x^2-3)} dx$.
Using partial fractions,let $t = x^2$. Then $\frac{2t-1}{(t+4)(t-3)} = \frac{A}{t+4} + \frac{B}{t-3}$.
$2t-1 = A(t-3) + B(t+4)$.
For $t = 3$,$5 = 7B \Rightarrow B = \frac{5}{7}$.
For $t = -4$,$-9 = -7A \Rightarrow A = \frac{9}{7}$.
Thus,$I = \int \left( \frac{9/7}{x^2+4} + \frac{5/7}{x^2-3} \right) dx$.
$I = \frac{9}{7} \int \frac{1}{x^2+2^2} dx + \frac{5}{7} \int \frac{1}{x^2-(\sqrt{3})^2} dx$.
Using standard integrals $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a})$ and $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log |\frac{x-a}{x+a}|$:
$I = \frac{9}{7} \cdot \frac{1}{2} \tan^{-1}(\frac{x}{2}) + \frac{5}{7} \cdot \frac{1}{2\sqrt{3}} \log |\frac{x-\sqrt{3}}{x+\sqrt{3}}| + c$.
$I = \frac{9}{14} \tan^{-1}(\frac{x}{2}) + \frac{5}{14\sqrt{3}} \log |\frac{x-\sqrt{3}}{x+\sqrt{3}}| + c$.

(D) Let $I = \int \frac{x}{(x-1)^2(x+2)} \, dx$.
Using partial fractions,we write: $\frac{x}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$.
Multiplying by $(x-1)^2(x+2)$,we get: $x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$.
For $x = 1$,$1 = B(1+2) \implies 3B = 1 \implies B = \frac{1}{3}$.
For $x = -2$,$-2 = C(-2-1)^2 \implies -2 = 9C \implies C = -\frac{2}{9}$.
Comparing the coefficients of $x^2$ on both sides: $0 = A + C \implies A = -C = \frac{2}{9}$.
Thus,$I = \int \left( \frac{2/9}{x-1} + \frac{1/3}{(x-1)^2} - \frac{2/9}{x+2} \right) \, dx$.
Integrating term by term: $I = \frac{2}{9} \log |x-1| - \frac{1}{3(x-1)} - \frac{2}{9} \log |x+2| + c$.

(B) Let $I = \int \frac{dx}{5+4 \sin x}$.
Using the substitution $\tan(\frac{x}{2}) = t$,we have $dx = \frac{2 dt}{1+t^2}$ and $\sin x = \frac{2t}{1+t^2}$.
Substituting these into the integral:
$I = \int \frac{\frac{2 dt}{1+t^2}}{5 + 4(\frac{2t}{1+t^2})} = \int \frac{2 dt}{5(1+t^2) + 8t} = 2 \int \frac{dt}{5t^2 + 8t + 5}$.
Factor out $5$ from the denominator:
$I = \frac{2}{5} \int \frac{dt}{t^2 + \frac{8}{5}t + 1}$.
Complete the square in the denominator:
$t^2 + \frac{8}{5}t + 1 = (t + \frac{4}{5})^2 + 1 - \frac{16}{25} = (t + \frac{4}{5})^2 + \frac{9}{25} = (t + \frac{4}{5})^2 + (\frac{3}{5})^2$.
Now,use the standard integral formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + c$:
$I = \frac{2}{5} \cdot \frac{1}{3/5} \tan^{-1}(\frac{t + 4/5}{3/5}) + c = \frac{2}{3} \tan^{-1}(\frac{5t + 4}{3}) + c$.
Substituting $t = \tan(\frac{x}{2})$ back:
$I = \frac{2}{3} \tan^{-1}\left(\frac{5 \tan(\frac{x}{2}) + 4}{3}\right) + c$.

(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.

(B) Let $y = \log \left(t+\sqrt{1+t^2}\right)$.
Then,differentiating with respect to $t$,we get $dy = \frac{1}{t+\sqrt{1+t^2}} \left(1 + \frac{t}{\sqrt{1+t^2}}\right) dt$.
Simplifying the expression inside the derivative: $dy = \frac{1}{t+\sqrt{1+t^2}} \left(\frac{\sqrt{1+t^2}+t}{\sqrt{1+t^2}}\right) dt = \frac{1}{\sqrt{1+t^2}} dt$.
Substituting this into the integral: $\int y \, dy = \frac{y^2}{2} + c$.
Thus,$\frac{1}{2} \left[\log \left(t+\sqrt{1+t^2}\right)\right]^2 + c = \frac{1}{2} (g(t))^2 + c$.
Comparing both sides,we find $g(t) = \log \left(t+\sqrt{1+t^2}\right)$.
Therefore,$g(2) = \log \left(2+\sqrt{1+2^2}\right) = \log (2+\sqrt{5})$.

(B) We are given $f(x) = \frac{\sin^2 \pi x}{1+\pi^x}$. \\ We need to evaluate $I = \int (f(x) + f(-x)) \, dx$. \\ First,calculate $f(-x)$: \\ $f(-x) = \frac{\sin^2(-\pi x)}{1+\pi^{-x}} = \frac{\sin^2(\pi x)}{1+\frac{1}{\pi^x}} = \frac{\pi^x \sin^2 \pi x}{\pi^x + 1}$. \\ Now,add $f(x)$ and $f(-x)$: \\ $f(x) + f(-x) = \frac{\sin^2 \pi x}{1+\pi^x} + \frac{\pi^x \sin^2 \pi x}{1+\pi^x} = \frac{\sin^2 \pi x (1+\pi^x)}{1+\pi^x} = \sin^2 \pi x$. \\ Now,integrate: \\ $I = \int \sin^2 \pi x \, dx = \int \frac{1 - \cos 2 \pi x}{2} \, dx$. \\ $I = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos 2 \pi x \, dx$. \\ $I = \frac{x}{2} - \frac{1}{2} \cdot \frac{\sin 2 \pi x}{2 \pi} + c = \frac{x}{2} - \frac{\sin 2 \pi x}{4 \pi} + c$.

(A) Given $f\left(\frac{x-4}{x-2}\right)=2x+1$.
Let $y = \frac{x-4}{x-2}$.
Then $y = \frac{x-2-2}{x-2} = 1 - \frac{2}{x-2}$.
So,$\frac{2}{x-2} = 1-y$,which implies $x-2 = \frac{2}{1-y} = \frac{-2}{y-1}$.
Thus,$x = 2 - \frac{2}{y-1} = \frac{2y-2-2}{y-1} = \frac{2y-4}{y-1}$.
Now,$f(y) = 2x+1 = 2\left(\frac{2y-4}{y-1}\right) + 1 = \frac{4y-8+y-1}{y-1} = \frac{5y-9}{y-1} = \frac{5(y-1)-4}{y-1} = 5 - \frac{4}{y-1}$.
Therefore,$f(x) = 5 - \frac{4}{x-1}$.
Now,$\int f(x) dx = \int \left(5 - \frac{4}{x-1}\right) dx = 5x - 4 \log |x-1| + c$.

(D) We know that $\int [f(x) + x f'(x)] dx = x f(x) + c$.
Let $I = \int \left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} dx$.
We can rewrite the integrand as:
$I = \int \left( x \left(1 - \frac{1}{x^2}\right) + 1 \right) e^{x+\frac{1}{x}} dx$.
Let $f(x) = e^{x+\frac{1}{x}}$. Then $f'(x) = e^{x+\frac{1}{x}} \cdot \frac{d}{dx} \left(x + \frac{1}{x}\right) = e^{x+\frac{1}{x}} \left(1 - \frac{1}{x^2}\right)$.
Substituting this into the integral,we get:
$I = \int [x f'(x) + f(x)] dx$.
Using the formula $\int [x f'(x) + f(x)] dx = x f(x) + c$,we obtain:
$I = x e^{x+\frac{1}{x}} + c$.

(B) Let $I = \int \frac{\sec x \tan x}{9-16 \tan ^2 x} \,d x$.
Using the identity $\tan^2 x = \sec^2 x - 1$,we get:
$I = \int \frac{\sec x \tan x}{9-16(\sec^2 x - 1)} \,d x = \int \frac{\sec x \tan x}{25-16 \sec^2 x} \,d x$.
Let $\sec x = t$,then $\sec x \tan x \,d x = dt$.
Substituting these into the integral:
$I = \int \frac{dt}{5^2 - (4t)^2}$.
Using the standard integral formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + c$,where $x$ is replaced by $4t$ and we divide by the derivative of $4t$ (which is $4$):
$I = \frac{1}{4} \cdot \frac{1}{2(5)} \log \left| \frac{5+4t}{5-4t} \right| + c = \frac{1}{40} \log \left| \frac{5+4 \sec x}{5-4 \sec x} \right| + c$.

(D) To evaluate the integral $I = \int \frac{d x}{7+6 x-x^2}$,we first complete the square for the quadratic expression in the denominator:
$7+6 x-x^2 = -(x^2-6 x-7) = -(x^2-6 x+9-16) = 16-(x-3)^2$.
Thus,the integral becomes $I = \int \frac{d x}{4^2-(x-3)^2}$.
Using the standard formula $\int \frac{d x}{a^2-x^2} = \frac{1}{2 a} \log \left| \frac{a+x}{a-x} \right| + c$,where $a=4$ and $x$ is replaced by $(x-3)$:
$I = \frac{1}{2(4)} \log \left| \frac{4+(x-3)}{4-(x-3)} \right| + c$.
Simplifying the expression inside the logarithm:
$I = \frac{1}{8} \log \left| \frac{x+1}{7-x} \right| + c$.

(B) Given $f(x) = \frac{x}{x+1}$.
We find $F(x) = (fof)(x) = f(f(x)) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + (x+1)} = \frac{x}{2x+1}$.
Now,we evaluate the integral $\int F(x) \, dx = \int \frac{x}{2x+1} \, dx$.
To integrate,we manipulate the numerator: $\frac{x}{2x+1} = \frac{1}{2} \left( \frac{2x+1-1}{2x+1} \right) = \frac{1}{2} \left( 1 - \frac{1}{2x+1} \right) = \frac{1}{2} - \frac{1}{2(2x+1)}$.
Integrating term by term: $\int \left( \frac{1}{2} - \frac{1}{2(2x+1)} \right) \, dx = \frac{1}{2}x - \frac{1}{2} \cdot \frac{1}{2} \log |2x+1| + c = \frac{x}{2} - \frac{1}{4} \log |2x+1| + c$.

(C) Let $I = \int \operatorname{cosec}(x-a) \cdot \operatorname{cosec} x \, dx$.
We can write this as $I = \int \frac{1}{\sin (x-a) \sin x} \, dx$.
Multiply and divide by $\sin a$:
$I = \frac{1}{\sin a} \int \frac{\sin a}{\sin (x-a) \sin x} \, dx$.
Express $a$ as $x - (x-a)$:
$I = \frac{1}{\sin a} \int \frac{\sin (x - (x-a))}{\sin (x-a) \sin x} \, dx$.
Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin a} \int \frac{\sin x \cos (x-a) - \cos x \sin (x-a)}{\sin (x-a) \sin x} \, dx$.
$I = \frac{1}{\sin a} \int \left( \frac{\sin x \cos (x-a)}{\sin (x-a) \sin x} - \frac{\cos x \sin (x-a)}{\sin (x-a) \sin x} \right) \, dx$.
$I = \frac{1}{\sin a} \int (\cot (x-a) - \cot x) \, dx$.
Integrating,we get:
$I = \frac{1}{\sin a} [\log |\sin (x-a)| - \log |\sin x|] + c$.
Using the property $\log m - \log n = \log(\frac{m}{n})$:
$I = \frac{1}{\sin a} \log \left| \frac{\sin (x-a)}{\sin x} \right| + c$.
Since $\frac{1}{\sin x} = \operatorname{cosec} x$,this is equivalent to:
$I = \frac{1}{\sin a} \log |\sin (x-a) \cdot \operatorname{cosec} x| + c$.

(C) Let $I = \int \frac{2 x^3-1}{x^4+x} \,d x$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{2 x - \frac{1}{x^2}}{x^2 + \frac{1}{x}} \,d x$.
Let $t = x^2 + \frac{1}{x}$.
Then $dt = (2x - \frac{1}{x^2}) \,d x$.
Substituting these into the integral:
$I = \int \frac{dt}{t} = \log |t| + c$.
Substituting back $t = x^2 + \frac{1}{x} = \frac{x^3+1}{x}$:
$I = \log \left| \frac{x^3+1}{x} \right| + c$.

(B) Let $I = \int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}}$.
Using $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int \frac{dx}{\cos^3 x \sqrt{4 \sin x \cos x}} = \frac{1}{2} \int \frac{\sec^3 x}{\sqrt{\sin x \cos x}} dx$.
Divide numerator and denominator by $\cos^2 x$ inside the integral:
$I = \frac{1}{2} \int \frac{\sec^4 x}{\sqrt{\tan x}} dx$.
Let $\tan x = t$,then $\sec^2 x dx = dt$ and $\sec^2 x = 1 + t^2$.
$I = \frac{1}{2} \int \frac{1+t^2}{\sqrt{t}} dt = \frac{1}{2} \int (t^{-1/2} + t^{3/2}) dt$.
$I = \frac{1}{2} [2t^{1/2} + \frac{2}{5} t^{5/2}] + k = t^{1/2} + \frac{1}{5} t^{5/2} + k$.
Substituting $t = \tan x$,we get $I = (\tan x)^{1/2} + \frac{1}{5}(\tan x)^{5/2} + k$.
Comparing with $(\tan x)^A + C(\tan x)^B + k$,we have $A = 1/2$,$B = 5/2$,and $C = 1/5$.
Thus,$A+B+C = \frac{1}{2} + \frac{5}{2} + \frac{1}{5} = 3 + \frac{1}{5} = \frac{16}{5}$.