Evaluation of various forms of integration Questions in English

(C) We know that $1 + \sin x = \sin^2(x/2) + \cos^2(x/2) + 2\sin(x/2)\cos(x/2) = (\sin(x/2) + \cos(x/2))^2$.
Alternatively,$1 + \sin x = 1 + \cos(\pi/2 - x) = 2\cos^2(\pi/4 - x/2)$.
Thus,$\sqrt{1 + \sin x} = \sqrt{2} \cos(\pi/4 - x/2) = \sqrt{2} \sin(\pi/4 + x/2)$.
The integral becomes $\int \frac{1}{\sqrt{2} \sin(\pi/4 + x/2)} \, dx$.
This is equal to $\frac{1}{\sqrt{2}} \int \csc(\pi/4 + x/2) \, dx$.
Using the formula $\int \csc \theta \, d\theta = \log \tan(\theta/2) + c$,we get:
$\frac{1}{\sqrt{2}} \cdot 2 \log \tan \left( \frac{\pi/4 + x/2}{2} \right) + c = \sqrt{2} \log \tan \left( \frac{\pi}{8} + \frac{x}{4} \right) + c$.

(A) We use the standard integration formula: $\int \sqrt{a^2 + x^2} \, dx = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\ln|x + \sqrt{a^2 + x^2}| + C$.
Substituting $a = 1$ into the formula:
$\int \sqrt{1 + x^2} \, dx = \frac{x}{2}\sqrt{1 + x^2} + \frac{1^2}{2}\ln|x + \sqrt{1 + x^2}| + C$.
Thus,the result is $\frac{x}{2}\sqrt{1 + x^2} + \frac{1}{2}\ln|x + \sqrt{1 + x^2}| + C$.

(B) Let $I = \int {\sqrt {{x^2} + {a^2}} \,dx} $.
Using integration by parts,taking $1$ as the second function:
$I = \sqrt {{x^2} + {a^2}} \cdot x - \int {\frac{x}{{\sqrt {{x^2} + {a^2}} }} \cdot x \,dx} $
$I = x\sqrt {{x^2} + {a^2}} - \int {\frac{{{x^2} + {a^2} - {a^2}}}{{\sqrt {{x^2} + {a^2}} }} \,dx} $
$I = x\sqrt {{x^2} + {a^2}} - \int {\sqrt {{x^2} + {a^2}} \,dx} + {a^2}\int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }}} $
$I = x\sqrt {{x^2} + {a^2}} - I + {a^2}\log \left| x + \sqrt {{x^2} + {a^2}} \right| + C$
$2I = x\sqrt {{x^2} + {a^2}} + {a^2}\log \left| x + \sqrt {{x^2} + {a^2}} \right| + C$
$I = \frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\log \left| x + \sqrt {{x^2} + {a^2}} \right| + C$.

(A) Let $I = \int \frac{e^{-x}}{1 + e^x} \, dx$.
Multiply the numerator and denominator by $e^{-x}$:
$I = \int \frac{e^{-x} \cdot e^{-x}}{e^{-x}(1 + e^x)} \, dx = \int \frac{e^{-2x}}{e^{-x} + 1} \, dx$.
Let $t = e^{-x} + 1$. Then $dt = -e^{-x} \, dx$,which implies $e^{-x} \, dx = -dt$.
Also,$e^{-x} = t - 1$.
Substituting these into the integral:
$I = \int \frac{t - 1}{t} (-dt) = -\int (1 - \frac{1}{t}) \, dt = \int (\frac{1}{t} - 1) \, dt$.
Integrating,we get:
$I = \log|t| - t + c$.
Substituting $t = e^{-x} + 1$ back:
$I = \log(e^{-x} + 1) - (e^{-x} + 1) + c$.
Since $\log(e^{-x} + 1) = \log(\frac{1 + e^x}{e^x}) = \log(1 + e^x) - \log(e^x) = \log(1 + e^x) - x$,
$I = \log(1 + e^x) - x - e^{-x} - 1 + c$.
Since $-1 + c$ is just another constant $C$,the final answer is $\log(1 + e^x) - x - e^{-x} + C$.

(A) We need to evaluate the integral $I = \int \sec x \tan^3 x \, dx$.
Rewrite $\tan^3 x$ as $\tan^2 x \cdot \tan x$:
$I = \int \sec x (\sec^2 x - 1) \tan x \, dx$
$I = \int \sec^2 x (\sec x \tan x) \, dx - \int \sec x \tan x \, dx$
Let $t = \sec x$,then $dt = \sec x \tan x \, dx$.
Substituting these into the integral:
$I = \int t^2 \, dt - \int 1 \, dt$
$I = \frac{t^3}{3} - t + c$
Substituting back $t = \sec x$:
$I = \frac{1}{3} \sec^3 x - \sec x + c$.

(C) Let $I = \int \frac{dx}{(x^2 - 1)\sqrt{x^2 + 1}}$.
Put $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
The integral becomes $\int \frac{\sec^2 \theta \, d\theta}{(\tan^2 \theta - 1)\sec \theta} = \int \frac{\sec \theta}{\tan^2 \theta - 1} \, d\theta = \int \frac{\cos \theta}{\sin^2 \theta - \cos^2 \theta} \, d\theta$.
Using $\cos^2 \theta = 1 - \sin^2 \theta$,we get $\int \frac{\cos \theta \, d\theta}{2\sin^2 \theta - 1}$.
Let $t = \sin \theta$,then $dt = \cos \theta \, d\theta$.
$I = \int \frac{dt}{2t^2 - 1} = \frac{1}{2} \int \frac{dt}{t^2 - (1/\sqrt{2})^2}$.
Using the formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right|$,we get $I = \frac{1}{2} \cdot \frac{1}{2(1/\sqrt{2})} \log \left| \frac{t - 1/\sqrt{2}}{t + 1/\sqrt{2}} \right| + c = \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{2}t - 1}{\sqrt{2}t + 1} \right| + c$.
Since $t = \sin \theta = \frac{x}{\sqrt{1+x^2}}$,we have $\sqrt{2}t = \frac{x\sqrt{2}}{\sqrt{1+x^2}}$.
Substituting this back,$I = \frac{1}{2\sqrt{2}} \log \left| \frac{\frac{x\sqrt{2}}{\sqrt{1+x^2}} - 1}{\frac{x\sqrt{2}}{\sqrt{1+x^2}} + 1} \right| + c = \frac{1}{2\sqrt{2}} \log \left| \frac{x\sqrt{2} - \sqrt{1+x^2}}{x\sqrt{2} + \sqrt{1+x^2}} \right| + c$.
Since $\log|a/b| = -\log|b/a|$,this is equivalent to $\frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{1+x^2} - x\sqrt{2}}{\sqrt{1+x^2} + x\sqrt{2}} \right| + c$.

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

(A) Let $I = \int \frac{1}{1 + \sin^2 x} \, dx$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x}{\sec^2 x + \tan^2 x} \, dx = \int \frac{\sec^2 x}{1 + \tan^2 x + \tan^2 x} \, dx = \int \frac{\sec^2 x}{1 + 2\tan^2 x} \, dx$.
Substitute $t = \tan x$,so $dt = \sec^2 x \, dx$:
$I = \int \frac{dt}{1 + 2t^2} = \frac{1}{2} \int \frac{dt}{\frac{1}{2} + t^2} = \frac{1}{2} \int \frac{dt}{(\frac{1}{\sqrt{2}})^2 + t^2}$.
Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = \frac{1}{2} \cdot \frac{1}{1/\sqrt{2}} \tan^{-1}(\frac{t}{1/\sqrt{2}}) + k = \frac{\sqrt{2}}{2} \tan^{-1}(\sqrt{2}t) + k = \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2} \tan x) + k$.

(D) To solve the integral $I = \int \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the numerator inside the square root:
$I = \int \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} \, dx = \int \sqrt{\frac{(1-x)^2}{1-x^2}} \, dx$
$I = \int \frac{1-x}{\sqrt{1-x^2}} \, dx = \int \frac{1}{\sqrt{1-x^2}} \, dx - \int \frac{x}{\sqrt{1-x^2}} \, dx$
For the first part,$\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x$.
For the second part,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$.
$-\int \frac{x}{\sqrt{1-x^2}} \, dx = -\int \frac{-1/2 \, du}{\sqrt{u}} = \frac{1}{2} \int u^{-1/2} \, du = \frac{1}{2} \cdot \frac{u^{1/2}}{1/2} = \sqrt{u} = \sqrt{1-x^2}$.
Combining these,we get $I = \sin^{-1} x + \sqrt{1-x^2} + c$.

(B) Let $I = \int \frac{\sin x}{\sin x - \cos x} \, dx$.
Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2\sin x}{\sin x - \cos x} \, dx$.
Rewrite the numerator: $2\sin x = (\sin x - \cos x) + (\sin x + \cos x)$.
So,$I = \frac{1}{2} \int \frac{(\sin x - \cos x) + (\sin x + \cos x)}{\sin x - \cos x} \, dx$.
$I = \frac{1}{2} \int \left( 1 + \frac{\sin x + \cos x}{\sin x - \cos x} \right) \, dx$.
$I = \frac{1}{2} \left[ \int 1 \, dx + \int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx \right]$.
For the second integral,let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) \, dx$.
$I = \frac{1}{2} [x + \log |\sin x - \cos x|] + c$.

(C) Let $I = \int \sqrt{\frac{1+x}{1-x}} \, dx$.
Multiply the numerator and denominator inside the square root by $(1+x)$:
$I = \int \sqrt{\frac{(1+x)^2}{(1-x)(1+x)}} \, dx = \int \frac{1+x}{\sqrt{1-x^2}} \, dx$.
Split the integral into two parts:
$I = \int \frac{1}{\sqrt{1-x^2}} \, dx + \int \frac{x}{\sqrt{1-x^2}} \, dx$.
The first integral is $\sin^{-1}x$.
For the second integral,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$.
$\int \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int u^{-1/2} \, du = -\frac{1}{2} (2u^{1/2}) = -\sqrt{1-x^2}$.
Combining these,we get $I = \sin^{-1}x - \sqrt{1-x^2} + c$.

(B) Given integral: $I = \int \frac{\cos 4x + 1}{\cot x - \tan x} dx$.
Using the identity $1 + \cos 4x = 2 \cos^2 2x$:
$I = \int \frac{2 \cos^2 2x}{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}} dx = \int \frac{2 \cos^2 2x}{\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}} dx$.
Since $\cos^2 x - \sin^2 x = \cos 2x$ and $\sin x \cos x = \frac{1}{2} \sin 2x$:
$I = \int \frac{2 \cos^2 2x}{\frac{\cos 2x}{\frac{1}{2} \sin 2x}} dx = \int \frac{2 \cos^2 2x \cdot \frac{1}{2} \sin 2x}{\cos 2x} dx$.
$I = \int \cos 2x \sin 2x dx = \frac{1}{2} \int \sin 4x dx$.
$I = \frac{1}{2} \left( -\frac{\cos 4x}{4} \right) + c = -\frac{1}{8} \cos 4x + c$.
Comparing with $k \cos 4x + c$,we get $k = -1/8$.

(A) Let $I = \int \frac{x^4}{x + x^5} dx$.
We can rewrite the numerator as $(x^4 + 1) - 1$.
$I = \int \frac{x^4 + 1 - 1}{x(1 + x^4)} dx$
$I = \int \frac{x^4 + 1}{x(1 + x^4)} dx - \int \frac{1}{x(1 + x^4)} dx$
$I = \int \frac{1}{x} dx - \int \frac{1}{x + x^5} dx$
Given that $\int \frac{1}{x + x^5} dx = f(x) + c_1$,we substitute this into the expression:
$I = \log |x| - (f(x) + c_1) + c_2$
$I = \log |x| - f(x) + c$,where $c = c_2 - c_1$ is a new constant.
Thus,the correct option is $A$.

(A) Let $I = \int \frac{dx}{\sin(x - a)\sin(x - b)}$.
Multiply and divide by $\sin(a - b)$:
$I = \frac{1}{\sin(a - b)} \int \frac{\sin((x - b) - (x - a))}{\sin(x - a)\sin(x - b)} dx$.
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin(a - b)} \int \frac{\sin(x - b)\cos(x - a) - \cos(x - b)\sin(x - a)}{\sin(x - a)\sin(x - b)} dx$.
$I = \frac{1}{\sin(a - b)} \left[ \int \frac{\cos(x - a)}{\sin(x - a)} dx - \int \frac{\cos(x - b)}{\sin(x - b)} dx \right]$.
$I = \frac{1}{\sin(a - b)} [ \log |\sin(x - a)| - \log |\sin(x - b)| ] + c$.
$I = \frac{1}{\sin(a - b)} \log \left| \frac{\sin(x - a)}{\sin(x - b)} \right| + c$.

(C) Let $I = \int \cos^{-3/7} x \sin^{-11/7} x \, dx$.
Here,$m = -3/7$ and $n = -11/7$.
$m + n = -3/7 - 11/7 = -14/7 = -2$,which is a negative even integer.
We can rewrite the integral as:
$I = \int \cos^{-3/7} x \sin^{-11/7} x \, dx = \int \left( \frac{\cos x}{\sin x} \right)^{-3/7} \sin^{-3/7 - 11/7} x \, dx$
$I = \int \cot^{-3/7} x \sin^{-2} x \, dx = \int \cot^{-3/7} x \csc^2 x \, dx$.
Let $\cot x = t$,then $-\csc^2 x \, dx = dt$,or $\csc^2 x \, dx = -dt$.
Substituting these into the integral:
$I = \int t^{-3/7} (-dt) = -\int t^{-3/7} \, dt$.
$I = -\frac{t^{-3/7 + 1}}{-3/7 + 1} + c = -\frac{t^{4/7}}{4/7} + c = -\frac{7}{4} t^{4/7} + c$.
Since $t = \cot x$,we have $I = -\frac{7}{4} \cot^{4/7} x + c$.
Using $\cot x = \frac{1}{\tan x}$,we get $I = -\frac{7}{4} (\tan x)^{-4/7} + c = -\frac{7}{4} \tan^{-4/7} x + c$.
Thus,option $(c)$ is correct.

(B) Let $I = \int {\left[ {\log (\log x) + \frac{1}{{{{(\log x)}^2}}}} \right]\,dx}$.
We can split the integral as $I = \int {\log (\log x)\,dx + \int {\frac{1}{{{{(\log x)}^2}}}} \,dx}$.
Applying integration by parts to the first term $\int {\log (\log x)\,dx}$,let $u = \log (\log x)$ and $dv = dx$. Then $du = \frac{1}{\log x} \cdot \frac{1}{x} dx$ and $v = x$.
So,$\int {\log (\log x)\,dx} = x \log (\log x) - \int {x \cdot \frac{1}{x \log x} \,dx} = x \log (\log x) - \int {\frac{1}{\log x} \,dx}$.
Now,apply integration by parts to $\int {\frac{1}{\log x} \,dx}$. Let $u = \frac{1}{\log x}$ and $dv = dx$. Then $du = -\frac{1}{(\log x)^2} \cdot \frac{1}{x} dx$ and $v = x$.
Thus,$\int {\frac{1}{\log x} \,dx} = \frac{x}{\log x} - \int {x \cdot \left( -\frac{1}{x(\log x)^2} \right) \,dx} = \frac{x}{\log x} + \int {\frac{1}{(\log x)^2} \,dx}$.
Substituting this back into the expression for $I$:
$I = x \log (\log x) - \left( \frac{x}{\log x} + \int {\frac{1}{(\log x)^2} \,dx} \right) + \int {\frac{1}{(\log x)^2} \,dx}$.
$I = x \log (\log x) - \frac{x}{\log x} + c$.

(B) We have $I = \int \frac{x - \sin x}{1 - \cos x} dx$.
Using the identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get:
$I = \int \frac{x}{2 \sin^2 \frac{x}{2}} dx - \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} dx$
$I = \frac{1}{2} \int x \csc^2 \frac{x}{2} dx - \int \cot \frac{x}{2} dx$.
Applying integration by parts to the first term:
Let $u = x$ and $dv = \csc^2 \frac{x}{2} dx$. Then $du = dx$ and $v = -2 \cot \frac{x}{2}$.
$\int x \csc^2 \frac{x}{2} dx = x(-2 \cot \frac{x}{2}) - \int (-2 \cot \frac{x}{2}) dx = -2x \cot \frac{x}{2} + 2 \int \cot \frac{x}{2} dx$.
Substituting this back into the expression for $I$:
$I = \frac{1}{2} [-2x \cot \frac{x}{2} + 2 \int \cot \frac{x}{2} dx] - \int \cot \frac{x}{2} dx$
$I = -x \cot \frac{x}{2} + \int \cot \frac{x}{2} dx - \int \cot \frac{x}{2} dx + c$
$I = -x \cot \frac{x}{2} + c$.

(C) We need to evaluate the integral $I = \int \sqrt{x^2 - 8x + 7} \, dx$.
First,complete the square for the quadratic expression: $x^2 - 8x + 7 = (x^2 - 8x + 16) - 16 + 7 = (x - 4)^2 - 9 = (x - 4)^2 - 3^2$.
Now the integral becomes $I = \int \sqrt{(x - 4)^2 - 3^2} \, dx$.
Using the standard formula $\int \sqrt{t^2 - a^2} \, dt = \frac{t}{2}\sqrt{t^2 - a^2} - \frac{a^2}{2}\log |t + \sqrt{t^2 - a^2}| + c$,where $t = x - 4$ and $a = 3$:
$I = \frac{x - 4}{2}\sqrt{(x - 4)^2 - 3^2} - \frac{3^2}{2}\log |(x - 4) + \sqrt{(x - 4)^2 - 3^2}| + c$.
Simplifying this,we get $I = \frac{1}{2}(x - 4)\sqrt{x^2 - 8x + 7} - \frac{9}{2}\log |x - 4 + \sqrt{x^2 - 8x + 7}| + c$.
Thus,the correct option is $C$.

(C) To evaluate the integral $I = \int \frac{1}{1 + \cos^2 x} dx$,divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x}{\sec^2 x + 1} dx$
Since $\sec^2 x = 1 + \tan^2 x$,we have:
$I = \int \frac{\sec^2 x}{1 + \tan^2 x + 1} dx = \int \frac{\sec^2 x}{\tan^2 x + 2} dx$
Let $t = \tan x$,then $dt = \sec^2 x dx$:
$I = \int \frac{dt}{t^2 + 2} = \int \frac{dt}{t^2 + (\sqrt{2})^2}$
Using the standard formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{t}{\sqrt{2}} \right) + c$
Substituting $t = \tan x$ back:
$I = \frac{1}{\sqrt{2}} \tan^{-1}\left( \frac{1}{\sqrt{2}} \tan x \right) + c$.

(A) To evaluate the integral $I = \int \frac{3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx$,we express the numerator as a linear combination of the denominator and its derivative:
$3\sin x + 2\cos x = A(3\cos x + 2\sin x) + B \frac{d}{dx}(3\cos x + 2\sin x)$
$3\sin x + 2\cos x = A(3\cos x + 2\sin x) + B(-3\sin x + 2\cos x)$
Comparing the coefficients of $\sin x$ and $\cos x$ on both sides:
For $\sin x$: $3 = 2A - 3B$
For $\cos x$: $2 = 3A + 2B$
Solving these simultaneous equations:
Multiply the first by $2$ and the second by $3$:
$6 = 4A - 6B$
$6 = 9A + 6B$
Adding them: $12 = 13A \implies A = \frac{12}{13}$
Substituting $A$ into the first equation: $3 = 2(\frac{12}{13}) - 3B \implies 3B = \frac{24}{13} - 3 = \frac{24 - 39}{13} = -\frac{15}{13} \implies B = -\frac{5}{13}$
Thus,the integral becomes:
$I = \int \left( \frac{12}{13} \frac{3\cos x + 2\sin x}{3\cos x + 2\sin x} - \frac{5}{13} \frac{-3\sin x + 2\cos x}{3\cos x + 2\sin x} \right) dx$
$I = \frac{12}{13} \int 1 \, dx - \frac{5}{13} \int \frac{d}{dx}(3\cos x + 2\sin x) \cdot \frac{1}{3\cos x + 2\sin x} \, dx$
$I = \frac{12}{13}x - \frac{5}{13} \log |3\cos x + 2\sin x| + C$

(D) To evaluate the integral $I = \int {\frac{{{x^2} + 1}}{{{x^4} + 1}}} \,dx$,divide the numerator and denominator by $x^2$:
$I = \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}} \,dx}$
Rewrite the denominator as a perfect square:
$I = \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{{\left( {x - \frac{1}{x}} \right)}^2} + 2}}} \,dx$
Let $t = x - \frac{1}{x}$. Then $dt = \left( {1 + \frac{1}{{{x^2}}}} \right)dx$.
Substituting these into the integral,we get:
$I = \int {\frac{{dt}}{{{t^2} + {{(\sqrt 2 )}^2}}}} = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{t}{{\sqrt 2 }}} \right) + c$
Substituting $t = x - \frac{1}{x} = \frac{{{x^2} - 1}}{x}$ back into the expression:
$I = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{{x^2} - 1}}{{\sqrt 2 x}}} \right) + c$.

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

(A) To evaluate the integral $I = \int \frac{3\cos x + 3\sin x}{4\sin x + 5\cos x} \, dx$,we express the numerator as $A(\text{denominator}) + B(\frac{d}{dx}(\text{denominator}))$.
Let $3\cos x + 3\sin x = A(4\sin x + 5\cos x) + B(4\cos x - 5\sin x)$.
Equating the coefficients of $\sin x$ and $\cos x$:
For $\sin x$: $4A - 5B = 3$
For $\cos x$: $5A + 4B = 3$
Solving these simultaneous equations:
Multiply the first by $4$ and the second by $5$: $16A - 20B = 12$ and $25A + 20B = 15$.
Adding them: $41A = 27 \implies A = \frac{27}{41}$.
Substituting $A$ into $5A + 4B = 3$: $5(\frac{27}{41}) + 4B = 3 \implies \frac{135}{41} + 4B = 3 \implies 4B = 3 - \frac{135}{41} = \frac{123 - 135}{41} = -\frac{12}{41} \implies B = -\frac{3}{41}$.
Thus,$I = \int \frac{A(4\sin x + 5\cos x) + B(4\cos x - 5\sin x)}{4\sin x + 5\cos x} \, dx = A \int 1 \, dx + B \int \frac{4\cos x - 5\sin x}{4\sin x + 5\cos x} \, dx$.
$I = \frac{27}{41}x - \frac{3}{41} \ln |4\sin x + 5\cos x| + C$.

(B) Let $I = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\;dx}$.
We know that $1 = (\sin^2 x + \cos^2 x)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x$,so the denominator $1 - 2\sin^2 x \cos^2 x = \sin^4 x + \cos^4 x$.
Thus,$I = \int {\frac{{(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)}}{{\sin^4 x + \cos^4 x}}\;dx}$.
$I = \int {(\sin^4 x - \cos^4 x)\,dx}$.
Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have $\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$.
Since $\sin^2 x + \cos^2 x = 1$,the integral becomes $I = \int {(\sin^2 x - \cos^2 x)\,dx}$.
Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have $\sin^2 x - \cos^2 x = -\cos 2x$.
Therefore,$I = \int {-\cos 2x\,dx} = -\frac{\sin 2x}{2} + c$.

(A) Let $I = \int x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \;dx$.
Multiply the numerator and denominator by $\sqrt{1-x^2}$:
$I = \int \frac{x(1-x^2)}{\sqrt{(1-x^2)(1+x^2)}} \;dx = \int \frac{x - x^3}{\sqrt{1-x^4}} \;dx$.
Split the integral:
$I = \int \frac{x}{\sqrt{1-x^4}} \;dx - \int \frac{x^3}{\sqrt{1-x^4}} \;dx$.
For the first part,let $x^2 = t$,then $2x \;dx = dt$,so $x \;dx = \frac{1}{2} dt$:
$\int \frac{x}{\sqrt{1-x^4}} \;dx = \frac{1}{2} \int \frac{dt}{\sqrt{1-t^2}} = \frac{1}{2} \sin^{-1}(t) = \frac{1}{2} \sin^{-1}(x^2)$.
For the second part,let $1-x^4 = u$,then $-4x^3 \;dx = du$,so $x^3 \;dx = -\frac{1}{4} du$:
$\int \frac{x^3}{\sqrt{1-x^4}} \;dx = -\frac{1}{4} \int u^{-1/2} \;du = -\frac{1}{4} (2\sqrt{u}) = -\frac{1}{2} \sqrt{1-x^4}$.
Combining these results:
$I = \frac{1}{2} \sin^{-1}(x^2) - (-\frac{1}{2} \sqrt{1-x^4}) + c = \frac{1}{2} [\sin^{-1}(x^2) + \sqrt{1-x^4}] + c$.

(C) Given $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log(f(x)) + c$.
Differentiating both sides with respect to $x$,we get:
$f(x) \sin x \cos x = \frac{1}{2(b^2 - a^2)} \cdot \frac{f'(x)}{f(x)}$.
Rearranging the terms:
$2(b^2 - a^2) \sin x \cos x = \frac{f'(x)}{f(x)^2}$.
Integrating both sides with respect to $x$:
$\int (2b^2 \sin x \cos x - 2a^2 \sin x \cos x) \, dx = \int \frac{f'(x)}{f(x)^2} \, dx$.
Using the substitution $u = \sin^2 x$ and $v = \cos^2 x$,we have $du = 2 \sin x \cos x \, dx$ and $dv = -2 \cos x \sin x \, dx$.
Thus,$b^2 \sin^2 x - a^2 \cos^2 x = -\frac{1}{f(x)} + C_1$.
Simplifying,we get $f(x) = \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x}$ (adjusting constants).
Comparing with the given options,the correct form is $f(x) = \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x}$.

(C) To evaluate the integral $I = \int \frac{dx}{4\sin^2 x + 5\cos^2 x}$,divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sec^2 x dx}{4\tan^2 x + 5}$.
Let $\tan x = t$,then $\sec^2 x dx = dt$.
The integral becomes $I = \int \frac{dt}{4t^2 + 5} = \frac{1}{4} \int \frac{dt}{t^2 + (\frac{\sqrt{5}}{2})^2}$.
Using the standard formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$,we get:
$I = \frac{1}{4} \cdot \frac{1}{\frac{\sqrt{5}}{2}} \tan^{-1} \left( \frac{t}{\frac{\sqrt{5}}{2}} \right) + c = \frac{1}{2\sqrt{5}} \tan^{-1} \left( \frac{2t}{\sqrt{5}} \right) + c$.
Substituting $t = \tan x$,we get $I = \frac{1}{2\sqrt{5}} \tan^{-1} \left( \frac{2\tan x}{\sqrt{5}} \right) + c$.

(D) Let $I = \int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx$. Multiply numerator and denominator by $e^x$:
$I = \int \frac{4e^{2x} + 6}{9e^{2x} - 4} dx$.
We express the numerator as $4e^{2x} + 6 = m(18e^{2x}) + n(9e^{2x} - 4)$.
Comparing coefficients: $18m + 9n = 4$ and $-4n = 6 \implies n = -\frac{3}{2}$.
$18m + 9(-\frac{3}{2}) = 4 \implies 18m = 4 + \frac{27}{2} = \frac{35}{2} \implies m = \frac{35}{36}$.
Thus,$I = \int \frac{\frac{35}{36}(18e^{2x}) - \frac{3}{2}(9e^{2x} - 4)}{9e^{2x} - 4} dx = \frac{35}{36} \int \frac{18e^{2x}}{9e^{2x} - 4} dx - \frac{3}{2} \int dx$.
$I = \frac{35}{36} \log |9e^{2x} - 4| - \frac{3}{2}x + C$.
Comparing with $Ax + B \log |9e^{2x} - 4| + C$,we get $A = -\frac{3}{2}$ and $B = \frac{35}{36}$.
Since none of the options match these values,the correct answer is $(d)$.

(C) Let $I = \int \frac{x^2}{(x\sin x + \cos x)^2} \, dx$.
We can rewrite the integrand as:
$I = \int \frac{x \cos x}{(x\sin x + \cos x)^2} \cdot \frac{x}{\cos x} \, dx$.
Let $u = \frac{x}{\cos x}$ and $dv = \frac{x \cos x}{(x\sin x + \cos x)^2} \, dx$.
Then $du = \frac{\cos x - x(-\sin x)}{\cos^2 x} \, dx = \frac{\cos x + x\sin x}{\cos^2 x} \, dx$.
Integrating $dv$,we get $v = \int \frac{x \cos x}{(x\sin x + \cos x)^2} \, dx$.
Let $t = x\sin x + \cos x$,then $dt = (\sin x + x\cos x - \sin x) \, dx = x\cos x \, dx$.
So,$v = \int \frac{dt}{t^2} = -\frac{1}{t} = -\frac{1}{x\sin x + \cos x}$.
Using integration by parts,$I = uv - \int v \, du$:
$I = -\frac{x}{\cos x(x\sin x + \cos x)} - \int \left( -\frac{1}{x\sin x + \cos x} \right) \cdot \frac{x\sin x + \cos x}{\cos^2 x} \, dx$.
$I = -\frac{x}{\cos x(x\sin x + \cos x)} + \int \sec^2 x \, dx$.
$I = -\frac{x}{\cos x(x\sin x + \cos x)} + \tan x$.
$I = \frac{-x + \tan x \cos x (x\sin x + \cos x)}{\cos x (x\sin x + \cos x)}$.
$I = \frac{-x + \sin x (x\sin x + \cos x)}{\cos x (x\sin x + \cos x)} = \frac{-x + x\sin^2 x + \sin x \cos x}{\cos x (x\sin x + \cos x)}$.
$I = \frac{\sin x \cos x - x(1 - \sin^2 x)}{\cos x (x\sin x + \cos x)} = \frac{\sin x \cos x - x \cos^2 x}{\cos x (x\sin x + \cos x)}$.
$I = \frac{\cos x (\sin x - x \cos x)}{\cos x (x\sin x + \cos x)} = \frac{\sin x - x \cos x}{x\sin x + \cos x} + C$.

(D) Let $I = \int {e^{x/2}} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \, dx$.
Using the formula $\int e^{ax} \sin(bx + c) \, dx = \frac{e^{ax}}{a^2 + b^2} [a \sin(bx + c) - b \cos(bx + c)] + C$.
Here $a = 1/2$ and $b = 1/2$.
$I = \frac{e^{x/2}}{(1/2)^2 + (1/2)^2} \left[ \frac{1}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) - \frac{1}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right] + C$.
$I = \frac{e^{x/2}}{1/4 + 1/4} \cdot \frac{1}{2} \left[ \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) - \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right] + C$.
$I = \frac{e^{x/2}}{1/2} \cdot \frac{1}{2} \left[ \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) - \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right] + C$.
$I = e^{x/2} \left[ \sin \frac{x}{2} \cos \frac{\pi}{4} + \cos \frac{x}{2} \sin \frac{\pi}{4} - (\cos \frac{x}{2} \cos \frac{\pi}{4} - \sin \frac{x}{2} \sin \frac{\pi}{4}) \right] + C$.
Since $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,
$I = e^{x/2} \left[ \frac{1}{\sqrt{2}} \sin \frac{x}{2} + \frac{1}{\sqrt{2}} \cos \frac{x}{2} - \frac{1}{\sqrt{2}} \cos \frac{x}{2} + \frac{1}{\sqrt{2}} \sin \frac{x}{2} \right] + C$.
$I = e^{x/2} \left[ \frac{2}{\sqrt{2}} \sin \frac{x}{2} \right] + C = \sqrt{2} {e^{x/2}} \sin \frac{x}{2} + C$.
Thus,the correct option is $D$.

(B) We know that $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$.
Substituting these into the integral,we get:
$\int \frac{dx}{2 + \cos x} = \int \frac{dx}{2(\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}) + (\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$
$= \int \frac{dx}{\sin^2 \frac{x}{2} + 3\cos^2 \frac{x}{2}}$
Divide numerator and denominator by $\cos^2 \frac{x}{2}$:
$= \int \frac{\sec^2 \frac{x}{2}}{\tan^2 \frac{x}{2} + 3} dx$
Let $\tan \frac{x}{2} = t$,then $\frac{1}{2} \sec^2 \frac{x}{2} dx = dt$,which implies $\sec^2 \frac{x}{2} dx = 2dt$.
Substituting this into the integral:
$= \int \frac{2 dt}{t^2 + 3} = 2 \int \frac{dt}{t^2 + (\sqrt{3})^2}$
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + c$:
$= 2 \cdot \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{t}{\sqrt{3}} \right) + c = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{\tan \frac{x}{2}}{\sqrt{3}} \right) + c$.

(A) Let $I = \int_0^{\pi /2} {\left( {\frac{\theta }{{\sin \theta }}} \right)^2 d\theta } $.
Using integration by parts,let $u = \theta^2$ and $dv = \csc^2 \theta \, d\theta$. Then $du = 2\theta \, d\theta$ and $v = -\cot \theta$.
$I = [-\theta^2 \cot \theta]_0^{\pi/2} + \int_0^{\pi/2} 2\theta \cot \theta \, d\theta$.
Evaluating the boundary term: $\lim_{\theta \to \pi/2} (-\theta^2 \cot \theta) = 0$ and $\lim_{\theta \to 0} (-\theta^2 \cot \theta) = \lim_{\theta \to 0} (-\theta^2 \frac{\cos \theta}{\sin \theta}) = 0$.
So,$I = 2 \int_0^{\pi/2} \theta \cot \theta \, d\theta$.
Using integration by parts again,let $u = \theta$ and $dv = \cot \theta \, d\theta$. Then $du = d\theta$ and $v = \ln(\sin \theta)$.
$I = 2 [\theta \ln(\sin \theta)]_0^{\pi/2} - 2 \int_0^{\pi/2} \ln(\sin \theta) \, d\theta$.
Since $\ln(\sin(\pi/2)) = \ln(1) = 0$ and $\lim_{\theta \to 0} \theta \ln(\sin \theta) = 0$,the first term is $0$.
We know that $\int_0^{\pi/2} \ln(\sin \theta) \, d\theta = -\frac{\pi}{2} \ln 2$.
Therefore,$I = 0 - 2(-\frac{\pi}{2} \ln 2) = \pi \ln 2$.

(D) Let $I(b) = \int_0^1 \frac{x^b - 1}{\log x} \, dx$.
Using Feynman's technique of differentiation under the integral sign,we differentiate $I(b)$ with respect to $b$:
$I'(b) = \frac{d}{db} \int_0^1 \frac{x^b - 1}{\log x} \, dx = \int_0^1 \frac{\partial}{\partial b} \left( \frac{x^b - 1}{\log x} \right) \, dx$.
Since $\frac{\partial}{\partial b} (x^b) = x^b \log x$,we have:
$I'(b) = \int_0^1 \frac{x^b \log x}{\log x} \, dx = \int_0^1 x^b \, dx$.
Evaluating the integral:
$I'(b) = \left[ \frac{x^{b+1}}{b+1} \right]_0^1 = \frac{1}{b+1}$.
Now,integrate $I'(b)$ with respect to $b$:
$I(b) = \int \frac{1}{b+1} \, db = \log(b+1) + C$.
When $b = 0$,$I(0) = \int_0^1 \frac{x^0 - 1}{\log x} \, dx = \int_0^1 0 \, dx = 0$.
Substituting $b = 0$ into our expression: $I(0) = \log(0+1) + C = 0 \implies C = 0$.
Thus,$I(b) = \log(b+1)$.

(A) The given formula is $\int_0^\infty \frac{x^2 \, dx}{(x^2 + a^2)(x^2 + b^2)(x^2 + c^2)} = \frac{\pi}{2(a + b)(b + c)(c + a)}$.
To find the value of $I = \int_0^\infty \frac{x^2 \, dx}{(x^2 + 4)(x^2 + 9)}$,we can rewrite the denominator as $(x^2 + 2^2)(x^2 + 3^2)(x^2 + 0^2)$.
Comparing this with the given formula,we have $a = 2$,$b = 3$,and $c = 0$.
Substituting these values into the right-hand side of the formula:
$I = \frac{\pi}{2(2 + 3)(3 + 0)(0 + 2)}$
$I = \frac{\pi}{2(5)(3)(2)}$
$I = \frac{\pi}{60}$.

(A) Let $I = \int \frac{dx}{x^2(x^4 + 1)^{3/4}}$.
Taking $x^4$ common from the term inside the bracket in the denominator:
$I = \int \frac{dx}{x^2 \left[ x^4(1 + \frac{1}{x^4}) \right]^{3/4}}$
$I = \int \frac{dx}{x^2 \cdot (x^4)^{3/4} \cdot (1 + \frac{1}{x^4})^{3/4}}$
$I = \int \frac{dx}{x^2 \cdot x^3 \cdot (1 + \frac{1}{x^4})^{3/4}}$
$I = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$
Let $t = 1 + \frac{1}{x^4}$. Then $dt = -\frac{4}{x^5} dx$,which implies $\frac{dt}{-4} = \frac{dx}{x^5}$.
Substituting these into the integral:
$I = \int \frac{1}{t^{3/4}} \cdot (-\frac{1}{4} dt)$
$I = -\frac{1}{4} \int t^{-3/4} dt$
$I = -\frac{1}{4} \left[ \frac{t^{1/4}}{1/4} \right] + c$
$I = -t^{1/4} + c$
Substituting $t = 1 + \frac{1}{x^4}$ back:
$I = -\left( 1 + \frac{1}{x^4} \right)^{1/4} + c$
$I = -\left( \frac{x^4 + 1}{x^4} \right)^{1/4} + c$.

(D) Let $I = \int \frac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}dx$.
Divide the numerator and denominator by ${x^{15}}$:
$I = \int \frac{{\frac{{2{x^{12}}}}{{{x^{15}}}} + \frac{{5{x^9}}}{{{x^{15}}}}}}{{{{\left( {\frac{{{x^5} + {x^3} + 1}}{{{x^5}}}} \right)}^3}}}dx = \int \frac{{2{x^{ - 3}} + 5{x^{ - 6}}}}{{{{\left( {1 + {x^{ - 2}} + {x^{ - 5}}} \right)}^3}}}dx$.
Let $t = 1 + {x^{ - 2}} + {x^{ - 5}}$.
Then $dt = ( - 2{x^{ - 3}} - 5{x^{ - 6}})dx$,which implies $-dt = (2{x^{ - 3}} + 5{x^{ - 6}})dx$.
Substituting these into the integral:
$I = \int \frac{{ - dt}}{{{t^3}}} = - \int {{t^{ - 3}}} dt = - \left( \frac{{{t^{ - 2}}}}{{ - 2}} \right) + C = \frac{1}{{2{t^2}}} + C$.
Substituting $t = 1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^5}}} = \frac{{{x^5} + {x^3} + 1}}{{{x^5}}}$:
$I = \frac{1}{{2{{\left( {\frac{{{x^5} + {x^3} + 1}}{{{x^5}}}} \right)}^2}}} + C = \frac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$.

(C) Given $I_n = \int \tan^n x dx$.
We need to evaluate $I_4 + I_6 = \int \tan^4 x dx + \int \tan^6 x dx$.
$I_4 + I_6 = \int (\tan^4 x + \tan^6 x) dx$.
Factor out $\tan^4 x$:
$I_4 + I_6 = \int \tan^4 x (1 + \tan^2 x) dx$.
Since $1 + \tan^2 x = \sec^2 x$,we have:
$I_4 + I_6 = \int \tan^4 x \sec^2 x dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
The integral becomes $\int u^4 du = \frac{u^5}{5} + C$.
Substituting back $u = \tan x$,we get:
$I_4 + I_6 = \frac{1}{5} \tan^5 x + C$.
Comparing this with $a \tan^5 x + b x^5 + C$,we get $a = \frac{1}{5}$ and $b = 0$.
Thus,the ordered pair $(a, b) = (\frac{1}{5}, 0)$.

(D) Let $I = \int {(1 + x - {x^{ - 1}}){e^{x + {x^{ - 1}}}}\,dx}$.
We can rewrite the integrand as:
$I = \int {e^{x + {x^{ - 1}}}} (1 + x - \frac{1}{x}) \,dx$.
Observe that $\frac{d}{dx}(x{e^{x + {x^{ - 1}}}}) = 1 \cdot {e^{x + {x^{ - 1}}}} + x \cdot {e^{x + {x^{ - 1}}}} \cdot (1 - \frac{1}{x^2})$.
$= {e^{x + {x^{ - 1}}}} [1 + x(1 - \frac{1}{x^2})] = {e^{x + {x^{ - 1}}}} [1 + x - \frac{1}{x}]$.
Thus,$\int {e^{x + {x^{ - 1}}}} (1 + x - \frac{1}{x}) \,dx = x{e^{x + {x^{ - 1}}}} + c$.

(B) Let $I = \int \frac{dx}{(1 + x^2)\sqrt{1 - x^2}}$.
Put $x = \sin \theta$,then $dx = \cos \theta \, d\theta$.
Substituting these into the integral:
$I = \int \frac{\cos \theta \, d\theta}{(1 + \sin^2 \theta)\sqrt{1 - \sin^2 \theta}} = \int \frac{\cos \theta \, d\theta}{(1 + \sin^2 \theta)\cos \theta} = \int \frac{d\theta}{1 + \sin^2 \theta}$.
Divide numerator and denominator by $\cos^2 \theta$:
$I = \int \frac{\sec^2 \theta \, d\theta}{\sec^2 \theta + \tan^2 \theta} = \int \frac{\sec^2 \theta \, d\theta}{1 + \tan^2 \theta + \tan^2 \theta} = \int \frac{\sec^2 \theta \, d\theta}{1 + 2\tan^2 \theta}$.
Now,put $t = \tan \theta$,so $dt = \sec^2 \theta \, d\theta$:
$I = \int \frac{dt}{1 + 2t^2} = \frac{1}{2} \int \frac{dt}{t^2 + (1/\sqrt{2})^2}$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = \frac{1}{2} \cdot \frac{1}{1/\sqrt{2}} \tan^{-1} \left( \frac{t}{1/\sqrt{2}} \right) + c = \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2}t) + c$.
Since $t = \tan \theta$ and $x = \sin \theta$,we have $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x}{\sqrt{1 - x^2}}$.
Therefore,$I = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x\sqrt{2}}{\sqrt{1 - x^2}} \right) + c$.

(D) To evaluate $I = \int \sqrt{\frac{a+x}{a-x}} \, dx$,multiply the numerator and denominator inside the square root by $\sqrt{a+x}$.
$I = \int \frac{a+x}{\sqrt{a^2-x^2}} \, dx = \int \frac{a}{\sqrt{a^2-x^2}} \, dx + \int \frac{x}{\sqrt{a^2-x^2}} \, dx$.
For the first integral,we use the standard formula $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}(\frac{x}{a}) + c$ or $-\cos^{-1}(\frac{x}{a}) + c$.
For the second integral,let $u = a^2-x^2$,then $du = -2x \, dx$,so $\int \frac{x}{\sqrt{a^2-x^2}} \, dx = -\sqrt{a^2-x^2} + c$.
Combining these,we get $I = a \sin^{-1}(\frac{x}{a}) - \sqrt{a^2-x^2} + c$.
Since $\sin^{-1}(\frac{x}{a}) = \frac{\pi}{2} - \cos^{-1}(\frac{x}{a})$,the expression becomes $a(\frac{\pi}{2} - \cos^{-1}(\frac{x}{a})) - \sqrt{a^2-x^2} + c = -a \cos^{-1}(\frac{x}{a}) - \sqrt{a^2-x^2} + C$.

(A) Let $x = \cos^2 \theta$. Then $dx = -2 \cos \theta \sin \theta \, d\theta$.
Substituting these into the integral:
$\int \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} (-2 \cos \theta \sin \theta) \, d\theta = \int \sqrt{\frac{2 \sin^2(\theta/2)}{2 \cos^2(\theta/2)}} (-2 \cos \theta \sin \theta) \, d\theta$
$= \int \tan(\theta/2) (-2 \cos \theta \sin \theta) \, d\theta = \int \frac{\sin(\theta/2)}{\cos(\theta/2)} (-4 \sin(\theta/2) \cos(\theta/2) \cos \theta) \, d\theta$
$= -4 \int \sin^2(\theta/2) \cos \theta \, d\theta = -2 \int (1 - \cos \theta) \cos \theta \, d\theta$
$= -2 \int (\cos \theta - \cos^2 \theta) \, d\theta = -2 \int \cos \theta \, d\theta + \int (1 + \cos 2\theta) \, d\theta$
$= -2 \sin \theta + \theta + \frac{1}{2} \sin 2\theta + c = \theta + \sin \theta \cos \theta - 2 \sin \theta + c$
Since $x = \cos^2 \theta$,we have $\cos \theta = \sqrt{x}$ and $\sin \theta = \sqrt{1 - x}$.
Substituting back: $\cos^{-1}\sqrt{x} + \sqrt{1 - x} \cdot \sqrt{x} - 2 \sqrt{1 - x} + c = \cos^{-1}\sqrt{x} + \sqrt{1 - x}(\sqrt{x} - 2) + c$.

(C) To evaluate the integral $I = \int \frac{dx}{(2\sin x + \cos x)^2}$,we can rewrite the denominator by factoring out $\cos x$ from the expression inside the square.
$I = \int \frac{dx}{(\cos x (2\tan x + 1))^2} = \int \frac{dx}{\cos^2 x (2\tan x + 1)^2}$
Since $\frac{1}{\cos^2 x} = \sec^2 x$,we have:
$I = \int \frac{\sec^2 x}{(2\tan x + 1)^2} dx$
Let $u = 2\tan x + 1$. Then $du = 2\sec^2 x dx$,which implies $\sec^2 x dx = \frac{du}{2}$.
Substituting these into the integral:
$I = \int \frac{1}{u^2} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-2} du$
$I = \frac{1}{2} \left( \frac{u^{-1}}{-1} \right) + c = -\frac{1}{2u} + c$
Substituting $u = 2\tan x + 1$ back:
$I = -\frac{1}{2(2\tan x + 1)} + c$
Alternatively,using the method provided in the prompt:
$I = \int \frac{dx}{\sin^2 x (2\csc x \sin x + \cot x)^2} = \int \frac{dx}{\sin^2 x (2 + \cot x)^2} = \int \frac{\csc^2 x}{(2 + \cot x)^2} dx$
Let $t = 2 + \cot x$,then $dt = -\csc^2 x dx$,so $\csc^2 x dx = -dt$.
$I = \int \frac{-dt}{t^2} = \frac{1}{t} + c = \frac{1}{2 + \cot x} + c$.

(D) We have the integral $I = \int \frac{1}{(\sin x + 4)(\sin x - 1)} dx$.
Using partial fractions,$\frac{1}{(\sin x + 4)(\sin x - 1)} = \frac{1}{5} \left( \frac{1}{\sin x - 1} - \frac{1}{\sin x + 4} \right)$.
Substituting $\sin x = \frac{2t}{1+t^2}$ where $t = \tan \frac{x}{2}$ and $dx = \frac{2dt}{1+t^2}$,we get:
$I = \frac{1}{5} \int \frac{1}{\frac{2t}{1+t^2} - 1} \frac{2dt}{1+t^2} - \frac{1}{5} \int \frac{1}{\frac{2t}{1+t^2} + 4} \frac{2dt}{1+t^2}$
$I = \frac{1}{5} \int \frac{2dt}{2t - 1 - t^2} - \frac{1}{5} \int \frac{2dt}{2t + 4 + 4t^2}$
$I = -\frac{2}{5} \int \frac{dt}{(t-1)^2} - \frac{1}{10} \int \frac{dt}{t^2 + \frac{1}{2}t + 1}$
$I = \frac{2}{5(t-1)} - \frac{1}{10} \int \frac{dt}{(t + \frac{1}{4})^2 + (\frac{\sqrt{15}}{4})^2}$
$I = \frac{2}{5(\tan \frac{x}{2} - 1)} - \frac{1}{10} \cdot \frac{4}{\sqrt{15}} \tan^{-1} \left( \frac{t + 1/4}{\sqrt{15}/4} \right) + C$
$I = \frac{2}{5(\tan \frac{x}{2} - 1)} - \frac{2}{5\sqrt{15}} \tan^{-1} \left( \frac{4\tan \frac{x}{2} + 1}{\sqrt{15}} \right) + C$.
Comparing this with the given form,we get $A = \frac{2}{5}, B = -\frac{2}{5\sqrt{15}}, f(x) = \frac{4\tan \frac{x}{2} + 1}{\sqrt{15}}$.

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

(D) We have $I = \int e^{\tan \theta} (\sec \theta - \sin \theta) d\theta$.
Rewrite the integrand: $I = \int e^{\tan \theta} \sec \theta (1 - \sin \theta \cos \theta) d\theta$.
Actually,let us simplify the expression: $I = \int e^{\tan \theta} \sec \theta d\theta - \int e^{\tan \theta} \sin \theta d\theta$.
Using the identity $\int e^{f(x)} (f'(x) g(x) + g'(x)) dx = e^{f(x)} g(x) + c$.
Let $f(\theta) = \tan \theta$,then $f'(\theta) = \sec^2 \theta$.
$I = \int e^{\tan \theta} \sec \theta d\theta - \int e^{\tan \theta} \sin \theta d\theta$.
Note that $\int e^{\tan \theta} \sec \theta d\theta$ is not standard,but we can rewrite the original integral as:
$I = \int e^{\tan \theta} \sec \theta (1 - \sin \theta \cos \theta) d\theta = \int e^{\tan \theta} (\sec \theta - \tan \theta \sin \theta) d\theta$.
This does not simplify easily. Let us re-evaluate:
$I = \int e^{\tan \theta} \sec \theta d\theta - \int e^{\tan \theta} \sin \theta d\theta$.
Actually,the standard form is $\int e^{\tan \theta} (\sec \theta + \sec \theta \tan \theta) d\theta = e^{\tan \theta} \sec \theta + c$.
Given the options,let us check the derivative of $e^{\tan \theta} \cos \theta$:
$\frac{d}{d\theta} (e^{\tan \theta} \cos \theta) = e^{\tan \theta} \sec^2 \theta \cos \theta + e^{\tan \theta} (-\sin \theta) = e^{\tan \theta} (\sec \theta - \sin \theta)$.
Thus,$\int e^{\tan \theta} (\sec \theta - \sin \theta) d\theta = e^{\tan \theta} \cos \theta + c$.

(B) Let $I = \int \frac{3x^4 - 1}{(x^4 + x + 1)^2} dx$.
Divide the numerator and denominator by $x^4$:
$I = \int \frac{3 - x^{-4}}{(1 + x^{-3} + x^{-4})^2} dx$.
This does not simplify easily. Let us rewrite the integrand by dividing numerator and denominator by $x^2$:
$I = \int \frac{3x^2 - x^{-2}}{(x^2 + 1 + x^{-1})^2} dx$.
Let $u = x^2 + 1 + x^{-1}$.
Then $du = (2x - x^{-2}) dx$. This is not matching.
Let us try $u = \frac{x}{x^4 + x + 1}$.
Using the quotient rule,$du = \frac{(x^4 + x + 1)(1) - x(4x^3 + 1)}{(x^4 + x + 1)^2} dx = \frac{x^4 + x + 1 - 4x^4 - x}{(x^4 + x + 1)^2} dx = \frac{1 - 3x^4}{(x^4 + x + 1)^2} dx$.
Thus,$du = - \frac{3x^4 - 1}{(x^4 + x + 1)^2} dx$.
Therefore,$I = - \int du = -u + c = -\frac{x}{x^4 + x + 1} + c$.

(C) Let $I = \int \frac{x^4 + 1}{x(x^2 + 1)^2} dx$.
Add and subtract $x^2$ in the numerator:
$I = \int \frac{x^4 + x^2 - x^2 + 1}{x(x^2 + 1)^2} dx$
$I = \int \frac{x^2(x^2 + 1) - (x^2 - 1)}{x(x^2 + 1)^2} dx$
$I = \int \frac{x^2(x^2 + 1)}{x(x^2 + 1)^2} dx - \int \frac{x^2 - 1}{x(x^2 + 1)^2} dx$
$I = \int \frac{x}{x^2 + 1} dx - \int \frac{x^2 - 1}{x(x^2 + 1)^2} dx$.
For the second integral,let $u = x^2 + 1$,then $du = 2x dx$. This does not simplify easily.
Alternatively,rewrite the integrand:
$\frac{x^4 + 1}{x(x^2 + 1)^2} = \frac{x^4 + 2x^2 + 1 - 2x^2}{x(x^2 + 1)^2} = \frac{(x^2 + 1)^2 - 2x^2}{x(x^2 + 1)^2} = \frac{1}{x} - \frac{2x}{(x^2 + 1)^2}$.
Now integrate:
$I = \int \frac{1}{x} dx - \int \frac{2x}{(x^2 + 1)^2} dx$.
Let $t = x^2 + 1$,then $dt = 2x dx$.
$I = \ln |x| - \int t^{-2} dt = \ln |x| - (\frac{t^{-1}}{-1}) + c = \ln |x| + \frac{1}{x^2 + 1} + c$.
Comparing this with $A \ln |x| + \frac{B}{1 + x^2} + c$,we get $A = 1$ and $B = 1$.

(C) We use the standard integration formula: $\int e^{ax} \cos(bx) \,dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) + c$.
Here,$a = 3$ and $b = 4$.
Substituting these values into the formula:
$\int e^{3x} \cos 4x \,dx = \frac{e^{3x}}{3^2 + 4^2} (3 \cos 4x + 4 \sin 4x) + c$
$= \frac{e^{3x}}{25} (4 \sin 4x + 3 \cos 4x) + c$
$= e^{3x} (\frac{4}{25} \sin 4x + \frac{3}{25} \cos 4x) + c$.
Comparing this with the given expression $e^{3x} (A \sin 4x + B \cos 4x) + c$,we get:
$A = \frac{4}{25}$ and $B = \frac{3}{25}$.
Now,checking the options:
$3A = 3(\frac{4}{25}) = \frac{12}{25}$ and $4B = 4(\frac{3}{25}) = \frac{12}{25}$.
Thus,$3A = 4B$ is correct.

(B) Given that $g(x)$ is an antiderivative of $f(x)$,we have $g'(x) = f(x)$.
We need to find the derivative of $\ln(1 + (g(x))^2)$ with respect to $x$.
Using the chain rule,let $y = \ln(1 + (g(x))^2)$.
Then $\frac{dy}{dx} = \frac{1}{1 + (g(x))^2} \cdot \frac{d}{dx}(1 + (g(x))^2)$.
$\frac{dy}{dx} = \frac{1}{1 + (g(x))^2} \cdot (2g(x) \cdot g'(x))$.
Since $g'(x) = f(x)$,we substitute this into the expression:
$\frac{dy}{dx} = \frac{2g(x)f(x)}{1 + (g(x))^2}$.
Thus,$\ln(1 + (g(x))^2)$ is an antiderivative for $\frac{2f(x)g(x)}{1 + (g(x))^2}$.

(C) The function $h(x)$ is the Dirac delta function, denoted as $\delta(x)$.
By definition, $\int_{-\infty}^{\infty} \delta'(x) \cdot f(x) \, dx = -f'(0)$.
Here, $f(x) = \sin x$.
Therefore, $f'(x) = \cos x$.
Evaluating at $x = 0$, we get $f'(0) = \cos(0) = 1$.
Thus, the integral is $\int_{-\infty}^{\infty} h'(x) \cdot \sin x \, dx = -f'(0) = -1$.