Evaluation of various forms of integration Questions in English

(A) To evaluate the integral $I = \int \frac{dx}{2+\cos x}$,we use the substitution $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$.
Substituting this into the integral,we get:
$I = \int \frac{dx}{2 + \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}} = \int \frac{(1+\tan^2(x/2)) dx}{2(1+\tan^2(x/2)) + 1 - \tan^2(x/2)} = \int \frac{\sec^2(x/2) dx}{3 + \tan^2(x/2)}$.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,which implies $\sec^2(x/2) dx = 2 dt$.
Substituting these into the integral:
$I = \int \frac{2 dt}{3 + t^2} = 2 \int \frac{dt}{(\sqrt{3})^2 + t^2}$.
Using the standard integral formula $\int \frac{dx}{a^2+x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$,we get:
$I = 2 \cdot \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + c = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{1}{\sqrt{3}} \tan \frac{x}{2}\right) + c$.

(B) To evaluate the integral $I = \int \sqrt{x^2-6x-16} \, dx$,we first complete the square for the quadratic expression inside the square root:
$x^2-6x-16 = (x^2-6x+9) - 9 - 16 = (x-3)^2 - 25 = (x-3)^2 - 5^2$.
Now,the integral becomes $I = \int \sqrt{(x-3)^2 - 5^2} \, dx$.
Using the standard integration formula $\int \sqrt{t^2-a^2} \, dt = \frac{t}{2} \sqrt{t^2-a^2} - \frac{a^2}{2} \log |t + \sqrt{t^2-a^2}| + c$,where $t = x-3$ and $a = 5$:
$I = \frac{x-3}{2} \sqrt{(x-3)^2 - 5^2} - \frac{5^2}{2} \log |(x-3) + \sqrt{(x-3)^2 - 5^2}| + c$.
Substituting back the original expression $x^2-6x-16$:
$I = \frac{x-3}{2} \sqrt{x^2-6x-16} - \frac{25}{2} \log |x-3 + \sqrt{x^2-6x-16}| + c$.

(C) We have $I = \int \left(\frac{x-3}{x^2+9}\right)^2 \, dx = \int \frac{x^2 - 6x + 9}{(x^2+9)^2} \, dx$.
Splitting the integral,we get $I = \int \frac{x^2+9}{(x^2+9)^2} \, dx - \int \frac{6x}{(x^2+9)^2} \, dx$.
$I = \int \frac{1}{x^2+9} \, dx - \int \frac{6x}{(x^2+9)^2} \, dx$.
The first part is $\int \frac{1}{x^2+3^2} \, dx = \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right)$.
For the second part,let $u = x^2+9$,then $du = 2x \, dx$,so $3 \, du = 6x \, dx$.
Thus,$\int \frac{6x}{(x^2+9)^2} \, dx = \int \frac{3}{u^2} \, du = 3 \left(-\frac{1}{u}\right) = -\frac{3}{x^2+9}$.
Combining these,$I = \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) - \left(-\frac{3}{x^2+9}\right) + c = \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) + \frac{3}{x^2+9} + c$.

(B) To evaluate the integral $I = \int \sqrt{x^2+3x} \, dx$,we complete the square inside the square root:
$x^2 + 3x = (x + \frac{3}{2})^2 - (\frac{3}{2})^2 = (x + \frac{3}{2})^2 - \frac{9}{4}$.
Now,the integral becomes $I = \int \sqrt{(x + \frac{3}{2})^2 - (\frac{3}{2})^2} \, dx$.
Using the standard formula $\int \sqrt{t^2 - a^2} \, dt = \frac{t}{2} \sqrt{t^2 - a^2} - \frac{a^2}{2} \log |t + \sqrt{t^2 - a^2}| + c$,where $t = x + \frac{3}{2}$ and $a = \frac{3}{2}$:
$I = \frac{x + \frac{3}{2}}{2} \sqrt{(x + \frac{3}{2})^2 - \frac{9}{4}} - \frac{9/4}{2} \log |(x + \frac{3}{2}) + \sqrt{(x + \frac{3}{2})^2 - \frac{9}{4}}| + c$.
Simplifying this,we get:
$I = \frac{2x+3}{4} \sqrt{x^2+3x} - \frac{9}{8} \log |x + \frac{3}{2} + \sqrt{x^2+3x}| + c$.

(C) Let $I = \int \frac{dx}{3 \cos 2x + 5}$.
Using the identity $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$,we get:
$I = \int \frac{dx}{3 \left(\frac{1 - \tan^2 x}{1 + \tan^2 x}\right) + 5}$
$I = \int \frac{1 + \tan^2 x}{3(1 - \tan^2 x) + 5(1 + \tan^2 x)} dx$
$I = \int \frac{\sec^2 x}{3 - 3 \tan^2 x + 5 + 5 \tan^2 x} dx$
$I = \int \frac{\sec^2 x}{8 + 2 \tan^2 x} dx$
Let $u = \tan x$,then $du = \sec^2 x dx$.
$I = \int \frac{du}{8 + 2u^2} = \frac{1}{2} \int \frac{du}{4 + u^2}$
Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c$:
$I = \frac{1}{2} \cdot \frac{1}{2} \tan^{-1}\left(\frac{u}{2}\right) + c$
$I = \frac{1}{4} \tan^{-1}\left(\frac{\tan x}{2}\right) + c$.

(D) Given $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.
We can rewrite the integral by taking $x^4$ common from the bracket:
$I = \int \frac{dx}{x^2 \left(x^4(1 + \frac{1}{x^4})\right)^{3/4}} = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$.
Let $t = 1 + \frac{1}{x^4}$.
Then $dt = -\frac{4}{x^5} dx$,which implies $\frac{dx}{x^5} = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \left( \frac{t^{1/4}}{1/4} \right) + c = -t^{1/4} + c$.
Substituting back $t = 1 + \frac{1}{x^4}$:
$I = -\left(1 + \frac{1}{x^4}\right)^{1/4} + c$.

(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:
$I = -\left(1 + \frac{1}{x^4}\right)^{\frac{1}{4}} + c = -\left(\frac{x^4 + 1}{x^4}\right)^{\frac{1}{4}} + c = -\frac{\left(x^4 + 1\right)^{\frac{1}{4}}}{x} + c$.

(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 $I = \int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}$.
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 $I = \int \frac{2x+5}{\sqrt{7-6x-x^2}} \, dx$.
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,$A+B = -2 + (-1) = -3$.

(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,$f(2) = 2 + \frac{4}{2} = 2 + 2 = 4$.

(B) Let $I = \int \frac{1}{\sin (x-a) \sin x} \,d x$.
Multiply and divide by $\sin a$:
$I = \frac{1}{\sin a} \int \frac{\sin a}{\sin (x-a) \sin x} \,d x$.
Express $\sin a$ as $\sin (x - (x-a))$:
$I = \frac{1}{\sin a} \int \frac{\sin (x - (x-a))}{\sin (x-a) \sin x} \,d x$.
Using the identity $\sin (A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin a} \int \frac{\sin x \cos (x-a) - \cos x \sin (x-a)}{\sin (x-a) \sin x} \,d x$.
Split the integral:
$I = \frac{1}{\sin a} \left[ \int \frac{\sin x \cos (x-a)}{\sin (x-a) \sin x} \,d x - \int \frac{\cos x \sin (x-a)}{\sin (x-a) \sin x} \,d x \right]$.
$I = \frac{1}{\sin a} \left[ \int \cot (x-a) \,d x - \int \cot x \,d x \right]$.
Integrating:
$I = \frac{1}{\sin a} [\log |\sin (x-a)| - \log |\sin x|] + c$.
$I = \operatorname{cosec} a \log |\frac{\sin (x-a)}{\sin x}| + c$.

(B) Given $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx$.
Using $\sin^2 x = 1 - \cos^2 x$,we have $\sin x(1 + \sin^2 x) = \sin x(1 + 1 - \cos^2 x) = \sin x(2 - \cos^2 x)$.
So,$I = \int \frac{\sin x(2 - \cos^2 x)}{2 \cos^2 x - 1} dx$.
Let $\cos x = t$,then $-\sin x dx = dt$,so $\sin x dx = -dt$.
$I = \int \frac{t^2 - 2}{2t^2 - 1} (-dt) = \int \frac{2 - t^2}{2t^2 - 1} dt$.
$I = \frac{1}{2} \int \frac{4 - 2t^2}{2t^2 - 1} dt = \frac{1}{2} \int \frac{-(2t^2 - 1) + 3}{2t^2 - 1} dt$.
$I = \frac{1}{2} \int (-1 + \frac{3}{2t^2 - 1}) dt = \frac{1}{2} [-t + \frac{3}{2\sqrt{2}} \log |\frac{\sqrt{2}t - 1}{\sqrt{2}t + 1}|] + c$.
$I = -\frac{1}{2} \cos x + \frac{3}{4\sqrt{2}} \log |\frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1}| + c$.
Comparing with $P \cos x + Q \log |\frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1}| + c$,we get $P = -\frac{1}{2}$ and $Q = \frac{3}{4\sqrt{2}}$.
Note: The provided options seem to have a sign discrepancy for $P$. Based on the standard integral form,$P = -1/2$ and $Q = 3/(4\sqrt{2})$. If $P$ is $1/2$,the numerator in the integral must be $\sin x(\cos^2 x - 2)$.

(B) Let $I = \int \frac{\tan x + \tan \alpha}{\tan x - \tan \alpha} dx$.
$= \int \frac{\frac{\sin x}{\cos x} + \frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x} - \frac{\sin \alpha}{\cos \alpha}} dx$
$= \int \frac{\sin x \cos \alpha + \sin \alpha \cos x}{\sin x \cos \alpha - \sin \alpha \cos x} dx$
$= \int \frac{\sin (x+\alpha)}{\sin (x-\alpha)} dx$.
Let $t = x - \alpha$,so $x = t + \alpha$ and $dx = dt$.
$I = \int \frac{\sin (t + 2\alpha)}{\sin t} dt$
$= \int \frac{\sin t \cos 2\alpha + \cos t \sin 2\alpha}{\sin t} dt$
$= \cos 2\alpha \int 1 dt + \sin 2\alpha \int \cot t dt$
$= t \cos 2\alpha + \sin 2\alpha \log |\sin t| + c$
$= (x - \alpha) \cos 2\alpha + \log |\sin (x - \alpha)| \sin 2\alpha + c$.
Comparing this with $A(x) \cos 2\alpha + B(x) \sin 2\alpha + c$,we get $A(x) = x - \alpha$ and $B(x) = \log |\sin (x - \alpha)|$.

(C) Let $I = \int \frac{5 \tan x}{\tan x - 2} \, dx = \int \frac{5 \sin x}{\sin x - 2 \cos x} \, dx$.
We express the numerator as $5 \sin x = A(\sin x - 2 \cos x) + B \frac{d}{dx}(\sin x - 2 \cos x)$.
$5 \sin x = A(\sin x - 2 \cos x) + B(\cos x + 2 \sin x)$.
Equating the coefficients of $\sin x$ and $\cos x$:
$A + 2B = 5$ and $-2A + B = 0$.
From the second equation,$B = 2A$. Substituting into the first: $A + 2(2A) = 5 \implies 5A = 5 \implies A = 1$.
Then $B = 2(1) = 2$.
Thus,$I = \int \left( 1 + 2 \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} \right) \, dx$.
$I = \int 1 \, dx + 2 \int \frac{d(\sin x - 2 \cos x)}{\sin x - 2 \cos x} = x + 2 \log |\sin x - 2 \cos x| + c$.
Comparing this with $x + a \log |\sin x - 2 \cos x| + c$,we get $a = 2$.

(C) Given $J-I = \int \left( \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} - \frac{e^x}{e^{4x} + e^{2x} + 1} \right) dx$.
Multiply the numerator and denominator of the first term by $e^{4x}$: $\frac{e^{-x} \cdot e^{4x}}{e^{-4x} \cdot e^{4x} + e^{-2x} \cdot e^{4x} + 1 \cdot e^{4x}} = \frac{e^{3x}}{1 + e^{2x} + e^{4x}}$.
Thus, $J-I = \int \frac{e^{3x} - e^x}{e^{4x} + e^{2x} + 1} dx = \int \frac{(e^{2x} - 1)e^x}{e^{4x} + e^{2x} + 1} dx$.
Let $e^x = t$, then $e^x dx = dt$.
The integral becomes $\int \frac{t^2 - 1}{t^4 + t^2 + 1} dt = \int \frac{1 - 1/t^2}{(t^2 + 1 + 1/t^2)} dt = \int \frac{1 - 1/t^2}{(t + 1/t)^2 - 1} dt$.
Let $u = t + 1/t$, then $du = (1 - 1/t^2) dt$.
The integral is $\int \frac{du}{u^2 - 1} = \frac{1}{2} \log \left| \frac{u-1}{u+1} \right| + c$.
Substituting $u = t + 1/t = e^x + e^{-x}$, we get $\frac{1}{2} \log \left| \frac{e^x + e^{-x} - 1}{e^x + e^{-x} + 1} \right| + c = \frac{1}{2} \log \left| \frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1} \right| + c$.

(A) Let $I = \int \frac{\sin x+\sin ^3 x}{\cos 2 x} \,d x$.
$= \int \frac{\sin x(1+\sin ^2 x)}{\cos 2 x} \,d x$.
$= \int \frac{\sin x(1+1-\cos ^2 x)}{2 \cos ^2 x-1} \,d x$.
$= \int \frac{\sin x(2-\cos ^2 x)}{2 \cos ^2 x-1} \,d x$.
Substitute $\cos x = t$,so $-\sin x \,d x = dt$ or $\sin x \,d x = -dt$.
$I = -\int \frac{2-t^2}{2 t^2-1} dt = \int \frac{t^2-2}{2 t^2-1} dt$.
$= \frac{1}{2} \int \frac{2 t^2-4}{2 t^2-1} dt = \frac{1}{2} \int \left(1 - \frac{3}{2 t^2-1}\right) dt$.
$= \frac{1}{2} \int dt - \frac{3}{2} \int \frac{dt}{(\sqrt{2} t)^2-1^2}$.
Using the formula $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log |\frac{x-a}{x+a}|$,we get:
$= \frac{1}{2} t - \frac{3}{2} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \log |\frac{\sqrt{2} t-1}{\sqrt{2} t+1}| + c$.
$= \frac{1}{2} \cos x - \frac{3}{4 \sqrt{2}} \log |\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}| + c$.
Comparing with $A \cos x + B \log |f(x)| + c$,we have $A = \frac{1}{2}$,$B = \frac{-3}{4 \sqrt{2}}$,and $f(x) = \frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}$.

(D) To evaluate $I = \int \frac{dx}{\sin(x-a) \sin(x-b)}$,multiply and divide by $\sin(b-a)$:
$I = \frac{1}{\sin(b-a)} \int \frac{\sin((x-a) - (x-b))}{\sin(x-a) \sin(x-b)} dx$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin(b-a)} \int \frac{\sin(x-a) \cos(x-b) - \cos(x-a) \sin(x-b)}{\sin(x-a) \sin(x-b)} dx$
$I = \frac{1}{\sin(b-a)} \int \left( \frac{\cos(x-b)}{\sin(x-b)} - \frac{\cos(x-a)}{\sin(x-a)} \right) dx$
$I = \frac{1}{\sin(b-a)} \int (\cot(x-b) - \cot(x-a)) dx$
Integrating,we get:
$I = \frac{1}{\sin(b-a)} [\log|\sin(x-b)| - \log|\sin(x-a)|] + c$
$I = \frac{1}{\sin(b-a)} \log \left| \frac{\sin(x-b)}{\sin(x-a)} \right| + c$

(D) Let $I = \int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}$.
Using $\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta}$ and $\sec 2\theta = \frac{1}{\cos 2\theta}$,we have:
$I = \int \frac{\sec ^2 \theta d \theta}{\frac{\sin 2 \theta+1}{\cos 2 \theta}} = \int \frac{\cos 2 \theta \sec ^2 \theta d \theta}{1+\sin 2 \theta}$.
Since $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ and $1 + \sin 2\theta = (\cos \theta + \sin \theta)^2$,we get:
$I = \int \frac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta) \sec^2 \theta d \theta}{(\cos \theta + \sin \theta)^2} = \int \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} \sec^2 \theta d \theta$.
Dividing the numerator and denominator by $\cos \theta$,we get:
$I = \int \frac{1-\tan \theta}{1+\tan \theta} \sec^2 \theta d \theta$.
Let $t = \tan \theta$,then $dt = \sec^2 \theta d\theta$:
$I = \int \frac{1-t}{1+t} dt = \int \left(-1 + \frac{2}{1+t}\right) dt = -t + 2 \log |1+t| + C$.
Substituting $t = \tan \theta$ back,we get:
$I = -\tan \theta + 2 \log |1+\tan \theta| + C$.
Comparing this with $\lambda \tan \theta + 2 \log |f(\theta)| + C$,we find $\lambda = -1$ and $f(\theta) = 1+\tan \theta$.

(B) Let $I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \, dx$.
Multiply numerator and denominator by $\sqrt{1-\sqrt{x}}$:
$I = \int \sqrt{\frac{(1-\sqrt{x})^2}{1-x}} \, dx = \int \frac{1-\sqrt{x}}{\sqrt{1-x}} \, dx$.
Split the integral:
$I = \int \frac{1}{\sqrt{1-x}} \, dx - \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx$.
The first part is $-2\sqrt{1-x}$.
For the second part,let $x = \cos^2 \theta$,so $dx = -2 \cos \theta \sin \theta \, d\theta$.
$\int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx = \int \frac{\cos \theta}{\sin \theta} (-2 \cos \theta \sin \theta) \, d\theta = -2 \int \cos^2 \theta \, d\theta = -\int (1 + \cos 2\theta) \, d\theta = -(\theta + \frac{\sin 2\theta}{2}) = -\theta - \sin \theta \cos \theta$.
Substituting back: $I = -2\sqrt{1-x} - (-\theta - \sin \theta \cos \theta) + C = -2\sqrt{1-x} + \theta + \sin \theta \cos \theta + C$.
Since $x = \cos^2 \theta$,$\theta = \cos^{-1} \sqrt{x}$,$\cos \theta = \sqrt{x}$,and $\sin \theta = \sqrt{1-x}$.
Thus,$I = -2\sqrt{1-x} + \cos^{-1} \sqrt{x} + \sqrt{x(1-x)} + C$.

(D) Given $f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \cdot \sec^2 x \, dx$.
We can rewrite the numerator as $x^2 + 1 - 1 + \sin^2 x = (1 + x^2) - (1 - \sin^2 x) = (1 + x^2) - \cos^2 x$.
Substituting this into the integral:
$f(x) = \int \frac{(1 + x^2) - \cos^2 x}{1 + x^2} \cdot \sec^2 x \, dx$
$f(x) = \int \left( 1 - \frac{\cos^2 x}{1 + x^2} \right) \sec^2 x \, dx$
$f(x) = \int \sec^2 x \, dx - \int \frac{\cos^2 x \cdot \sec^2 x}{1 + x^2} \, dx$
Since $\cos^2 x \cdot \sec^2 x = 1$,we have:
$f(x) = \int \sec^2 x \, dx - \int \frac{1}{1 + x^2} \, dx$
$f(x) = \tan x - \tan^{-1} x + C$.
Given $f(0) = 0$:
$0 = \tan(0) - \tan^{-1}(0) + C \implies 0 = 0 - 0 + C \implies C = 0$.
Thus,$f(x) = \tan x - \tan^{-1} x$.
Therefore,$f(1) = \tan(1) - \tan^{-1}(1) = \tan(1) - \frac{\pi}{4}$.

(A) Let $I = \int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x$.
Multiply numerator and denominator by $e^x$:
$I = \int \frac{2 e^{2x}+3}{3 e^{2x}+4} d x$.
We want to express the numerator in terms of the denominator and its derivative.
Let $2 e^{2x} + 3 = A_1(3 e^{2x} + 4) + B_1 \frac{d}{dx}(3 e^{2x} + 4)$.
$2 e^{2x} + 3 = 3 A_1 e^{2x} + 4 A_1 + 6 B_1 e^{2x}$.
Comparing coefficients:
$3 A_1 + 6 B_1 = 2$ and $4 A_1 = 3 \implies A_1 = \frac{3}{4}$.
Substituting $A_1$: $3(\frac{3}{4}) + 6 B_1 = 2 \implies \frac{9}{4} + 6 B_1 = 2 \implies 6 B_1 = 2 - \frac{9}{4} = -\frac{1}{4} \implies B_1 = -\frac{1}{24}$.
Thus,$I = \int \left( \frac{3}{4} - \frac{1}{24} \frac{6 e^{2x}}{3 e^{2x} + 4} \right) d x$.
$I = \frac{3}{4} x - \frac{1}{24} \log(3 e^{2x} + 4) + C$.
Comparing with $Ax + B \log(3 e^{2x} + 4) + C$,we get $A = \frac{3}{4}$ and $B = -\frac{1}{24}$.

(D) To solve the integral $I = \int \frac{dx}{2+\cos x}$,we use the substitution $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$.
Substituting this into the integral,we get:
$I = \int \frac{dx}{2 + \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}} = \int \frac{1+\tan^2(x/2)}{2(1+\tan^2(x/2)) + 1-\tan^2(x/2)} dx$
$I = \int \frac{\sec^2(x/2)}{3+\tan^2(x/2)} dx$.
Let $u = \tan(x/2)$,then $du = \frac{1}{2} \sec^2(x/2) dx$,which implies $\sec^2(x/2) dx = 2 du$.
Substituting these into the integral:
$I = \int \frac{2 du}{3+u^2} = 2 \int \frac{du}{(\sqrt{3})^2 + u^2}$.
Using the standard formula $\int \frac{dx}{a^2+x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = 2 \cdot \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{u}{\sqrt{3}}\right) + C = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{\tan(x/2)}{\sqrt{3}}\right) + C$.

(B) To solve the integral $I = \int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \,d x$, we factor out $x^5$ from the denominator term inside the cube:
$I = \int \frac{2 x^{12}+5 x^9}{\left[x^5\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)\right]^3} \,d x$
$I = \int \frac{2 x^{12}+5 x^9}{x^{15}\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} \,d x$
$I = \int \frac{\frac{2}{x^3}+\frac{5}{x^6}}{\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} \,d x$
Let $t = 1+\frac{1}{x^2}+\frac{1}{x^5}$. Then $dt = \left(-\frac{2}{x^3}-\frac{5}{x^6}\right) dx$, which implies $-dt = \left(\frac{2}{x^3}+\frac{5}{x^6}\right) dx$.
Substituting these into the integral:
$I = \int \frac{-dt}{t^3} = -\int t^{-3} dt = -\frac{t^{-2}}{-2} + C = \frac{1}{2t^2} + C$
Substituting $t$ back:
$I = \frac{1}{2\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^2} + C = \frac{1}{2\left(\frac{x^5+x^3+1}{x^5}\right)^2} + C = \frac{x^{10}}{2\left(x^5+x^3+1\right)^2} + C$

(B) Let $I = \int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \, dx$.
Multiply the numerator and denominator by $e^{\frac{x}{2}}$:
$I = \int \frac{e^{\frac{x}{2}} \cdot e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x} \cdot e^{\frac{x}{2}}} \, dx = \int \frac{e^x}{\sqrt{e^{-x} \cdot e^{\frac{x}{2}} - e^x \cdot e^{\frac{x}{2}}}} \, dx$.
Wait,let us simplify the denominator differently:
$I = \int \frac{e^{\frac{x}{2}}}{\sqrt{\frac{1}{e^x} - e^x}} \, dx = \int \frac{e^{\frac{x}{2}}}{\sqrt{\frac{1 - e^{2x}}{e^x}}} \, dx = \int \frac{e^{\frac{x}{2}} \cdot \sqrt{e^x}}{\sqrt{1 - e^{2x}}} \, dx$.
Since $\sqrt{e^x} = e^{\frac{x}{2}}$,we have:
$I = \int \frac{e^{\frac{x}{2}} \cdot e^{\frac{x}{2}}}{\sqrt{1 - (e^x)^2}} \, dx = \int \frac{e^x}{\sqrt{1 - (e^x)^2}} \, dx$.
Let $t = e^x$,then $dt = e^x \, dx$.
$I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + C = \sin^{-1}(e^x) + C$.
Comparing this with $\sin^{-1}(f(x)) + C$,we get $f(x) = e^x$.
Therefore,$f(2) = e^2$.

(D) Let $I = \int \frac{1+x^2}{1+x^4} dx$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{\frac{1}{x^2} + 1}{\frac{1}{x^2} + x^2} dx = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx$.
We can write the denominator as $(x - \frac{1}{x})^2 + 2$:
$I = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 2} dx$.
Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2 + 2} = \int \frac{dt}{t^2 + (\sqrt{2})^2}$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{t}{\sqrt{2}}) + c$.
Substituting $t = x - \frac{1}{x}$ back:
$I = \frac{1}{\sqrt{2}} \tan^{-1}\left[\frac{x - \frac{1}{x}}{\sqrt{2}}\right] + c$.
Comparing this with the given expression $\frac{1}{\sqrt{2}} \tan^{-1}\left[\frac{f(x)}{\sqrt{2}}\right] + c$,we get $f(x) = x - \frac{1}{x}$.

(B) We need to evaluate the integral $I = \int \frac{dx}{\sqrt{(x-1)(x-2)}}$.
First,expand the denominator: $(x-1)(x-2) = x^2 - 3x + 2$.
Now,complete the square for the quadratic expression: $x^2 - 3x + 2 = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4} = (x - \frac{3}{2})^2 - (\frac{1}{2})^2$.
The integral becomes $I = \int \frac{dx}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + c$,we get:
$I = \log |(x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}| + c$.
Simplifying the expression inside the square root back to the original form,we get:
$I = \log |(x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2}| + c$.

(A) Let $I = \int \frac{1}{1-\cot x} dx$.
Substitute $\cot x = \frac{\cos x}{\sin x}$,so $I = \int \frac{1}{1-\frac{\cos x}{\sin x}} dx = \int \frac{\sin x}{\sin x - \cos x} dx$.
Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2 \sin x}{\sin x - \cos x} dx$.
Rewrite the numerator: $2 \sin x = (\sin x - \cos x) + (\sin x + \cos x)$.
Thus,$I = \frac{1}{2} \int \frac{(\sin x - \cos x) + (\sin x + \cos x)}{\sin x - \cos x} dx = \frac{1}{2} \int \left( 1 + \frac{\sin x + \cos x}{\sin x - \cos x} \right) dx$.
Let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) dx$.
$I = \frac{1}{2} \left( \int 1 dx + \int \frac{1}{u} du \right) = \frac{1}{2} (x + \log |u|) + C = \frac{1}{2} x + \frac{1}{2} \log |\sin x - \cos x| + C$.
Comparing with $Ax + B \log |\sin x - \cos x| + C$,we get $A = \frac{1}{2}$ and $B = \frac{1}{2}$.
Therefore,$A + B = \frac{1}{2} + \frac{1}{2} = 1$.

(D) Let $I = \int \frac{x^2+1}{x^4-x^2+1} \, dx$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{1 + \frac{1}{x^2}}{x^2 - 1 + \frac{1}{x^2}} \, dx$.
Rewrite the denominator as $(x - \frac{1}{x})^2 + 1$:
$I = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 1} \, dx$.
Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) \, dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2 + 1} = \tan^{-1}(t) + c$.
Substituting $t = x - \frac{1}{x}$ back:
$I = \tan^{-1}\left(x - \frac{1}{x}\right) + c = \tan^{-1}\left(\frac{x^2-1}{x}\right) + c$.

(A) We have,$\tan 2 x = \tan ((x - \alpha) + (x + \alpha))$.
Using the identity $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan 2 x = \frac{\tan (x - \alpha) + \tan (x + \alpha)}{1 - \tan (x - \alpha) \tan (x + \alpha)}$.
Rearranging the terms:
$\tan 2 x (1 - \tan (x - \alpha) \tan (x + \alpha)) = \tan (x - \alpha) + \tan (x + \alpha)$.
$\tan 2 x - \tan (x - \alpha) \tan (x + \alpha) \tan 2 x = \tan (x - \alpha) + \tan (x + \alpha)$.
Thus,$\tan (x - \alpha) \tan (x + \alpha) \tan 2 x = \tan 2 x - \tan (x - \alpha) - \tan (x + \alpha)$.
Now,integrating both sides:
$\int \tan (x - \alpha) \tan (x + \alpha) \tan 2 x \ d x = \int (\tan 2 x - \tan (x - \alpha) - \tan (x + \alpha)) \ d x$.
$= \frac{1}{2} \log |\sec 2 x| - \log |\sec (x - \alpha)| - \log |\sec (x + \alpha)| + C$.
Comparing this with $p \log |\sec 2 x| + q \log |\sec (x + \alpha)| + r \log |\sec (x - \alpha)| + c$,we get:
$p = \frac{1}{2}$,$q = -1$,and $r = -1$.
Therefore,$p + q + r = \frac{1}{2} - 1 - 1 = \frac{1}{2} - 2 = \frac{-3}{2}$.

(B) Let $I = \int \frac{dx}{(x^2+1)^2}$.
Substitute $x = \tan \theta$,then $dx = \sec^2 \theta d \theta$.
$I = \int \frac{\sec^2 \theta d \theta}{(\tan^2 \theta + 1)^2} = \int \frac{\sec^2 \theta d \theta}{\sec^4 \theta} = \int \cos^2 \theta d \theta$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2 \theta}{2}$,we get:
$I = \frac{1}{2} \int (1 + \cos 2 \theta) d \theta = \frac{1}{2} (\theta + \frac{\sin 2 \theta}{2}) + c$.
Since $\sin 2 \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$ and $\theta = \tan^{-1} x$,we have:
$I = \frac{1}{2} \tan^{-1} x + \frac{1}{4} \cdot \frac{2x}{1 + x^2} + c = \frac{1}{2} \tan^{-1} x + \frac{x}{2(1 + x^2)} + c$.

(D) To evaluate $\int \sqrt{\frac{x - 5}{x - 7}} dx$,we multiply the numerator and denominator by $\sqrt{x - 5}$:
$\int \frac{x - 5}{\sqrt{(x - 5)(x - 7)}} dx = \int \frac{x - 5}{\sqrt{x^2 - 12x + 35}} dx$
Next,we express the numerator in terms of the derivative of the quadratic expression $x^2 - 12x + 35$,which is $2x - 12$:
$= \frac{1}{2} \int \frac{2x - 10}{\sqrt{x^2 - 12x + 35}} dx = \frac{1}{2} \int \frac{2x - 12 + 2}{\sqrt{x^2 - 12x + 35}} dx$
$= \frac{1}{2} \int \frac{2x - 12}{\sqrt{x^2 - 12x + 35}} dx + \int \frac{1}{\sqrt{x^2 - 12x + 35}} dx$
$= \sqrt{x^2 - 12x + 35} + \int \frac{1}{\sqrt{(x - 6)^2 - 1}} dx$
Using the standard integral $\int \frac{1}{\sqrt{t^2 - a^2}} dt = \log |t + \sqrt{t^2 - a^2}|$,we get:
$= \sqrt{x^2 - 12x + 35} + \log |x - 6 + \sqrt{(x - 6)^2 - 1}| + C$
$= 1 \cdot \sqrt{x^2 - 12x + 35} + \log |x - 6 + \sqrt{x^2 - 12x + 35}| + C$
Comparing this with the given form,we find $A = 1$.

(A) Let $I = \int \frac{x^{2}+1}{x^{4}+x^{2}+1} dx$.
Divide the numerator and denominator by $x^{2}$:
$I = \int \frac{1 + 1/x^{2}}{x^{2} + 1 + 1/x^{2}} dx$.
Rewrite the denominator as $(x - 1/x)^{2} + 3$:
$I = \int \frac{1 + 1/x^{2}}{(x - 1/x)^{2} + (\sqrt{3})^{2}} dx$.
Let $t = x - 1/x$,then $dt = (1 + 1/x^{2}) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^{2} + (\sqrt{3})^{2}}$.
Using the standard formula $\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a} \tan^{-1}(x/a) + C$:
$I = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + C$.
Substituting $t = x - 1/x$ back:
$I = \frac{1}{\sqrt{3}} \tan^{-1}\left\{\frac{x - 1/x}{\sqrt{3}}\right\} + C$.

(B) Let $I = \int \sqrt{1+\sec x} \, dx$.
We know that $1+\sec x = 1 + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x} = \frac{2 \cos^2 \frac{x}{2}}{\cos x}$.
So,$I = \int \frac{\sqrt{2} \cos \frac{x}{2}}{\sqrt{\cos x}} \, dx$.
Using $\cos x = 1 - 2 \sin^2 \frac{x}{2}$,we get $I = \int \frac{\sqrt{2} \cos \frac{x}{2}}{\sqrt{1 - 2 \sin^2 \frac{x}{2}}} \, dx$.
Let $t = \sqrt{2} \sin \frac{x}{2}$. Then $dt = \sqrt{2} \cdot \frac{1}{2} \cos \frac{x}{2} \, dx = \frac{1}{\sqrt{2}} \cos \frac{x}{2} \, dx$.
This implies $\sqrt{2} \cos \frac{x}{2} \, dx = 2 \, dt$.
Substituting these into the integral,we get $I = \int \frac{2 \, dt}{\sqrt{1 - t^2}} = 2 \sin^{-1}(t) + C$.
Substituting back $t = \sqrt{2} \sin \frac{x}{2}$,we obtain $I = 2 \sin^{-1}(\sqrt{2} \sin \frac{x}{2}) + C$.

(B) Let $I = \int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$.
We can write this as $I = \int \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right) \, dx = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$.
Multiply numerator and denominator by $\sqrt{2}$:
$I = \int \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx$.
Since $\sin 2x = 1 - (\sin x - \cos x)^2$,let $t = \sin x - \cos x$.
Then $dt = (\cos x + \sin x) \, dx$.
Substituting these into the integral:
$I = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} \sin^{-1}(t) + C = \sqrt{2} \sin^{-1}(\sin x - \cos x) + C$.
Using the identity $\sin^{-1}(u) = \tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)$,we get:
$I = \sqrt{2} \tan^{-1}\left(\frac{\sin x - \cos x}{\sqrt{1 - (\sin x - \cos x)^2}}\right) + C = \sqrt{2} \tan^{-1}\left(\frac{\sin x - \cos x}{\sqrt{\sin 2x}}\right) + C$.
Dividing numerator and denominator by $\sqrt{\cos x}$,we obtain:
$I = \sqrt{2} \tan^{-1}\left(\frac{\tan x - 1}{\sqrt{2 \tan x}}\right) + C$.

(A) Let $I = \int \operatorname{cosec}(x-a) \operatorname{cosec} x \, dx = \int \frac{dx}{\sin(x-a) \sin x}$.
Multiply and divide by $\sin a$:
$I = \frac{1}{\sin a} \int \frac{\sin a}{\sin(x-a) \sin x} \, dx$.
Since $a = x - (x-a)$,we have:
$I = \frac{1}{\sin a} \int \frac{\sin(x - (x-a))}{\sin(x-a) \sin x} \, dx$.
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin a} \int \frac{\sin x \cos(x-a) - \cos x \sin(x-a)}{\sin(x-a) \sin x} \, dx$.
$I = \frac{1}{\sin a} \left( \int \frac{\cos(x-a)}{\sin(x-a)} \, dx - \int \frac{\cos x}{\sin x} \, dx \right)$.
$I = \frac{1}{\sin a} (\log |\sin(x-a)| - \log |\sin x|) + c$.
$I = \operatorname{cosec} a \cdot \log \left| \frac{\sin(x-a)}{\sin x} \right| + c$.

(C) Let $I = \int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$.
We know that $\frac{d}{d x}(x \sin x+\cos x) = \sin x + x \cos x - \sin x = x \cos x$.
We can rewrite the integral as:
$I = \int \left( \frac{x}{\cos x} \right) \left( \frac{x \cos x}{(x \sin x+\cos x)^{2}} \right) d x$.
Using integration by parts,let $u = \frac{x}{\cos x}$ and $dv = \frac{x \cos x}{(x \sin x+\cos x)^{2}} d x$.
Then $du = \frac{\cos x - x(-\sin x)}{\cos^{2} x} d x = \frac{\cos x + x \sin x}{\cos^{2} x} d x$ and $v = \frac{-1}{x \sin x + \cos x}$.
$I = \left( \frac{x}{\cos x} \right) \left( \frac{-1}{x \sin x + \cos x} \right) - \int \left( \frac{\cos x + x \sin x}{\cos^{2} x} \right) \left( \frac{-1}{x \sin x + \cos x} \right) d x$.
$I = \frac{-x}{\cos x(x \sin x + \cos x)} + \int \frac{1}{\cos^{2} x} d x$.
$I = \frac{-x}{\cos x(x \sin x + \cos x)} + \tan x + C$.
$I = \frac{-x + \tan x \cos x (x \sin x + \cos x)}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{-x + \sin x (x \sin x + \cos x)}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{-x + x \sin^{2} x + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{-x(1 - \sin^{2} x) + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{-x \cos^{2} x + \sin x \cos x}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{\cos x(\sin x - x \cos x)}{\cos x(x \sin x + \cos x)} + C$.
$I = \frac{\sin x - x \cos x}{x \sin x + \cos x} + C$.

(B) We are given the integral $I = \int \frac{d x}{\cot ^2 x-1}$.
First,rewrite the integrand using $\cot x = \frac{\cos x}{\sin x}$:
$I = \int \frac{d x}{\frac{\cos ^2 x}{\sin ^2 x}-1} = \int \frac{\sin ^2 x}{\cos ^2 x-\sin ^2 x} d x$.
Using the identity $\cos 2x = \cos ^2 x - \sin ^2 x$,we get:
$I = \int \frac{\sin ^2 x}{\cos 2 x} d x$.
Using the identity $\sin ^2 x = \frac{1-\cos 2 x}{2}$,we have:
$I = \int \frac{1-\cos 2 x}{2 \cos 2 x} d x = \frac{1}{2} \int (\sec 2 x - 1) d x$.
Integrating term by term:
$I = \frac{1}{2} \left( \frac{\log |\sec 2 x + \tan 2 x|}{2} - x \right) + c$.
$I = \frac{1}{4} \log |\sec 2 x + \tan 2 x| - \frac{x}{2} + c$.
Comparing this with the given form $\frac{1}{A} \log |\sec 2 x + \tan 2 x| - \frac{x}{B} + c$,we find $A = 4$ and $B = 2$.
Therefore,$A + B = 4 + 2 = 6$.

(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,$A+B+C = \frac{1}{2} + \frac{5}{2} + \frac{1}{5} = 3 + \frac{1}{5} = \frac{16}{5}$.

(C) Let $I = \int \sqrt{\frac{x-7}{x-9}} ~dx$
Multiply numerator and denominator by $\sqrt{x-7}$:
$I = \int \frac{x-7}{\sqrt{(x-9)(x-7)}} ~dx = \int \frac{x-7}{\sqrt{x^2-16x+63}} ~dx$
Let $x-7 = A \frac{d}{dx}(x^2-16x+63) + B$
$x-7 = A(2x-16) + B$
Comparing coefficients of $x$ and constant terms:
$2A = 1 \implies A = \frac{1}{2}$
$-16A + B = -7 \implies -16(\frac{1}{2}) + B = -7 \implies -8 + B = -7 \implies B = 1$
Now,$I = \int \frac{\frac{1}{2}(2x-16) + 1}{\sqrt{x^2-16x+63}} ~dx$
$I = \frac{1}{2} \int \frac{2x-16}{\sqrt{x^2-16x+63}} ~dx + \int \frac{1}{\sqrt{(x-8)^2 - 1^2}} ~dx$
Using $\int \frac{f'(x)}{\sqrt{f(x)}} ~dx = 2\sqrt{f(x)} + c$ and $\int \frac{1}{\sqrt{x^2-a^2}} ~dx = \log |x + \sqrt{x^2-a^2}| + c$:
$I = \frac{1}{2} \cdot 2\sqrt{x^2-16x+63} + \log |(x-8) + \sqrt{(x-8)^2 - 1^2}| + c$
$I = 1 \cdot \sqrt{x^2-16x+63} + \log |(x-8) + \sqrt{x^2-16x+63}| + c$
Comparing this with the given expression $A \sqrt{x^2-16x+63} + \log |(x-8) + \sqrt{x^2-16x+63}| + c$,we get $A = 1$.

(D) Let $I = \int \frac{5 \tan x}{\tan x-2} d x$.
Expressing in terms of $\sin x$ and $\cos x$:
$I = \int \frac{5 \sin x}{\sin x-2 \cos x} d x$.
Let the numerator be $5 \sin x = A(\sin x - 2 \cos x) + B \frac{d}{dx}(\sin x - 2 \cos x)$.
$5 \sin x = A(\sin x - 2 \cos x) + B(\cos x + 2 \sin x)$.
$5 \sin x = (A + 2B) \sin x + (B - 2A) \cos x$.
Comparing coefficients:
$A + 2B = 5$ and $B - 2A = 0 \implies B = 2A$.
Substituting $B = 2A$ into the first equation: $A + 2(2A) = 5 \implies 5A = 5 \implies A = 1$.
Thus,$B = 2(1) = 2$.
So,$5 \sin x = 1(\sin x - 2 \cos x) + 2(\cos x + 2 \sin x)$.
$I = \int \frac{1(\sin x - 2 \cos x) + 2(\cos x + 2 \sin x)}{\sin x - 2 \cos x} d x$.
$I = \int 1 d x + 2 \int \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} d x$.
$I = x + 2 \log |\sin x - 2 \cos x| + c$.
Comparing this with the given expression $x + a \log |\sin x - 2 \cos x| + c$,we get $a = 2$.

(A) Let $I = \int \frac{x^{2}+1}{x^{4}+x^{2}+1} dx$.
Divide the numerator and denominator by $x^{2}$:
$I = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} dx$.
Rewrite the denominator as $(x^{2} + \frac{1}{x^{2}} - 2) + 3 = (x - \frac{1}{x})^{2} + (\sqrt{3})^{2}$.
So,$I = \int \frac{1 + \frac{1}{x^{2}}}{(x - \frac{1}{x})^{2} + (\sqrt{3})^{2}} dx$.
Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^{2}}) dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{t^{2} + (\sqrt{3})^{2}}$.
Using the standard integral formula $\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = \frac{1}{\sqrt{3}} \tan^{-1}(\frac{t}{\sqrt{3}}) + c$.
Replacing $t$ with $x - \frac{1}{x}$,we get:
$I = \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x - \frac{1}{x}}{\sqrt{3}}) + c$.

(D) Let $I = \int \left[ \log (1+\cos x) - x \tan \left( \frac{x}{2} \right) \right] dx$.
Using integration by parts on the first term,$\int u \cdot v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = \log(1+\cos x)$ and $v = 1$.
$I = x \log(1+\cos x) - \int x \cdot \frac{d}{dx} [\log(1+\cos x)] dx - \int x \tan \left( \frac{x}{2} \right) dx$.
Since $\frac{d}{dx} [\log(1+\cos x)] = \frac{-\sin x}{1+\cos x} = \frac{-2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} = -\tan(x/2)$.
$I = x \log(1+\cos x) - \int x \left( -\tan \frac{x}{2} \right) dx - \int x \tan \frac{x}{2} dx$.
$I = x \log(1+\cos x) + \int x \tan \frac{x}{2} dx - \int x \tan \frac{x}{2} dx$.
$I = x \log(1+\cos x) + c$.

(A) We have the integral $I = \int \frac{\sin \theta}{\sin 3 \theta} d \theta$.
Using the identity $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$,we get:
$I = \int \frac{\sin \theta}{\sin \theta(3 - 4 \sin^2 \theta)} d \theta = \int \frac{1}{3 - 4 \sin^2 \theta} d \theta$.
Dividing the numerator and denominator by $\cos^2 \theta$:
$I = \int \frac{\sec^2 \theta}{3 \sec^2 \theta - 4 \tan^2 \theta} d \theta = \int \frac{\sec^2 \theta}{3(1 + \tan^2 \theta) - 4 \tan^2 \theta} d \theta = \int \frac{\sec^2 \theta}{3 - \tan^2 \theta} d \theta$.
Let $\tan \theta = t$,then $\sec^2 \theta d \theta = dt$.
$I = \int \frac{1}{(\sqrt{3})^2 - t^2} dt$.
Using the formula $\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + c$:
$I = \frac{1}{2 \sqrt{3}} \log \left| \frac{\sqrt{3} + t}{\sqrt{3} - t} \right| + c = \frac{1}{2 \sqrt{3}} \log \left| \frac{\sqrt{3} + \tan \theta}{\sqrt{3} - \tan \theta} \right| + c$.
Comparing this with the given expression $\frac{1}{2 k} \log \left|\frac{k+\tan \theta}{k-\tan \theta}\right|+c$,we find $k = \sqrt{3}$.

(A) Let $I = \int \frac{x + \sin x}{1 + \cos x} dx$.
Using the half-angle formulas,$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$.
Substituting these into the integral:
$I = \int \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} dx$
$I = \int \left( \frac{x}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) dx$
$I = \int \left( \frac{x}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2} \right) dx$
Now,apply integration by parts to the first term $\int \frac{x}{2} \sec^2 \frac{x}{2} dx$ taking $u = \frac{x}{2}$ and $dv = \sec^2 \frac{x}{2} dx$:
$\int \frac{x}{2} \sec^2 \frac{x}{2} dx = \frac{x}{2} \cdot 2 \tan \frac{x}{2} - \int (1) \cdot \tan \frac{x}{2} dx = x \tan \frac{x}{2} - \int \tan \frac{x}{2} dx$.
Substituting this back into the expression for $I$:
$I = x \tan \frac{x}{2} - \int \tan \frac{x}{2} dx + \int \tan \frac{x}{2} dx + c$
$I = x \tan \frac{x}{2} + c$.

(A) To solve the integral $I = \int \frac{\sin x}{\sin (x-\alpha)} dx$,we substitute $u = x - \alpha$,so $x = u + \alpha$ and $dx = du$.
Substituting these into the integral,we get:
$I = \int \frac{\sin (u + \alpha)}{\sin u} du$
Using the identity $\sin (A+B) = \sin A \cos B + \cos A \sin B$:
$I = \int \frac{\sin u \cos \alpha + \cos u \sin \alpha}{\sin u} du$
$I = \int (\cos \alpha + \cot u \sin \alpha) du$
$I = \cos \alpha \int du + \sin \alpha \int \cot u du$
$I = u \cos \alpha + \sin \alpha \log |\sin u| + c$
Substituting back $u = x - \alpha$:
$I = (x - \alpha) \cos \alpha + \sin \alpha \log |\sin (x - \alpha)| + c$
$I = x \cos \alpha - \alpha \cos \alpha + \sin \alpha \log |\sin (x - \alpha)| + c$
Comparing this with the given form $px - q \log |\sin (x - \alpha)| + c$:
$p = \cos \alpha$
$-q = \sin \alpha \implies q = -\sin \alpha$
Therefore,$pq = (\cos \alpha)(-\sin \alpha) = -\sin \alpha \cos \alpha = -\frac{1}{2} (2 \sin \alpha \cos \alpha) = -\frac{1}{2} \sin 2\alpha$.

(D) To solve the integral $\int \sqrt{x^2-8 x+7} \, dx$,we first complete the square for the quadratic expression inside the square root:
$x^2 - 8x + 7 = (x^2 - 8x + 16) - 16 + 7 = (x-4)^2 - 9 = (x-4)^2 - 3^2$.
Now the integral becomes $\int \sqrt{(x-4)^2 - 3^2} \, dx$.
Using the standard formula $\int \sqrt{t^2 - a^2} \, dt = \frac{t}{2} \sqrt{t^2 - a^2} - \frac{a^2}{2} \log \left|t + \sqrt{t^2 - a^2}\right| + C$,where $t = x-4$ and $a = 3$:
$= \frac{x-4}{2} \sqrt{(x-4)^2 - 3^2} - \frac{3^2}{2} \log \left|(x-4) + \sqrt{(x-4)^2 - 3^2}\right| + C$
$= \frac{x-4}{2} \sqrt{x^2-8x+7} - \frac{9}{2} \log \left|x-4 + \sqrt{x^2-8x+7}\right| + C$.
Comparing this with the given options,the correct option is $D$.

(A) To evaluate $\int \sqrt{x^2+4x+1} \, dx$,first complete the square inside the square root:
$x^2+4x+1 = (x^2+4x+4) - 3 = (x+2)^2 - (\sqrt{3})^2$.
Now the integral becomes $\int \sqrt{(x+2)^2 - (\sqrt{3})^2} \, dx$.
Using the standard formula $\int \sqrt{t^2-a^2} \, dt = \frac{t}{2} \sqrt{t^2-a^2} - \frac{a^2}{2} \log |t + \sqrt{t^2-a^2}| + C$,where $t = x+2$ and $a = \sqrt{3}$:
$= \frac{x+2}{2} \sqrt{(x+2)^2 - 3} - \frac{3}{2} \log |(x+2) + \sqrt{(x+2)^2 - 3}| + C$.
$= \frac{x+2}{2} \sqrt{x^2+4x+1} - \frac{3}{2} \log |x+2 + \sqrt{x^2+4x+1}| + C$.
Comparing this with the given options,the correct option is $A$.