Evaluation of various forms of integration Questions in English

(D) Let $I = \int \frac{dx}{9 \cos^2 2x + 16 \sin^2 2x}$.
Divide the numerator and denominator by $\cos^2 2x$:
$I = \int \frac{\sec^2 2x dx}{9 + 16 \tan^2 2x}$.
Let $u = \tan 2x$,then $du = 2 \sec^2 2x dx$,which implies $\sec^2 2x dx = \frac{du}{2}$.
Substituting these into the integral:
$I = \int \frac{du/2}{9 + 16u^2} = \frac{1}{2} \int \frac{du}{9 + 16u^2}$.
Factor out $16$ from the denominator:
$I = \frac{1}{2 \times 16} \int \frac{du}{\frac{9}{16} + u^2} = \frac{1}{32} \int \frac{du}{(\frac{3}{4})^2 + u^2}$.
Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = \frac{1}{32} \times \frac{1}{3/4} \tan^{-1}(\frac{u}{3/4}) + C = \frac{1}{32} \times \frac{4}{3} \tan^{-1}(\frac{4u}{3}) + C$.
$I = \frac{1}{24} \tan^{-1}(\frac{4}{3} \tan 2x) + C$.

(A) Let $I = \int \sqrt{\operatorname{cosec} x + 1} dx = \int \sqrt{\frac{1}{\sin x} + 1} dx = \int \sqrt{\frac{1 + \sin x}{\sin x}} dx$.
Using $\sin x = \frac{2 \tan(x/2)}{1 + \tan^2(x/2)}$ and $1 + \sin x = (\sin(x/2) + \cos(x/2))^2$,we substitute $u = \tan(x/2)$,$dx = \frac{2}{1+u^2} du$.
The integral becomes $I = \int \sqrt{\frac{(1+u)^2}{2u}} \cdot \frac{2}{1+u^2} du = \sqrt{2} \int \frac{1+u}{\sqrt{u}(1+u^2)} du$.
Let $v = \sqrt{u}$,then $dv = \frac{1}{2\sqrt{u}} du$,so $du = 2v dv$.
$I = 2\sqrt{2} \int \frac{1+v^2}{1+v^4} dv = \sqrt{2} \int \frac{1+1/v^2}{v^2+1/v^2} dv = \sqrt{2} \int \frac{d(v-1/v)}{(v-1/v)^2 + 2} + \sqrt{2} \int \frac{d(v+1/v)}{(v+1/v)^2 - 2}$.
Evaluating this leads to $I = 2 \tan^{-1}\left(\frac{\sqrt{2\tan(x/2)}}{1-\tan(x/2)}\right) + c$.
Thus,$k = 2$ and $f(x) = \frac{\sqrt{2\tan(x/2)}}{1-\tan(x/2)}$.
For $x = \pi/6$,$\tan(x/2) = \tan(\pi/12) = 2 - \sqrt{3}$.
$f(\pi/6) = \frac{\sqrt{2(2-\sqrt{3})}}{1-(2-\sqrt{3})} = \frac{\sqrt{4-2\sqrt{3}}}{\sqrt{3}-1} = \frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{3}-1} = 1$.
Therefore,$\frac{1}{k} f(\pi/6) = \frac{1}{2} \cdot 1 = \frac{1}{2}$.

(B) To solve the integral $I = \int \frac{1}{x^m (x^m+1)^{1/m}} dx$,factor out $x^m$ from the radical:
$I = \int \frac{1}{x^m \cdot x (1 + x^{-m})^{1/m}} dx = \int \frac{1}{x^{m+1} (1 + x^{-m})^{1/m}} dx$.
Let $u = 1 + x^{-m}$. Then $du = -m x^{-m-1} dx = -m x^{-(m+1)} dx$.
Thus,$dx / x^{m+1} = -du / m$.
Substituting these into the integral:
$I = \int \frac{1}{u^{1/m}} \cdot \left(-\frac{du}{m}\right) = -\frac{1}{m} \int u^{-1/m} du$.
Integrating with respect to $u$:
$I = -\frac{1}{m} \cdot \frac{u^{1 - 1/m}}{1 - 1/m} + C = -\frac{1}{m} \cdot \frac{u^{(m-1)/m}}{(m-1)/m} + C = -\frac{1}{m-1} u^{(m-1)/m} + C$.
Substituting $u = 1 + x^{-m} = \frac{x^m+1}{x^m}$ back:
$I = -\frac{1}{m-1} \left(\frac{x^m+1}{x^m}\right)^{(m-1)/m} + C = -\frac{1}{m-1} \left(\frac{(x^m+1)^{1/m}}{x}\right)^{m-1} + C$.
This matches option $B$.

(D) We have $I = \int \frac{1}{x^4+8x^2+9} dx$. Dividing numerator and denominator by $x^2$,we get $I = \int \frac{1+3/x^2}{x^2+8+9/x^2} dx - \int \frac{3/x^2-1}{x^2+8+9/x^2} dx$ is not direct,so we use:
$I = \frac{1}{6} \int \frac{(1+3/x^2)}{(x-3/x)^2+14} dx - \frac{1}{6} \int \frac{(1-3/x^2)}{(x+3/x)^2+2} dx$.
Let $t = x-3/x$ and $u = x+3/x$. Then $dt = (1+3/x^2) dx$ and $du = (1-3/x^2) dx$.
$I = \frac{1}{6} \int \frac{dt}{t^2+(\sqrt{14})^2} - \frac{1}{6} \int \frac{du}{u^2+(\sqrt{2})^2}$.
$I = \frac{1}{6} \left[ \frac{1}{\sqrt{14}} \tan^{-1} \left( \frac{x-3/x}{\sqrt{14}} \right) - \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x+3/x}{\sqrt{2}} \right) \right] + c$.
Comparing with the given form,$k=6$,$f(x) = \frac{x-3/x}{\sqrt{14}}$,and $g(x) = \frac{x+3/x}{\sqrt{2}}$.
Now,$f(\sqrt{3}) = \frac{\sqrt{3}-3/\sqrt{3}}{\sqrt{14}} = 0$ and $g(1) = \frac{1+3/1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
Thus,$\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)} = \sqrt{\frac{6}{2} + 0 + 2\sqrt{2}} = \sqrt{3+2\sqrt{2}} = \sqrt{(\sqrt{2}+1)^2} = \sqrt{2}+1$.

(A) Let $I = \int \frac{\sec x}{3(\sec x+\tan x)+2} dx$.
Multiply the numerator and denominator by $(\sec x - \tan x)$:
$I = \int \frac{\sec x(\sec x - \tan x)}{3(\sec^2 x - \tan^2 x) + 2(\sec x - \tan x)} dx$.
Since $\sec^2 x - \tan^2 x = 1$,we have:
$I = \int \frac{\sec^2 x - \sec x \tan x}{3 + 2(\sec x - \tan x)} dx$.
Let $u = \sec x - \tan x$. Then $du = (\sec x \tan x - \sec^2 x) dx$,so $-du = (\sec^2 x - \sec x \tan x) dx$.
$I = \int \frac{-du}{3 + 2u} = -\frac{1}{2} \ln|3 + 2u| + C = -\frac{1}{2} \ln|3 + 2(\sec x - \tan x)| + C$.
Using half-angle identities $\sec x = \frac{1+\tan^2(x/2)}{1-\tan^2(x/2)}$ and $\tan x = \frac{2\tan(x/2)}{1-\tan^2(x/2)}$:
$I = -\frac{1}{2} \ln \left| \frac{3(1-\tan^2(x/2)) + 2(1+\tan^2(x/2)) - 4\tan(x/2)}{1-\tan^2(x/2)} \right| + C$.
$I = -\frac{1}{2} \ln \left| \frac{5 - 4\tan(x/2) - \tan^2(x/2)}{(1-\tan(x/2))(1+\tan(x/2))} \right| + C$.
$I = -\frac{1}{2} \ln \left| \frac{(5+\tan(x/2))(1-\tan(x/2))}{(1-\tan(x/2))(1+\tan(x/2))} \right| + C$.
$I = -\frac{1}{2} \ln \left| \frac{5+\tan(x/2)}{1+\tan(x/2)} \right| + C = \frac{1}{2} \ln \left| \frac{1+\tan(x/2)}{5+\tan(x/2)} \right| + C$.

(C) We use the substitution $\tan \frac{x}{2} = t$,which implies $dx = \frac{2 dt}{1 + t^2}$,$\cos x = \frac{1 - t^2}{1 + t^2}$,and $\sin x = \frac{2t}{1 + t^2}$.
Substituting these into the integral:
$\int \frac{1}{3(\frac{1 - t^2}{1 + t^2}) - 4(\frac{2t}{1 + t^2}) + 5} \cdot \frac{2 dt}{1 + t^2}$
$= \int \frac{2 dt}{3(1 - t^2) - 8t + 5(1 + t^2)}$
$= \int \frac{2 dt}{3 - 3t^2 - 8t + 5 + 5t^2}$
$= \int \frac{2 dt}{2t^2 - 8t + 8} = \int \frac{dt}{t^2 - 4t + 4}$
$= \int \frac{dt}{(t - 2)^2} = -(t - 2)^{-1} + c$
$= \frac{-1}{t - 2} + c = \frac{1}{2 - t} + c$
Substituting back $t = \tan \frac{x}{2}$,we get $\frac{1}{2 - \tan \frac{x}{2}} + c$.

(A) To evaluate the integral $I = \int \frac{1}{16-7 \sin^2 x} dx$,divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x}{16 \sec^2 x - 7 \tan^2 x} dx$
Using $\sec^2 x = 1 + \tan^2 x$,we get:
$I = \int \frac{\sec^2 x}{16(1 + \tan^2 x) - 7 \tan^2 x} dx = \int \frac{\sec^2 x}{16 + 9 \tan^2 x} dx$
Let $\tan x = t$,then $\sec^2 x dx = dt$:
$I = \int \frac{dt}{16 + 9t^2} = \frac{1}{9} \int \frac{dt}{\frac{16}{9} + t^2} = \frac{1}{9} \int \frac{dt}{(\frac{4}{3})^2 + t^2}$
Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = \frac{1}{9} \times \frac{1}{4/3} \tan^{-1}\left(\frac{t}{4/3}\right) + C = \frac{1}{9} \times \frac{3}{4} \tan^{-1}\left(\frac{3t}{4}\right) + C$
$I = \frac{1}{12} \tan^{-1}\left(\frac{3 \tan x}{4}\right) + C$

(A) To solve $\int \frac{2x+3}{\sqrt{3x^2-2x+1}} dx$,we express the numerator as a multiple of the derivative of the quadratic expression inside the square root. The derivative of $3x^2-2x+1$ is $6x-2$.
We write $2x+3 = \frac{1}{3}(6x-2) + \frac{11}{3}$.
Thus,the integral becomes:
$\int \frac{2x+3}{\sqrt{3x^2-2x+1}} dx = \frac{1}{3} \int \frac{6x-2}{\sqrt{3x^2-2x+1}} dx + \frac{11}{3} \int \frac{dx}{\sqrt{3(x^2 - \frac{2}{3}x + \frac{1}{3})}}$.
For the first part,let $u = 3x^2-2x+1$,then $du = (6x-2)dx$. The integral is $\frac{1}{3} \int u^{-1/2} du = \frac{2}{3} \sqrt{3x^2-2x+1}$.
For the second part,complete the square: $x^2 - \frac{2}{3}x + \frac{1}{3} = (x-\frac{1}{3})^2 + \frac{2}{9}$.
So,$\frac{11}{3\sqrt{3}} \int \frac{dx}{\sqrt{(x-\frac{1}{3})^2 + (\frac{\sqrt{2}}{3})^2}} = \frac{11}{3\sqrt{3}} \sinh^{-1}\left(\frac{x-1/3}{\sqrt{2}/3}\right) = \frac{11}{3\sqrt{3}} \sinh^{-1}\left(\frac{3x-1}{\sqrt{2}}\right)$.
Combining these,the result is $\frac{2}{3} \sqrt{3x^2-2x+1} + \frac{11}{3\sqrt{3}} \sinh^{-1}\left(\frac{3x-1}{\sqrt{2}}\right) + C$.

(B) We have $I = \int(\sqrt{1-\sin x} + \sqrt{1+\sin x}) \, dx$.
Using $1 \pm \sin x = \left(\cos \frac{x}{2} \pm \sin \frac{x}{2}\right)^2$,we get:
$I = \int \left( \sqrt{(\cos \frac{x}{2} - \sin \frac{x}{2})^2} + \sqrt{(\cos \frac{x}{2} + \sin \frac{x}{2})^2} \right) \, dx = \int \left( |\cos \frac{x}{2} - \sin \frac{x}{2}| + |\cos \frac{x}{2} + \sin \frac{x}{2}| \right) \, dx$.
Given $\frac{3 \pi}{2} < x < \frac{5 \pi}{2}$,we have $\frac{3 \pi}{4} < \frac{x}{2} < \frac{5 \pi}{4}$.
In this interval,$\sin \frac{x}{2} > \cos \frac{x}{2}$ and $\sin \frac{x}{2} + \cos \frac{x}{2} < 0$ (since $\frac{x}{2}$ is in the third quadrant).
Thus,$|\cos \frac{x}{2} - \sin \frac{x}{2}| = \sin \frac{x}{2} - \cos \frac{x}{2}$ and $|\cos \frac{x}{2} + \sin \frac{x}{2}| = -(\cos \frac{x}{2} + \sin \frac{x}{2})$.
$I = \int (\sin \frac{x}{2} - \cos \frac{x}{2} - \cos \frac{x}{2} - \sin \frac{x}{2}) \, dx = \int -2 \cos \frac{x}{2} \, dx = -4 \sin \frac{x}{2} + c$.
So,$f(x) = -4 \sin \frac{x}{2}$.
Then $f\left(\frac{\pi}{3}\right) - f(0) = -4 \sin \frac{\pi}{6} - (-4 \sin 0) = -4 \left(\frac{1}{2}\right) + 0 = -2$.

(A) Let $I = \int \frac{d x}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}$.
We can rewrite the integrand as:
$I = \int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{3}{4}}(x+2)^{\frac{2}{4}}} d x = \int \left(\frac{x-1}{x+2}\right)^{-\frac{3}{4}} \cdot \frac{1}{(x+2)^2} d x$.
Let $t = \frac{x-1}{x+2}$.
Then,$dt = \frac{(x+2)(1) - (x-1)(1)}{(x+2)^2} d x = \frac{3}{(x+2)^2} d x$.
Thus,$\frac{1}{(x+2)^2} d x = \frac{dt}{3}$.
Substituting these into the integral:
$I = \int t^{-\frac{3}{4}} \cdot \frac{dt}{3} = \frac{1}{3} \int t^{-\frac{3}{4}} dt$.
$I = \frac{1}{3} \cdot \frac{t^{\frac{1}{4}}}{\frac{1}{4}} + C = \frac{4}{3} t^{\frac{1}{4}} + C$.
Substituting back $t = \frac{x-1}{x+2}$,we get:
$I = \frac{4}{3} \left(\frac{x-1}{x+2}\right)^{\frac{1}{4}} + C$.

(C) Let $I = \int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{(3-2x)^2} \, dx$.
Substitute $t = \log \left(\frac{2x+3}{3-2x}\right)$,so $\frac{2x+3}{3-2x} = e^t$.
Differentiating both sides with respect to $x$: $\frac{d}{dx} \left( \frac{2x+3}{3-2x} \right) = \frac{(3-2x)(2) - (2x+3)(-2)}{(3-2x)^2} = \frac{6-4x+4x+6}{(3-2x)^2} = \frac{12}{(3-2x)^2}$.
Thus,$\frac{12}{(3-2x)^2} \, dx = e^t \, dt$,which implies $\frac{1}{(3-2x)^2} \, dx = \frac{1}{12} e^t \, dt$.
Substituting into the integral: $I = \int \cos(t) \cdot \frac{1}{12} e^t \, dt = \frac{1}{12} \int e^t \cos(t) \, dt$.
Using the given formula $\int e^t \cos(t) \, dt = \frac{e^t}{2}(\cos t + \sin t)$,we get $I = \frac{1}{12} \cdot \frac{e^t}{2}(\cos t + \sin t) = \frac{e^t}{24}(\cos t + \sin t)$.
Substituting back $t = \log \left(\frac{2x+3}{3-2x}\right)$ and $e^t = \frac{2x+3}{3-2x}$,we have $I = \frac{\frac{2x+3}{3-2x}}{24} \left[ \cos \left( \log \left( \frac{2x+3}{3-2x} \right) \right) + \sin \left( \log \left( \frac{2x+3}{3-2x} \right) \right) \right] + c$.
Comparing with the given form,$f(x) = \frac{2x+3}{3-2x}$ and $g(x) = \log \left( \frac{2x+3}{3-2x} \right)$.
Therefore,$g(x) = \log(f(x))$,which implies $g(1) = \log(f(1))$.

(C) We need to evaluate the integral $I = \int \frac{3 x^6-2 x^4+x^2-2}{x^2+1} d x$.
Performing polynomial long division of the numerator $3x^6 - 2x^4 + x^2 - 2$ by the denominator $x^2 + 1$:
$3x^6 - 2x^4 + x^2 - 2 = (x^2 + 1)(3x^4 - 5x^2 + 6) - 8$.
Thus,the integral becomes:
$I = \int \left( 3x^4 - 5x^2 + 6 - \frac{8}{x^2+1} \right) d x$.
Integrating term by term:
$I = 3 \int x^4 d x - 5 \int x^2 d x + 6 \int 1 d x - 8 \int \frac{1}{x^2+1} d x$.
Using the standard integral formulas:
$I = 3 \left( \frac{x^5}{5} \right) - 5 \left( \frac{x^3}{3} \right) + 6x - 8 \tan^{-1} x + c$.
$I = \frac{3}{5} x^5 - \frac{5}{3} x^3 + 6x - 8 \tan^{-1} x + c$.

(A) Given $f(x) = \int \left[ \tan^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{(2 \sin x \cos x)^2} \right] dx$.
Since $\sin^2 2x = 4 \sin^2 x \cos^2 x$,the expression becomes $\int \left[ \tan^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{4 \sin^2 x \cos^2 x} \right] dx$.
$= \int \left[ \tan^2 x + \cot^2 x + \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} \right] dx$.
$= \int \left[ (\sec^2 x - 1) + (\csc^2 x - 1) + \sec x \tan x + \csc x \cot x \right] dx$.
$= \tan x - x - \cot x - x + \sec x - \csc x + C$.
$f(x) = \tan x - \cot x + \sec x - \csc x - 2x + C$.
Given $f\left(\frac{\pi}{4}\right) = 0$: $\tan\frac{\pi}{4} - \cot\frac{\pi}{4} + \sec\frac{\pi}{4} - \csc\frac{\pi}{4} - 2\left(\frac{\pi}{4}\right) + C = 0$.
$1 - 1 + \sqrt{2} - \sqrt{2} - \frac{\pi}{2} + C = 0 \implies C = \frac{\pi}{2}$.
So $f(x) = \tan x - \cot x + \sec x - \csc x - 2x + \frac{\pi}{2}$.
Now $f\left(\frac{\pi}{6}\right) = \tan\frac{\pi}{6} - \cot\frac{\pi}{6} + \sec\frac{\pi}{6} - \csc\frac{\pi}{6} - 2\left(\frac{\pi}{6}\right) + \frac{\pi}{2}$.
$= \frac{1}{\sqrt{3}} - \sqrt{3} + \frac{2}{\sqrt{3}} - 2 - \frac{\pi}{3} + \frac{\pi}{2} = \frac{3}{\sqrt{3}} - \sqrt{3} - 2 + \frac{\pi}{6} = \sqrt{3} - \sqrt{3} - 2 + \frac{\pi}{6} = \frac{\pi}{6} - 2$.
Then $3 \left[ f\left(\frac{\pi}{6}\right) + 2 \right] = 3 \left[ \frac{\pi}{6} - 2 + 2 \right] = 3 \left( \frac{\pi}{6} \right) = \frac{\pi}{2}$.

(A) Let $I = \int \frac{d x}{\sin x + \cos x}$.
Multiply and divide by $\sqrt{2}$:
$I = \frac{1}{\sqrt{2}} \int \frac{d x}{\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x}$.
Using $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we get:
$I = \frac{1}{\sqrt{2}} \int \frac{d x}{\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(x + \frac{\pi}{4}\right)}$.
Using $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
$I = \frac{1}{\sqrt{2}} \int \frac{d x}{2 \sin \left(\frac{x}{2} + \frac{\pi}{8}\right) \cos \left(\frac{x}{2} + \frac{\pi}{8}\right)}$.
Divide numerator and denominator by $\cos^2 \left(\frac{x}{2} + \frac{\pi}{8}\right)$:
$I = \frac{1}{2 \sqrt{2}} \int \frac{\sec^2 \left(\frac{x}{2} + \frac{\pi}{8}\right) d x}{\tan \left(\frac{x}{2} + \frac{\pi}{8}\right)}$.
Let $t = \tan \left(\frac{x}{2} + \frac{\pi}{8}\right)$,then $dt = \frac{1}{2} \sec^2 \left(\frac{x}{2} + \frac{\pi}{8}\right) d x$,so $\sec^2 \left(\frac{x}{2} + \frac{\pi}{8}\right) d x = 2 dt$.
$I = \frac{1}{2 \sqrt{2}} \int \frac{2 dt}{t} = \frac{1}{\sqrt{2}} \int \frac{dt}{t} = \frac{1}{\sqrt{2}} \log |t| + C$.
Substituting $t$ back:
$I = \frac{1}{\sqrt{2}} \log \left| \tan \left(\frac{x}{2} + \frac{\pi}{8}\right) \right| + C$.

(D) Let $I = \int \left( \frac{\sqrt{1+x+x^2}}{1+x} + \frac{1}{2 \sqrt{1+x+x^2}} - \frac{1}{(1+x) \sqrt{1+x+x^2}} \right) dx$.
Combine the first and third terms:
$I = \int \left( \frac{1+x+x^2-1}{(1+x) \sqrt{1+x+x^2}} \right) dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx$.
Simplify the numerator:
$I = \int \frac{x(1+x)}{(1+x) \sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx = \int \frac{x}{\sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx$.
Multiply and divide the first integral by $2$:
$I = \int \frac{2x}{2 \sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx = \int \frac{2x+1-1}{2 \sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx$.
Split the integral:
$I = \int \frac{2x+1}{2 \sqrt{1+x+x^2}} dx - \int \frac{1}{2 \sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx = \int \frac{2x+1}{2 \sqrt{1+x+x^2}} dx$.
Let $t = x^2+x+1$,then $dt = (2x+1) dx$.
$I = \frac{1}{2} \int \frac{1}{\sqrt{t}} dt = \sqrt{t} + C = \sqrt{x^2+x+1} + C$.

(B) Given the integral $I = \int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} d x$.
Divide the numerator and denominator by $x^{15}$ inside the integral:
$I = \int \frac{2 x^{-3} + 5 x^{-6}}{(1 + x^{-2} + x^{-5})^3} d x$.
Let $t = 1 + x^{-2} + x^{-5}$.
Then $dt = (-2x^{-3} - 5x^{-6}) dx$,which implies $-(2x^{-3} + 5x^{-6}) dx = dt$.
Substituting this into the integral:
$I = -\int \frac{dt}{t^3} = -\int t^{-3} dt = -\frac{t^{-2}}{-2} + C = \frac{1}{2t^2} + C$.
Comparing this with $\frac{1}{2} f(x) + C$,we get $f(x) = \frac{1}{t^2} = \frac{1}{(1 + x^{-2} + x^{-5})^2} = \frac{x^{10}}{(x^5 + x^3 + 1)^2}$.
Now,calculate $f(1) - f(0)$.
$f(1) = \frac{1^10}{(1^5 + 1^3 + 1)^2} = \frac{1}{3^2} = \frac{1}{9}$.
For $f(0)$,the expression $\frac{x^{10}}{(x^5 + x^3 + 1)^2}$ at $x=0$ is $0$.
Thus,$f(1) - f(0) = \frac{1}{9} - 0 = \frac{1}{9}$.

(C) Let $I = \int \frac{dx}{(x^2 - a^2)^{3/2}}$.
Substitute $x = a \sec \theta$,then $dx = a \sec \theta \tan \theta d\theta$.
Substituting these into the integral:
$I = \int \frac{a \sec \theta \tan \theta d\theta}{(a^2 \tan^2 \theta)^{3/2}} = \int \frac{a \sec \theta \tan \theta d\theta}{a^3 \tan^3 \theta} = \frac{1}{a^2} \int \frac{\sec \theta}{\tan^2 \theta} d\theta$.
Using $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get:
$I = \frac{1}{a^2} \int \frac{1}{\cos \theta} \cdot \frac{\cos^2 \theta}{\sin^2 \theta} d\theta = \frac{1}{a^2} \int \frac{\cos \theta}{\sin^2 \theta} d\theta$.
Let $u = \sin \theta$,then $du = \cos \theta d\theta$. So,
$I = \frac{1}{a^2} \int u^{-2} du = \frac{1}{a^2} (-u^{-1}) + C = -\frac{1}{a^2 \sin \theta} + C$.
Since $x = a \sec \theta$,$\sec \theta = \frac{x}{a}$,so $\cos \theta = \frac{a}{x}$.
Then $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{a^2}{x^2}} = \frac{\sqrt{x^2 - a^2}}{x}$.
Substituting $\sin \theta$ back,we get:
$I = -\frac{1}{a^2 \left( \frac{\sqrt{x^2 - a^2}}{x} \right)} + C = -\frac{x}{a^2 \sqrt{x^2 - a^2}} + C$.

(D) We have $5+2 x+x^2 = (x+1)^2 + 4$.
Let $x+1 = z$,then $dx = dz$.
Substituting this into the integral,we get $I = \int \frac{dz}{(z^2+4)^{3/2}}$.
Let $z = 2 \tan \theta$,then $dz = 2 \sec^2 \theta \ d\theta$.
Also,$(z^2+4)^{3/2} = (4 \tan^2 \theta + 4)^{3/2} = (4 \sec^2 \theta)^{3/2} = 8 \sec^3 \theta$.
Therefore,$I = \int \frac{2 \sec^2 \theta \ d\theta}{8 \sec^3 \theta} = \frac{1}{4} \int \cos \theta \ d\theta = \frac{1}{4} \sin \theta + C$.
Since $\tan \theta = \frac{z}{2}$,we have $\sin \theta = \frac{z}{\sqrt{z^2+4}}$.
Thus,$I = \frac{1}{4} \cdot \frac{z}{\sqrt{z^2+4}} + C = \frac{x+1}{4 \sqrt{(x+1)^2+4}} + C = \frac{x+1}{4 \sqrt{5+2x+x^2}} + C$.

(D) Let $I = \int \frac{1-(\cot x)^{2019}}{\tan x+(\cot x)^{2020}} dx$.
Multiplying numerator and denominator by $\sin x \cos^{2019} x$,we simplify the integrand.
Alternatively,notice that $\frac{d}{dx}(\sin^{2021} x + \cos^{2021} x) = 2021 \sin^{2020} x \cos x - 2021 \cos^{2020} x \sin x = 2021 \sin x \cos x (\sin^{2019} x - \cos^{2019} x)$.
Thus,the integral evaluates to $\frac{1}{2021} \ln |\sin^{2021} x + \cos^{2021} x| + c$.
Comparing with the given form,$n = 2021$,$f(x) = \sin x$,and $g(x) = \cos x$.
We need to evaluate $n[(f(x))^4 + (g(x))^4]_{x=\frac{\pi}{3}} = 2021 [\sin^4(\frac{\pi}{3}) + \cos^4(\frac{\pi}{3})]$.
$= 2021 [(\frac{\sqrt{3}}{2})^4 + (\frac{1}{2})^4] = 2021 [\frac{9}{16} + \frac{1}{16}] = 2021 [\frac{10}{16}] = 2021 [\frac{5}{8}] = \frac{10105}{8}$.

(B) We have,$\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41} \log |b \sin x+5 \cos x|+C$.
Let $a \cos x-2 \sin x = \lambda(b \sin x+5 \cos x) + \mu(b \cos x-5 \sin x)$.
Equating coefficients of $\sin x$ and $\cos x$:
$a = 5\lambda + \mu b$ and $-2 = b\lambda - 5\mu$.
Given $\lambda = \frac{7}{41}$ and $\mu = \frac{22}{41}$,we have:
$a = 5(\frac{7}{41}) + b(\frac{22}{41}) = \frac{35+22b}{41}$ and $-2 = b(\frac{7}{41}) - 5(\frac{22}{41}) \Rightarrow -82 = 7b - 110 \Rightarrow 7b = 28 \Rightarrow b = 4$.
Substituting $b=4$ into $a$: $a = \frac{35+22(4)}{41} = \frac{35+88}{41} = \frac{123}{41} = 3$.
Now,$\int \frac{d x}{b+a \cos x} = \int \frac{d x}{4+3 \cos x}$.
Using the substitution $\tan \frac{x}{2} = t$,$\cos x = \frac{1-t^2}{1+t^2}$ and $dx = \frac{2 dt}{1+t^2}$:
$\int \frac{2 dt}{(1+t^2)(4+3(\frac{1-t^2}{1+t^2}))} = \int \frac{2 dt}{4+4t^2+3-3t^2} = \int \frac{2 dt}{7+t^2}$.
$= \frac{2}{\sqrt{7}} \tan^{-1}(\frac{t}{\sqrt{7}}) + C = \frac{2}{\sqrt{7}} \tan^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{7}}\right) + C$.

(D) Let $I = \int e^{\sin ^2 x}(\sin x \cos x + \cos ^3 x \sin x) d x$.
Factor out $\sin x \cos x$:
$I = \int e^{\sin ^2 x}(1 + \cos ^2 x) \sin x \cos x d x$.
Since $\cos ^2 x = 1 - \sin ^2 x$,we have:
$I = \int e^{\sin ^2 x}(1 + 1 - \sin ^2 x) \sin x \cos x d x = \int e^{\sin ^2 x}(2 - \sin ^2 x) \sin x \cos x d x$.
Let $t = \sin ^2 x$,then $dt = 2 \sin x \cos x d x$,so $\sin x \cos x d x = \frac{dt}{2}$.
Substituting into the integral:
$I = \frac{1}{2} \int e^t(2 - t) d t = \int e^t d t - \frac{1}{2} \int t e^t d t$.
Using integration by parts for $\int t e^t d t = t e^t - e^t$:
$I = e^t - \frac{1}{2}(t e^t - e^t) + C = e^t(1 - \frac{t}{2} + \frac{1}{2}) + C = e^t(\frac{3}{2} - \frac{t}{2}) + C$.
Comparing with $e^{\sin ^2 x}(1 + f(x)) + C$,we have $1 + f(x) = \frac{3}{2} - \frac{\sin ^2 x}{2}$.
Thus,$f(x) = \frac{1}{2} - \frac{\sin ^2 x}{2}$.
Taking the derivative: $f^{\prime}(x) = 0 - \frac{1}{2}(2 \sin x \cos x) = -\sin x \cos x = -\frac{1}{2} \sin 2 x$.

(B) Let $I = \int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}}$.
Substitute $x-2 = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$ and $x = 2 + \frac{1}{t}$.
Substituting these into the integral:
$I = \int \frac{-1/t^2 dt}{(1/t) \sqrt{(2+1/t)^2 - 3(2+1/t) + 5}}$
$I = -\int \frac{dt/t}{\sqrt{4 + 4/t + 1/t^2 - 6 - 3/t + 5}}$
$I = -\int \frac{dt}{\sqrt{1/t^2 + 1/t + 3}} = -\int \frac{dt}{\sqrt{(1 + t + 3t^2)/t^2}} = -\int \frac{dt}{\sqrt{3t^2 + t + 1}}$
$I = -\frac{1}{\sqrt{3}} \int \frac{dt}{\sqrt{t^2 + t/3 + 1/3}} = -\frac{1}{\sqrt{3}} \int \frac{dt}{\sqrt{(t + 1/6)^2 + 11/36}}$
Using the formula $\int \frac{du}{\sqrt{u^2 + a^2}} = \sinh^{-1}(\frac{u}{a}) + C$:
$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{t + 1/6}{\sqrt{11}/6} \right) + C = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6t + 1}{\sqrt{11}} \right) + C$
Since $t = \frac{1}{x-2}$,we have:
$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6/(x-2) + 1}{\sqrt{11}} \right) + C = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{6 + x - 2}{\sqrt{11}(x-2)} \right) + C$
$I = -\frac{1}{\sqrt{3}} \sinh^{-1}\left( \frac{x + 4}{\sqrt{11}(x-2)} \right) + C$.

(B) We have,$I = \int \frac{(x-1) dx}{(x+1) \sqrt{x^3+x^2+x}}$.
Dividing numerator and denominator by $x$,we get $I = \int \frac{(1 - 1/x) dx}{(1 + 1/x) \sqrt{x + 1 + 1/x}}$.
Let $t = \sqrt{x + 1 + 1/x}$. Then $t^2 = x + 1 + 1/x$.
Differentiating both sides,$2t dt = (1 - 1/x^2) dx = \frac{x^2-1}{x^2} dx = \frac{(x-1)(x+1)}{x^2} dx$.
Also,$1 + 1/x = \frac{x+1}{x}$.
Substituting these into the integral,we get $I = \int \frac{2t dt}{t^2 \cdot (t^2-1) \cdot \frac{x}{x+1} \cdot \frac{x+1}{x}} = \int \frac{2 dt}{t^2+1} = 2 \tan^{-1}(t) + C$.
Thus,$I = 2 \tan^{-1} \sqrt{x + 1 + 1/x} + C$.
Comparing with $A \tan^{-1} \sqrt{f(x)} + C$,we get $A = 2$ and $f(x) = x + 1 + 1/x$.
Then $f(-1) = -1 + 1 + (1/-1) = -1$.
Therefore,the ordered pair $(A, f(-1)) = (2, -1)$.

(B) Let $I = \int \frac{\sqrt{\cos 2x}}{\sin x} dx$.
We know that $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$.
So,$I = \int \frac{\sqrt{1-\tan^2 x}}{\sec x \sin x} dx = \int \frac{\sqrt{1-\tan^2 x}}{\tan x} dx$.
Let $1-\tan^2 x = t^2$,then $-2\tan x \sec^2 x dx = 2t dt$,which implies $dx = \frac{-t dt}{\tan x (1+\tan^2 x)} = \frac{-t dt}{\sqrt{1-t^2}(2-t^2)}$.
Substituting these,$I = \int \frac{t}{\sqrt{1-t^2}} \cdot \frac{-t dt}{\sqrt{1-t^2}(2-t^2)} = -\int \frac{t^2}{(1-t^2)(2-t^2)} dt$.
Using partial fractions,$\frac{t^2}{(1-t^2)(2-t^2)} = \frac{2}{t^2-2} - \frac{1}{t^2-1}$.
Thus,$I = -\int \left( \frac{2}{t^2-2} - \frac{1}{t^2-1} \right) dt = 2 \int \frac{1}{2-t^2} dt + \int \frac{1}{t^2-1} dt$.
Using standard integrals,$I = 2 \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{2}+t}{\sqrt{2}-t} \right| + \frac{1}{2} \log \left| \frac{t-1}{t+1} \right| + c$.
Substituting $t = \sqrt{1-\tan^2 x}$,we get $I = \frac{1}{\sqrt{2}} \log \left| \frac{\sqrt{2}+\sqrt{1-\tan^2 x}}{\sqrt{2}-\sqrt{1-\tan^2 x}} \right| - \frac{1}{2} \log \left| \frac{1+\sqrt{1-\tan^2 x}}{1-\sqrt{1-\tan^2 x}} \right| + c$.

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

(A) Let $I = \int \frac{dx}{(1+x) \sqrt{8+7x-x^2}}$.
Factor the quadratic expression: $8+7x-x^2 = (8-x)(1+x)$.
So,$I = \int \frac{dx}{(1+x) \sqrt{(8-x)(1+x)}} = \int \frac{dx}{(1+x)^{3/2} \sqrt{8-x}} = \int \frac{dx}{(1+x)^2 \sqrt{\frac{8-x}{1+x}}}$.
Let $t^2 = \frac{8-x}{1+x}$.
Differentiating both sides with respect to $x$:
$2t \frac{dt}{dx} = \frac{(1+x)(-1) - (8-x)(1)}{(1+x)^2} = \frac{-1-x-8+x}{(1+x)^2} = \frac{-9}{(1+x)^2}$.
Thus,$\frac{dx}{(1+x)^2} = -\frac{2}{9} t \, dt$.
Substituting these into the integral:
$I = \int \frac{1}{t} \left(-\frac{2}{9} t \, dt\right) = -\frac{2}{9} \int dt = -\frac{2}{9} t + c$.
Substituting $t = \sqrt{\frac{8-x}{1+x}}$ back:
$I = -\frac{2}{9} \sqrt{\frac{8-x}{1+x}} + c$.

(B) Let $I = \int \frac{3 \sin x + 5 \cos x + 4}{\sin x + \cos x + 2} dx$.
We express the numerator as $A(\sin x + \cos x + 2) + B \frac{d}{dx}(\sin x + \cos x + 2) + C$.
$3 \sin x + 5 \cos x + 4 = A(\sin x + \cos x + 2) + B(\cos x - \sin x) + C$.
Equating coefficients:
$A - B = 3$
$A + B = 5$
$2A + C = 4$
Solving these,we get $2A = 8 \Rightarrow A = 4$,$B = 1$,and $C = 4 - 2(4) = -4$.
Thus,$I = \int 4 dx + \int \frac{\cos x - \sin x}{\sin x + \cos x + 2} dx - 4 \int \frac{dx}{\sin x + \cos x + 2}$.
$I = 4x + \log|\sin x + \cos x + 2| - 4 \int \frac{dx}{\sin x + \cos x + 2}$.
Using the substitution $\tan(x/2) = t$,$dx = \frac{2 dt}{1+t^2}$,$\sin x = \frac{2t}{1+t^2}$,$\cos x = \frac{1-t^2}{1+t^2}$.
$\int \frac{dx}{\sin x + \cos x + 2} = \int \frac{2 dt / (1+t^2)}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} + 2} = \int \frac{2 dt}{2t + 1 - t^2 + 2 + 2t^2} = \int \frac{2 dt}{t^2 + 2t + 3} = \int \frac{2 dt}{(t+1)^2 + 2}$.
$= 2 \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t+1}{\sqrt{2}}\right) = \sqrt{2} \tan^{-1}\left(\frac{\tan(x/2) + 1}{\sqrt{2}}\right)$.
Therefore,$I = 4x + \log|\sin x + \cos x + 2| - 4\sqrt{2} \tan^{-1}\left(\frac{\tan(x/2) + 1}{\sqrt{2}}\right) + c$.

(C) $I = \int \frac{dx}{4 \sin x + 3 \cos x}$
Let $3 = r \cos \theta$ and $4 = r \sin \theta$.
Squaring and adding,we get $r^2 = 3^2 + 4^2 = 25$,so $r = 5$.
Dividing,we get $\tan \theta = \frac{4}{3}$,so $\theta = \tan^{-1} \frac{4}{3}$.
The integral becomes $I = \int \frac{dx}{r(\sin x \cos \theta + \cos x \sin \theta)} = \int \frac{dx}{r \sin(x + \theta)}$.
Alternatively,using $4 \sin x + 3 \cos x = 5(\frac{4}{5} \sin x + \frac{3}{5} \cos x) = 5 \cos(x - \theta)$ where $\cos \theta = \frac{3}{5}$ and $\sin \theta = \frac{4}{5}$,which implies $\theta = \tan^{-1} \frac{4}{3}$.
Thus,$I = \int \frac{dx}{5 \cos(x - \theta)} = \frac{1}{5} \int \sec(x - \theta) dx$.
Using the formula $\int \sec u du = \log |\sec u + \tan u| + c$,we get:
$I = \frac{1}{5} \log |\sec(x - \theta) + \tan(x - \theta)| + c$.
Substituting $\theta = \tan^{-1} \frac{4}{3}$,we get $I = \frac{1}{5} \log |\sec(x - \tan^{-1} \frac{4}{3}) + \tan(x - \tan^{-1} \frac{4}{3})| + c$.

(A) Let $I = \int \frac{dx}{(2ax+x^2)^{3/2}}$.
We can rewrite the expression inside the integral as:
$I = \int \frac{dx}{(x^2+2ax)^{3/2}} = \int \frac{dx}{[x(x+2a)]^{3/2}} = \int \frac{dx}{x^{3/2}(x+2a)^{3/2}}$.
Alternatively,complete the square inside the bracket:
$2ax+x^2 = (x+a)^2 - a^2$.
Let $x+a = a \sec \theta$,then $dx = a \sec \theta \tan \theta d\theta$.
Also,$(x+a)^2 - a^2 = a^2 \sec^2 \theta - a^2 = a^2 \tan^2 \theta$.
Substituting these into the integral:
$I = \int \frac{a \sec \theta \tan \theta d\theta}{(a^2 \tan^2 \theta)^{3/2}} = \int \frac{a \sec \theta \tan \theta d\theta}{a^3 \tan^3 \theta} = \frac{1}{a^2} \int \frac{\sec \theta}{\tan^2 \theta} d\theta$.
$I = \frac{1}{a^2} \int \frac{1/\cos \theta}{\sin^2 \theta / \cos^2 \theta} d\theta = \frac{1}{a^2} \int \frac{\cos \theta}{\sin^2 \theta} d\theta$.
Let $u = \sin \theta$,then $du = \cos \theta d\theta$.
$I = \frac{1}{a^2} \int u^{-2} du = \frac{1}{a^2} (-u^{-1}) + c = -\frac{1}{a^2 \sin \theta} + c$.
Since $\sec \theta = \frac{x+a}{a}$,we have $\cos \theta = \frac{a}{x+a}$.
Then $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{a^2}{(x+a)^2}} = \sqrt{\frac{(x+a)^2 - a^2}{(x+a)^2}} = \frac{\sqrt{x^2+2ax}}{x+a}$.
Therefore,$I = -\frac{1}{a^2} \cdot \frac{x+a}{\sqrt{x^2+2ax}} + c$.

(D) Let $I = \int(\sqrt{\tan x}+\sqrt{\cot x}) d x$.
Expressing in terms of $\sin x$ and $\cos x$:
$I = \int \left(\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}}\right) d x = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} d x$.
Multiply and divide by $\sqrt{2}$:
$I = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} d x$.
Since $1 - (\sin x - \cos x)^2 = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x) = 2 \sin x \cos x$:
$I = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} d x$.
Let $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) d x$.
Substituting these into the integral:
$I = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} \sin^{-1}(t) + c$.
Substituting back $t = \sin x - \cos x$:
$I = \sqrt{2} \sin^{-1}(\sin x - \cos x) + c$.

(A) We have,$\int \frac{x^4+1}{x^6+1} dx = A \tan^{-1} x + B \tan^{-1} x^3 + c$.
Let $I = \int \frac{x^4+1}{x^6+1} dx = \int \frac{(x^4-x^2+1) + x^2}{(x^2)^3 + 1^3} dx$.
Using the identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$,we have $x^6+1 = (x^2+1)(x^4-x^2+1)$.
Thus,$I = \int \frac{x^4-x^2+1}{(x^2+1)(x^4-x^2+1)} dx + \int \frac{x^2}{x^6+1} dx$.
$I = \int \frac{1}{x^2+1} dx + \int \frac{x^2}{(x^3)^2+1} dx$.
$I = \tan^{-1} x + \int \frac{x^2}{(x^3)^2+1} dx$.
Let $x^3 = t$,then $3x^2 dx = dt$,so $x^2 dx = \frac{1}{3} dt$.
$I = \tan^{-1} x + \frac{1}{3} \int \frac{1}{t^2+1} dt = \tan^{-1} x + \frac{1}{3} \tan^{-1} t + c$.
$I = \tan^{-1} x + \frac{1}{3} \tan^{-1} x^3 + c$.
Comparing with $A \tan^{-1} x + B \tan^{-1} x^3 + c$,we get $A=1$ and $B=\frac{1}{3}$.

(A) Let $I = \int \frac{x^2+1}{x^4+7 x^2+1} d x$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}} d x = \int \frac{1+\frac{1}{x^2}}{(x^2+\frac{1}{x^2}+2)+5} d x$.
$I = \int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+9} d x$.
Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^2}) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2 + 3^2} = \frac{1}{3} \tan^{-1}(\frac{t}{3}) + C$.
Substituting $t = x - \frac{1}{x}$ back:
$I = \frac{1}{3} \tan^{-1}(\frac{x - \frac{1}{x}}{3}) + C = \frac{1}{3} \tan^{-1}(\frac{x^2-1}{3x}) + C$.

(D) Let $I = \int \sqrt{\frac{2+x}{2-x}} \, dx$.
Multiplying the numerator and denominator inside the square root by $\sqrt{2+x}$:
$I = \int \frac{2+x}{\sqrt{(2-x)(2+x)}} \, dx = \int \frac{2+x}{\sqrt{4-x^2}} \, dx$.
Splitting the integral into two parts:
$I = 2 \int \frac{1}{\sqrt{2^2-x^2}} \, dx + \int \frac{x}{\sqrt{4-x^2}} \, dx$.
For the first part,use the standard integral $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}(\frac{x}{a})$.
For the second part,let $u = 4-x^2$,so $du = -2x \, dx$ or $x \, dx = -\frac{1}{2} du$.
$I = 2 \sin^{-1}\left(\frac{x}{2}\right) - \frac{1}{2} \int u^{-1/2} \, du = 2 \sin^{-1}\left(\frac{x}{2}\right) - \frac{1}{2} (2u^{1/2}) + C$.
$I = 2 \sin^{-1}\left(\frac{x}{2}\right) - \sqrt{4-x^2} + C$.

(C) Let $I = \int \frac{dx}{x^2 \sqrt{4+x^2}}$.
Substitute $x = 2 \tan \theta$,so $dx = 2 \sec^2 \theta \ d\theta$.
The integral becomes $I = \int \frac{2 \sec^2 \theta \ d\theta}{(4 \tan^2 \theta) \sqrt{4 + 4 \tan^2 \theta}}$.
Since $\sqrt{4(1+\tan^2 \theta)} = 2 \sec \theta$,we have:
$I = \int \frac{2 \sec^2 \theta \ d\theta}{4 \tan^2 \theta \cdot 2 \sec \theta} = \frac{1}{4} \int \frac{\sec \theta}{\tan^2 \theta} \ d\theta$.
$I = \frac{1}{4} \int \frac{1/\cos \theta}{\sin^2 \theta / \cos^2 \theta} \ d\theta = \frac{1}{4} \int \frac{\cos \theta}{\sin^2 \theta} \ d\theta$.
Let $u = \sin \theta$,then $du = \cos \theta \ d\theta$.
$I = \frac{1}{4} \int u^{-2} \ du = \frac{1}{4} (-u^{-1}) + C = -\frac{1}{4 \sin \theta} + C$.
Since $\tan \theta = \frac{x}{2}$,we have $\sin \theta = \frac{x}{\sqrt{x^2+4}}$.
Thus,$I = -\frac{1}{4} \cdot \frac{\sqrt{x^2+4}}{x} + C = -\frac{\sqrt{4+x^2}}{4x} + C$.

(D) Let $I = \int (1 + x - x^{-1}) e^{x + x^{-1}} dx$.
We can rewrite the integral as:
$I = \int e^{x + x^{-1}} dx + \int (x - x^{-1}) e^{x + x^{-1}} dx$.
Using integration by parts on the first term $\int e^{x + x^{-1}} dx$:
Let $u = x$ and $dv = e^{x + x^{-1}} dx$. This is not straightforward.
Instead,consider the derivative of $x e^{x + x^{-1}}$:
$\frac{d}{dx} (x e^{x + x^{-1}}) = 1 \cdot e^{x + x^{-1}} + x \cdot e^{x + x^{-1}} \cdot (1 - x^{-2}) = e^{x + x^{-1}} + x e^{x + x^{-1}} - x^{-1} e^{x + x^{-1}} = (1 + x - x^{-1}) e^{x + x^{-1}}$.
Therefore,$\int (1 + x - x^{-1}) e^{x + x^{-1}} dx = x e^{x + x^{-1}} + C$.

(A) To evaluate the integral $I = \int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}$,divide both the numerator and the denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x dx}{a^2 \tan^2 x + b^2}$
Now,let $u = \tan x$,then $du = \sec^2 x dx$:
$I = \int \frac{du}{a^2 u^2 + b^2} = \frac{1}{a^2} \int \frac{du}{u^2 + (b/a)^2}$
Using the standard integral formula $\int \frac{dx}{x^2 + k^2} = \frac{1}{k} \tan^{-1}(\frac{x}{k}) + C$:
$I = \frac{1}{a^2} \cdot \frac{1}{b/a} \tan^{-1}\left(\frac{u}{b/a}\right) + C$
$I = \frac{1}{ab} \tan^{-1}\left(\frac{a \tan x}{b}\right) + C$

(B) Let $I = \int \frac{dx}{4+3 \cot x} = \int \frac{\sin x dx}{4 \sin x + 3 \cos x}$.
We express the numerator as $A(4 \cos x - 3 \sin x) + B(4 \sin x + 3 \cos x)$,where $4 \cos x - 3 \sin x$ is the derivative of the denominator $4 \sin x + 3 \cos x$.
Equating coefficients of $\sin x$ and $\cos x$:
$4B - 3A = 1$ and $3B + 4A = 0$.
From $3B + 4A = 0$,we get $A = -\frac{3B}{4}$.
Substituting into the first equation: $4B - 3(-\frac{3B}{4}) = 1 \Rightarrow 4B + \frac{9B}{4} = 1 \Rightarrow \frac{25B}{4} = 1 \Rightarrow B = \frac{4}{25}$.
Then $A = -\frac{3}{25}$.
Thus,$I = \int \frac{-\frac{3}{25}(4 \cos x - 3 \sin x) + \frac{4}{25}(4 \sin x + 3 \cos x)}{4 \sin x + 3 \cos x} dx$.
$I = -\frac{3}{25} \int \frac{4 \cos x - 3 \sin x}{4 \sin x + 3 \cos x} dx + \frac{4}{25} \int dx$.
$I = -\frac{3}{25} \ln |4 \sin x + 3 \cos x| + \frac{4}{25} x + C$.

(A) We are given the integral $I = \int \sin (101 x)(\sin x)^{99} d x$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we have $\sin(100x + x) = \sin(100x)\cos x + \cos(100x)\sin x$.
Substituting this into the integral:
$I = \int (\sin(100x)\cos x + \cos(100x)\sin x)(\sin x)^{99} d x$
$I = \int \sin(100x) \cos x (\sin x)^{99} d x + \int \cos(100x) (\sin x)^{100} d x$.
Applying integration by parts to the first integral $\int \sin(100x) \cdot (\cos x (\sin x)^{99}) d x$:
Let $u = \sin(100x)$ and $dv = \cos x (\sin x)^{99} d x$.
Then $du = 100 \cos(100x) d x$ and $v = \frac{(\sin x)^{100}}{100}$.
Using $\int u dv = uv - \int v du$:
$I = \sin(100x) \cdot \frac{(\sin x)^{100}}{100} - \int \frac{(\sin x)^{100}}{100} \cdot 100 \cos(100x) d x + \int \cos(100x) (\sin x)^{100} d x$.
$I = \frac{\sin(100x)(\sin x)^{100}}{100} - \int \cos(100x)(\sin x)^{100} d x + \int \cos(100x)(\sin x)^{100} d x + C$.
$I = \frac{\sin(100x)(\sin x)^{100}}{100} + C$.
Comparing this with the given form $\frac{\sin (100 x)(\sin x)^\lambda}{\mu}+c$,we get $\lambda = 100$ and $\mu = 100$.
Therefore,$\frac{\lambda}{\mu} = \frac{100}{100} = 1$.

(C) Let $I = \int \frac{\sin x \sec^2 x - \tan x \sin x + \cos x}{1 - \cos 2x} dx$.
Using $1 - \cos 2x = 2 \sin^2 x$,we get:
$I = \int \frac{\sin x \sec^2 x - \tan x \sin x + \cos x}{2 \sin^2 x} dx$
$I = \frac{1}{2} \int \left( \frac{\sin x \sec^2 x}{\sin^2 x} - \frac{\tan x \sin x}{\sin^2 x} + \frac{\cos x}{\sin^2 x} \right) dx$
$I = \frac{1}{2} \int \left( \sec x \tan x - \sec x + \operatorname{cosec} x \cot x \right) dx$
Integrating term by term:
$\int \sec x \tan x dx = \sec x$
$\int \sec x dx = \log |\sec x + \tan x| = \log |\tan(\frac{\pi}{4} + \frac{x}{2})|$
$\int \operatorname{cosec} x \cot x dx = -\operatorname{cosec} x$
Thus,$I = \frac{1}{2} [\sec x - \log |\tan(\frac{\pi}{4} + \frac{x}{2})| - \operatorname{cosec} x] + c$
$I = \frac{1}{2} [\sec x - \operatorname{cosec} x - \log |\tan(\frac{\pi}{4} + \frac{x}{2})|] + c$
Since $\tan(\frac{x}{2}) = \frac{\sin x}{1 + \cos x}$,the expression simplifies to option $(C)$.

(D) Let $I = \int x[\log (1+x)]^3 dx$.
Put $\log (1+x) = t$,so $1+x = e^t$ and $dx = e^t dt$.
$I = \int (e^t - 1) t^3 e^t dt = \int (e^{2t} t^3 - e^t t^3) dt$.
Using integration by parts,$\int e^{at} t^n dt = \frac{e^{at}}{a} [t^n - \frac{n t^{n-1}}{a} + \frac{n(n-1) t^{n-2}}{a^2} - \dots]$.
$I = \frac{e^{2t}}{2} [t^3 - \frac{3t^2}{2} + \frac{6t}{4} - \frac{6}{8}] - e^t [t^3 - 3t^2 + 6t - 6] + C$.
Substituting $e^t = 1+x$ and $t = \log (1+x)$:
$I = \frac{(1+x)^2}{16} [8t^3 - 12t^2 + 12t - 6] - (1+x) [t^3 - 3t^2 + 6t - 6] + C$.
Comparing with $\frac{(1+x)^2}{16} f(x) + (1+x) g(x)$,we get $f(x) = 8t^3 - 12t^2 + 12t - 6$ and $g(x) = -(t^3 - 3t^2 + 6t - 6) = -t^3 + 3t^2 - 6t + 6$.
Thus,$f(x) + g(x) = 7t^3 - 9t^2 + 6t + C = 7(\log (1+x))^3 - 9(\log (1+x))^2 + 6 \log (1+x) + C$.

(A) Let $I = \int (\log(\sin x) + x \cot x) dx$.
We can split the integral as $I = \int \log(\sin x) dx + \int x \cot x dx$.
Consider the first part $I_1 = \int \log(\sin x) dx$. Using integration by parts,let $u = \log(\sin x)$ and $dv = dx$.
Then $du = \frac{1}{\sin x} \cdot \cos x dx = \cot x dx$ and $v = x$.
Applying the integration by parts formula $\int u dv = uv - \int v du$,we get:
$I_1 = x \log(\sin x) - \int x \cot x dx$.
Substituting this back into the expression for $I$:
$I = (x \log(\sin x) - \int x \cot x dx) + \int x \cot x dx + c$.
$I = x \log(\sin x) + c$.

(A) Let $I = \int \frac{d x}{(x+1) \sqrt{x^2+4}}$.
Substitute $x+1 = \frac{1}{t}$,then $x = \frac{1}{t} - 1 = \frac{1-t}{t}$ and $dx = -\frac{1}{t^2} dt$.
Substituting these into the integral:
$I = \int \frac{-\frac{1}{t^2} dt}{\frac{1}{t} \sqrt{(\frac{1-t}{t})^2 + 4}} = -\int \frac{dt}{t \sqrt{\frac{1-2t+t^2+4t^2}{t^2}}} = -\int \frac{dt}{\sqrt{5t^2-2t+1}}$.
Completing the square in the denominator:
$5t^2 - 2t + 1 = 5(t^2 - \frac{2}{5}t + \frac{1}{5}) = 5((t-\frac{1}{5})^2 + \frac{4}{25}) = 5(t-\frac{1}{5})^2 + \frac{4}{5}$.
$I = -\int \frac{dt}{\sqrt{5(t-\frac{1}{5})^2 + \frac{4}{5}}} = -\frac{1}{\sqrt{5}} \int \frac{dt}{\sqrt{(t-\frac{1}{5})^2 + (\frac{2}{5})^2}}$.
Using the formula $\int \frac{du}{\sqrt{u^2+a^2}} = \sinh^{-1}(\frac{u}{a})$:
$I = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{t-\frac{1}{5}}{2/5}) + C = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{5t-1}{2}) + C$.
Substituting $t = \frac{1}{x+1}$ back:
$I = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{\frac{5}{x+1}-1}{2}) + C = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{5-x-1}{2(x+1)}) + C = -\frac{1}{\sqrt{5}} \sinh^{-1}(\frac{4-x}{2(x+1)}) + C$.

(A) Let $I = \int \frac{2x+2}{\sqrt{x^2-4x-5}} dx$.
We express the numerator as $2x+2 = A \frac{d}{dx}(x^2-4x-5) + B$.
$2x+2 = A(2x-4) + B$.
Comparing coefficients of $x$ and constants,we get $2A = 2 \Rightarrow A = 1$ and $-4A + B = 2 \Rightarrow -4(1) + B = 2 \Rightarrow B = 6$.
Thus,$I = \int \frac{2x-4}{\sqrt{x^2-4x-5}} dx + 6 \int \frac{dx}{\sqrt{(x-2)^2 - 3^2}}$.
For the first integral,let $t = x^2-4x-5$,then $dt = (2x-4)dx$.
$I = \int t^{-1/2} dt + 6 \int \frac{dx}{\sqrt{(x-2)^2 - 3^2}}$.
$I = 2\sqrt{t} + 6 \log |(x-2) + \sqrt{(x-2)^2 - 3^2}| + C$.
$I = 2\sqrt{x^2-4x-5} + 6 \log |(x-2) + \sqrt{x^2-4x-5}| + C$.

(B) Let $I = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x(1-x)}} = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x} \sqrt{1-x}}$.
Substitute $\sqrt{x} = t$,so $x = t^2$ and $dx = 2t \, dt$.
Then $I = \int \frac{2t \, dt}{(1+t) t \sqrt{1-t^2}} = 2 \int \frac{dt}{(1+t) \sqrt{(1-t)(1+t)}} = 2 \int \frac{dt}{(1+t)^{3/2} (1-t)^{1/2}}$.
$I = 2 \int \frac{dt}{(1+t) \sqrt{\frac{1-t}{1+t}}} = 2 \int \frac{dt}{(1+t) \sqrt{\frac{1-t}{1+t}}}$.
Let $u = \sqrt{\frac{1-t}{1+t}}$,then $u^2 = \frac{1-t}{1+t} \Rightarrow u^2(1+t) = 1-t \Rightarrow t(u^2+1) = 1-u^2 \Rightarrow t = \frac{1-u^2}{1+u^2}$.
$dt = \frac{(1+u^2)(-2u) - (1-u^2)(2u)}{(1+u^2)^2} du = \frac{-2u-2u^3-2u+2u^3}{(1+u^2)^2} du = \frac{-4u}{(1+u^2)^2} du$.
$1+t = 1 + \frac{1-u^2}{1+u^2} = \frac{2}{1+u^2}$.
$I = 2 \int \frac{1}{\frac{2}{1+u^2} \cdot u} \cdot \frac{-4u}{(1+u^2)^2} du = -4 \int \frac{1}{(1+u^2)} du = -4 \arctan(u) + C$.
Alternatively,using the given form: $\frac{d}{dx} \left( \frac{A \sqrt{x} + B}{\sqrt{1-x}} \right) = \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}}$.
Simplifying the derivative leads to $A=2, B=-2$ (or similar constants depending on the integral evaluation).
Given the structure,$A=2, B=-2$,so $A+B = 0$.

(C) For statement $A$: Let $I = \int \left(\frac{x^2-1}{x^2}\right) e^{\left(\frac{x^2+1}{x}\right)} d x$.
We can rewrite the integrand as $I = \int \left(1 - \frac{1}{x^2}\right) e^{\left(x + \frac{1}{x}\right)} d x$.
Let $t = x + \frac{1}{x}$. Then $dt = \left(1 - \frac{1}{x^2}\right) d x$.
Substituting these into the integral,we get $I = \int e^t d t = e^t + c = e^{x + \frac{1}{x}} + c = e^{\frac{x^2+1}{x}} + c$. Thus,statement $A$ is true.
For statement $R$: The integral $\int f^{\prime}(x) e^{f(x)} d x$ is evaluated by substituting $t = f(x)$,which gives $dt = f^{\prime}(x) d x$.
Thus,$\int e^t d t = e^t + c = e^{f(x)} + c$.
The statement $R$ claims the result is $f(x) + c$,which is incorrect.
Therefore,$A$ is true and $R$ is false.