Evaluation of various forms of integration Questions in English

(D) Let $I = \int \frac{2x + 1}{(x^2 + 4x + 1)^{3/2}} \, dx$.
Rewrite the numerator as $2x + 1 = (2x + 4) - 3$.
Then $I = \int \frac{2x + 4}{(x^2 + 4x + 1)^{3/2}} \, dx - 3 \int \frac{1}{(x^2 + 4x + 1)^{3/2}} \, dx$.
For the first part,let $u = x^2 + 4x + 1$,then $du = (2x + 4) \, dx$.
$\int u^{-3/2} \, du = -2u^{-1/2} = -\frac{2}{\sqrt{x^2 + 4x + 1}}$.
For the second part,complete the square: $x^2 + 4x + 1 = (x + 2)^2 - 3$.
Let $x + 2 = \sqrt{3} \sec \theta$,then $dx = \sqrt{3} \sec \theta \tan \theta \, d\theta$.
The integral becomes $\int \frac{\sqrt{3} \sec \theta \tan \theta \, d\theta}{(3 \tan^2 \theta)^{3/2}} = \int \frac{\sqrt{3} \sec \theta \tan \theta}{3\sqrt{3} \tan^3 \theta} \, d\theta = \frac{1}{3} \int \frac{\cos \theta}{\sin^2 \theta} \, d\theta = -\frac{1}{3 \sin \theta} = -\frac{x+2}{3\sqrt{x^2 + 4x + 1}}$.
Combining these,$I = -\frac{2}{\sqrt{x^2 + 4x + 1}} + \frac{x+2}{\sqrt{x^2 + 4x + 1}} + C = \frac{x-2}{\sqrt{x^2 + 4x + 1}} + C$.

(B) Let $I = \int_{e}^{e^4} \sqrt{\ln x} dx$.
Substitute $\ln x = t^2$,which implies $x = e^{t^2}$.
Then $dx = 2t e^{t^2} dt$.
When $x = e$,$t^2 = \ln e = 1$,so $t = 1$.
When $x = e^4$,$t^2 = \ln e^4 = 4$,so $t = 2$.
Thus,$I = \int_{1}^{2} t (2t e^{t^2}) dt = \int_{1}^{2} 2t^2 e^{t^2} dt$.
Using integration by parts,let $u = t$ and $dv = 2t e^{t^2} dt$.
Then $du = dt$ and $v = e^{t^2}$.
Applying the formula $\int u dv = uv - \int v du$:
$I = [t e^{t^2}]_{1}^{2} - \int_{1}^{2} e^{t^2} dt$.
$I = (2 e^{2^2} - 1 e^{1^2}) - \alpha$.
$I = 2e^4 - e - \alpha$.

(D) To evaluate $I = \int \frac{x^2 + 2}{x^4 + 4} dx$,divide the numerator and denominator by $x^2$:
$I = \int \frac{1 + \frac{2}{x^2}}{x^2 + \frac{4}{x^2}} dx$
$= \int \frac{1 + \frac{2}{x^2}}{(x^2 + \frac{4}{x^2})} dx$
$= \int \frac{1 + \frac{2}{x^2}}{(x - \frac{2}{x})^2 + 4} dx$
Let $t = x - \frac{2}{x}$. Then $dt = (1 + \frac{2}{x^2}) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2 + 2^2}$
Using the formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} (\frac{x}{a}) + C$:
$I = \frac{1}{2} \tan^{-1} (\frac{t}{2}) + C$
Substituting $t = x - \frac{2}{x} = \frac{x^2 - 2}{x}$ back:
$I = \frac{1}{2} \tan^{-1} \left( \frac{x^2 - 2}{2x} \right) + C$
Thus,the correct option is $D$.

(C) Let $I = \int (x^3 - 2x^2 + 5)e^{3x} dx$. Using the method of differentiation of the assumed form $e^{3x}(Ax^3 + Bx^2 + Cx + D)$,we have:
$\frac{d}{dx} [e^{3x}(Ax^3 + Bx^2 + Cx + D)] = e^{3x}(x^3 - 2x^2 + 5)$.
Applying the product rule:
$3e^{3x}(Ax^3 + Bx^2 + Cx + D) + e^{3x}(3Ax^2 + 2Bx + C) = e^{3x}(x^3 - 2x^2 + 5)$.
Comparing coefficients of powers of $x$:
$x^3: 3A = 1 \implies A = 1/3$.
$x^2: 3B + 3A = -2 \implies 3B + 1 = -2 \implies 3B = -3 \implies B = -1$.
$x^1: 3C + 2B = 0 \implies 3C - 2 = 0 \implies C = 2/3$.
$x^0: 3D + C = 5 \implies 3D + 2/3 = 5 \implies 3D = 13/3 \implies D = 13/9$.
Now check the options:
$A) C + 3D = 2/3 + 13/3 = 15/3 = 5$ (Correct).
$B) A + B + 2/3 = 1/3 - 1 + 2/3 = 0$ (Correct).
$C) C + 2B = 2/3 + 2(-1) = 2/3 - 2 = -4/3 \neq 0$ (Incorrect).
$D) A + B + C = 1/3 - 1 + 2/3 = 0$ (Correct).
Thus,the incorrect statement is $C$.

(A) We are given the integral $I = \int \left( \frac{x^2 + \cos^2 x}{1 + x^2} \right) \operatorname{cosec}^2 x \, dx$.
Rewrite the numerator as $(x^2 + 1 - 1 + \cos^2 x)$:
$I = \int \left( \frac{x^2 + 1}{1 + x^2} - \frac{1}{1 + x^2} + \frac{\cos^2 x}{1 + x^2} \right) \operatorname{cosec}^2 x \, dx$
Simplify the expression:
$I = \int \left( 1 - \frac{1}{1 + x^2} + \frac{\cos^2 x}{1 + x^2} \right) \operatorname{cosec}^2 x \, dx$
Distribute $\operatorname{cosec}^2 x$:
$I = \int \left( \operatorname{cosec}^2 x - \frac{\operatorname{cosec}^2 x}{1 + x^2} + \frac{\cos^2 x}{(1 + x^2) \sin^2 x} \right) \, dx$
Since $\frac{\cos^2 x}{\sin^2 x} = \cot^2 x$:
$I = \int \left( \operatorname{cosec}^2 x + \frac{\cot^2 x - \operatorname{cosec}^2 x}{1 + x^2} \right) \, dx$
Using the identity $\cot^2 x - \operatorname{cosec}^2 x = -1$:
$I = \int \left( \operatorname{cosec}^2 x - \frac{1}{1 + x^2} \right) \, dx$
Integrating term by term:
$I = -\cot x - \tan^{-1} x + c$.

(A) Let $I = \int \frac{e^x + 9\cos x - 2\sin x + 7}{e^x + 7\sin x + 11\cos x + 14} dx$.
We express the numerator as $A(e^x + 7\sin x + 11\cos x + 14) + B(e^x + 7\cos x - 11\sin x) + K$.
Comparing coefficients:
$A + B = 1$ (for $e^x$)
$7A - 11B = -2$ (for $\sin x$)
$11A + 7B = 9$ (for $\cos x$)
$14A = 7 \implies A = \frac{1}{2}$.
Substituting $A = \frac{1}{2}$ into $A + B = 1$,we get $B = \frac{1}{2}$.
Thus,the numerator is $\frac{1}{2}(e^x + 7\sin x + 11\cos x + 14) + \frac{1}{2}(e^x + 7\cos x - 11\sin x)$.
$I = \int \left( \frac{1}{2} + \frac{1}{2} \frac{e^x + 7\cos x - 11\sin x}{e^x + 7\sin x + 11\cos x + 14} \right) dx$.
$I = \frac{1}{2}x + \frac{1}{2} \ln |e^x + 7\sin x + 11\cos x + 14| + C$.

(D) Let $I = \int {\frac{{\csc^2 x}}{{{{\left( {\csc x + \cot x} \right)}^{9/2}}}}} dx$.
Let $z = \csc x + \cot x$.
Then $\frac{dz}{dx} = -\csc x \cot x - \csc^2 x = -\csc x(\cot x + \csc x) = -\csc x \cdot z$.
Also,$\csc x - \cot x = \frac{1}{z}$.
Adding the two equations: $2\csc x = z + \frac{1}{z} \implies \csc x = \frac{z^2 + 1}{2z}$.
Substituting into the derivative: $dz = -\left( \frac{z^2 + 1}{2z} \right) z dx = -\frac{z^2 + 1}{2} dx$.
Thus,$dx = -\frac{2}{z^2 + 1} dz$.
Substituting into $I$: $I = \int \frac{(\csc x)^2}{z^{9/2}} \left( -\frac{2}{z^2 + 1} \right) dz = \int \frac{(\frac{z^2+1}{2z})^2}{z^{9/2}} \left( -\frac{2}{z^2 + 1} \right) dz$.
$I = -\frac{1}{2} \int \frac{z^2+1}{z^{13/2}} dz = -\frac{1}{2} \int (z^{-9/2} + z^{-13/2}) dz$.
$I = -\frac{1}{2} \left[ \frac{z^{-7/2}}{-7/2} + \frac{z^{-11/2}}{-11/2} \right] + C = \frac{z^{-7/2}}{7} + \frac{z^{-11/2}}{11} + C$.
Since $z^{-1} = \csc x - \cot x$,we have $I = (\csc x - \cot x)^{7/2} \left( \frac{1}{7} + \frac{(\csc x - \cot x)^2}{11} \right) + C$.
Comparing this with the given expression,we find $\alpha = 7$.

(B) Let $I = \int \frac{\log x - (\log x)^2 + x^2}{x^3} dx$.
We can split the integral as:
$I = \int \frac{\log x - (\log x)^2}{x^3} dx + \int \frac{x^2}{x^3} dx$.
$I = \int \frac{\log x(1 - \log x)}{x^3} dx + \int \frac{1}{x} dx$.
Let $u = \frac{\log x}{x}$. Then $du = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} dx = \frac{1 - \log x}{x^2} dx$.
Substituting this into the first part:
$I = \int u \cdot \frac{1}{x} dx$ is not quite right,let's re-evaluate.
Actually,$\int \frac{\log x(1 - \log x)}{x^3} dx = \int \frac{\log x}{x} \cdot \frac{1 - \log x}{x^2} dx$.
Let $t = \frac{\log x}{x}$,then $dt = \frac{1 - \log x}{x^2} dx$.
So,$\int t dt = \frac{t^2}{2} = \frac{(\log x)^2}{2x^2}$.
Adding the second part $\int \frac{1}{x} dx = \log x$.
Thus,$I = \frac{(\log x)^2}{2x^2} + \log x + C = \frac{(\log x)^2 + 2x^2 \log x}{2x^2} + C$.

(C) Let $I = \int \frac{\sin 2\theta d\theta}{\cos^2 \theta \cos 3\theta} = \int \frac{2 \sin \theta \cos \theta d\theta}{\cos^2 \theta (4 \cos^3 \theta - 3 \cos \theta)} = \int \frac{2 \sin \theta d\theta}{\cos^2 \theta (4 \cos^2 \theta - 3)}$.
Let $\cos \theta = x$,then $-\sin \theta d\theta = dx$.
$I = -2 \int \frac{dx}{x^2(4x^2 - 3)} = -\frac{2}{3} \int \frac{4x^2 - (4x^2 - 3)}{x^2(4x^2 - 3)} dx = -\frac{2}{3} \int \left( \frac{1}{x^2 - 3/4} - \frac{1}{x^2} \right) dx$.
$I = -\frac{2}{3} \left( \frac{1}{2(\sqrt{3}/2)} \ln \left| \frac{x - \sqrt{3}/2}{x + \sqrt{3}/2} \right| + \frac{1}{x} \right) + C = -\frac{2}{3} \left( \frac{1}{\sqrt{3}} \ln \left| \frac{2x - \sqrt{3}}{2x + \sqrt{3}} \right| + \frac{1}{x} \right) + C$.
Since $\tan(\pi/6) = 1/\sqrt{3}$ and $\cos(\pi/6) = \sqrt{3}/2$,we have $I = \frac{2}{3} \ln \left| \left( \frac{2x + \sqrt{3}}{2x - \sqrt{3}} \right)^{\tan(\pi/6)} \cdot e^{-1/x} \right| + C$.
Substituting $x = \cos \theta$,$I = \frac{2}{3} \ln \left| \left( \frac{\cos \theta + \cos(\pi/6)}{\cos \theta - \cos(\pi/6)} \right)^{\tan(\pi/6)} \cdot e^{-\sec \theta} \right| + C$.
Since $-\sec \theta = \sec(\pi - \theta)$,the result matches option $C$.

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

(A) Given the integral $I = \int {{\sec }^{ - 1}}\left[ { - {\sin }^2x} \right]dx$.
Since $0 \le {\sin }^2x \le 1$,we have $-1 \le -{\sin }^2x \le 0$.
For $x \neq 0$,$-{\sin }^2x$ is in the interval $[-1, 0)$.
The greatest integer function $[ - {\sin }^2x ]$ is $-1$ for all $x \neq n\pi$.
Thus,the integral becomes $\int {{\sec }^{ - 1}}( - 1)dx = \int \pi dx = \pi x + C$.
Given $f(0) = 0$,we have $f(x) = \pi x$.
Now,we need to find the value of ${\left( {f\left( {\frac{8}{{\pi x}}} \right)} \right)''}$ at $x = 2$.
$f\left( {\frac{8}{{\pi x}}} \right) = \pi \left( {\frac{8}{{\pi x}}} \right) = \frac{8}{x} = 8x^{-1}$.
First derivative: $\frac{d}{dx}(8x^{-1}) = -8x^{-2}$.
Second derivative: $\frac{d^2}{dx^2}(8x^{-1}) = 16x^{-3} = \frac{16}{x^3}$.
At $x = 2$,the value is $\frac{16}{2^3} = \frac{16}{8} = 2$.

(A) We know that $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 \cdot ((\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x) = 1 - 3 \sin^2 x \cos^2 x$.
Dividing numerator and denominator by $\cos^6 x$,we get:
$\int \frac{\sec^6 x dx}{\tan^6 x + 1} = \int \frac{\sec^4 x \cdot \sec^2 x dx}{\tan^6 x + 1} = \int \frac{(1 + \tan^2 x)^2 \sec^2 x dx}{\tan^6 x + 1}$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
The integral becomes $\int \frac{(1 + u^2)^2}{u^6 + 1} du = \int \frac{u^4 + 2u^2 + 1}{u^6 + 1} du$.
Dividing numerator and denominator by $u^3$:
$\int \frac{u + \frac{2}{u} + \frac{1}{u^3}}{u^3 + \frac{1}{u^3}} du = \int \frac{u + \frac{1}{u^3} + \frac{2}{u}}{u^3 + \frac{1}{u^3}} du$.
Using the substitution $t = \tan x - \cot x = u - \frac{1}{u}$,we have $dt = (1 + \frac{1}{u^2}) du = \frac{u^2 + 1}{u^2} du$.
After simplification,the integral evaluates to $\tan^{-1}(\tan x - \cot x) + C$.

(A) Let $I = \int \frac{3\cos x + 2\sin x}{4\sin x + 5\cos x} dx$.
We express the numerator as $A(\text{denominator}) + B(\text{derivative of denominator})$.
Let $3\cos x + 2\sin x = \lambda(4\sin x + 5\cos x) + \mu(4\cos x - 5\sin x)$.
Comparing coefficients of $\sin x$ and $\cos x$:
$2 = 4\lambda - 5\mu$ and $3 = 5\lambda + 4\mu$.
Solving these equations:
Multiply the first by $4$ and second by $5$: $8 = 16\lambda - 20\mu$ and $15 = 25\lambda + 20\mu$.
Adding them gives $23 = 41\lambda$,so $\lambda = 23/41$.
Substituting $\lambda$ into $2 = 4(23/41) - 5\mu$:
$5\mu = 92/41 - 82/41 = 10/41$,so $\mu = 2/41$.
Thus,$I = \int \frac{\lambda(4\sin x + 5\cos x) + \mu(4\cos x - 5\sin x)}{4\sin x + 5\cos x} dx = \lambda x + \mu \ln |4\sin x + 5\cos x| + c$.
$I = \frac{23}{41}x + \frac{2}{41} \ln |4\sin x + 5\cos x| + c = \frac{1}{41} \{23x + 2 \ln |4\sin x + 5\cos x|\} + c$.
Comparing with the given form,$A = 1/41$ and $f(x) = \ln |x|$.

(A) Let $I = \int {\frac{{\sec x \cdot \csc x}}{{2\cot x - \sec x \cdot \csc x}}} dx$.
Expressing the integrand in terms of $\sin x$ and $\cos x$:
$\sec x \cdot \csc x = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$.
$2\cot x - \sec x \cdot \csc x = \frac{2\cos x}{\sin x} - \frac{1}{\sin x \cos x} = \frac{2\cos^2 x - 1}{\sin x \cos x} = \frac{\cos 2x}{\sin x \cos x}$.
Substituting these into the integral:
$I = \int \frac{1/(\sin x \cos x)}{\cos 2x / (\sin x \cos x)} dx = \int \frac{1}{\cos 2x} dx = \int \sec 2x dx$.
Using the standard integral $\int \sec u du = \ln |\sec u + \tan u| + c$,we get:
$I = \frac{1}{2} \ln |\sec 2x + \tan 2x| + c$.

(A) To solve the integral $I = \int \frac{2x + 5}{\sqrt{7 - 6x - x^2}} dx$,we express the numerator in terms of the derivative of the quadratic expression inside the square root.
Let $f(x) = 7 - 6x - x^2$. Then $f'(x) = -6 - 2x$.
We write $2x + 5 = -( -2x - 6 ) - 1$.
Thus,$I = \int \frac{-( -2x - 6 ) - 1}{\sqrt{7 - 6x - x^2}} dx = -\int \frac{-2x - 6}{\sqrt{7 - 6x - x^2}} dx - \int \frac{1}{\sqrt{7 - 6x - x^2}} dx$.
For the first integral,let $u = 7 - 6x - x^2$,so $du = (-6 - 2x) dx$. Then $\int \frac{du}{\sqrt{u}} = 2\sqrt{u} = 2\sqrt{7 - 6x - x^2}$.
So,$-\int \frac{-2x - 6}{\sqrt{7 - 6x - x^2}} dx = -2\sqrt{7 - 6x - x^2}$.
For the second integral,complete the square: $7 - 6x - x^2 = 16 - (x^2 + 6x + 9) = 4^2 - (x + 3)^2$.
So,$\int \frac{1}{\sqrt{4^2 - (x + 3)^2}} dx = \sin^{-1}\left(\frac{x + 3}{4}\right)$.
Combining these,$I = -2\sqrt{7 - 6x - x^2} - \sin^{-1}\left(\frac{x + 3}{4}\right) + C$.
Comparing with $A\sqrt{7 - 6x - x^2} + B\sin^{-1}\left(\frac{x + 3}{4}\right) + C$,we get $A = -2$ and $B = -1$.

(A) Let $I = \int \frac{\tan x}{1 + \tan x + \tan^2 x} dx$.
We can rewrite the numerator as $\frac{1}{2} [ (1 + \tan x + \tan^2 x) - (1 - \tan x + \tan^2 x) ]$ is not ideal,so let's use the identity: $\tan x = \frac{1}{2} (1 + \tan x + \tan^2 x) - \frac{1}{2} (1 - \tan x + \tan^2 x)$.
Alternatively,note that $\int \frac{\tan x}{1 + \tan x + \tan^2 x} dx = \int \frac{\tan x + 1 + \tan^2 x - (1 + \tan^2 x)}{1 + \tan x + \tan^2 x} dx = \int 1 dx - \int \frac{\sec^2 x}{1 + \tan x + \tan^2 x} dx$.
Let $I = x - \int \frac{\sec^2 x}{1 + \tan x + \tan^2 x} dx$.
Substitute $\tan x = t$,so $\sec^2 x dx = dt$.
$I = x - \int \frac{dt}{t^2 + t + 1} = x - \int \frac{dt}{(t + 1/2)^2 + 3/4}$.
Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = x - \frac{1}{\sqrt{3}/2} \tan^{-1} \left( \frac{t + 1/2}{\sqrt{3}/2} \right) + C = x - \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2t + 1}{\sqrt{3}} \right) + C$.
Substituting $t = \tan x$,we have $I = x - \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2 \tan x + 1}{\sqrt{3}} \right) + C$.
Comparing this with the given form $x - \frac{K}{\sqrt{A}} \tan^{-1} \left( \frac{K \tan x + 1}{\sqrt{A}} \right) + C$,we find $K = 2$ and $A = 3$.
Thus,the ordered pair $(K, A)$ is $(2, 3)$.

(C) Let $I = \int \frac{dx}{(1 + \sqrt{x}) \sqrt{x} \sqrt{1 - x}}$.
Substitute $\sqrt{x} = u$,then $dx = 2u \, du$.
$I = \int \frac{2u \, du}{(1 + u) u \sqrt{1 - u^2}} = 2 \int \frac{du}{(1 + u) \sqrt{(1 - u)(1 + u)}} = 2 \int \frac{du}{(1 + u) \sqrt{1 - u} \sqrt{1 + u}} = 2 \int \frac{du}{(1 + u)^{3/2} (1 - u)^{1/2}}$.
Alternatively,let $1 + \sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,so $\frac{dx}{\sqrt{x}} = 2 dt$.
Since $\sqrt{x} = t - 1$,then $x = (t - 1)^2$,so $1 - x = 1 - (t - 1)^2 = 1 - (t^2 - 2t + 1) = 2t - t^2$.
$I = \int \frac{2 dt}{t \sqrt{2t - t^2}} = 2 \int \frac{dt}{t \sqrt{t(2 - t)}} = 2 \int \frac{dt}{t \sqrt{t} \sqrt{2 - t}} = 2 \int \frac{dt}{t^{3/2} \sqrt{2 - t}}$.
Let $t = \frac{1}{z}$,then $dt = -\frac{1}{z^2} dz$.
$I = 2 \int \frac{-dz/z^2}{(1/z)^{3/2} \sqrt{2 - 1/z}} = 2 \int \frac{-dz/z^2}{z^{-3/2} \sqrt{(2z - 1)/z}} = 2 \int \frac{-dz/z^2}{z^{-3/2} \cdot z^{-1/2} \sqrt{2z - 1}} = 2 \int \frac{-dz}{\sqrt{2z - 1}}$.
$I = -2 \int (2z - 1)^{-1/2} dz = -2 \cdot \frac{(2z - 1)^{1/2}}{1/2 \cdot 2} = -2 \sqrt{2z - 1} + c$.
Substituting $z = 1/t = 1/(1 + \sqrt{x})$:
$I = -2 \sqrt{\frac{2}{1 + \sqrt{x}} - 1} + c = -2 \sqrt{\frac{2 - 1 - \sqrt{x}}{1 + \sqrt{x}}} + c = -2 \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$.

(B) Let $I = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$
Using the identity $a^2 - b^2 = (a - b)(a + b)$,we have:
$I = \int {\frac{{({{\sin }^4}x - {{\cos }^4}x)({{\sin }^4}x + {{\cos }^4}x)}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx$
Since ${{\sin }^4}x + {{\cos }^4}x = ({{\sin }^2}x + {{\cos }^2}x)^2 - 2{{\sin }^2}x{{\cos }^2}x = 1 - 2{{\sin }^2}x{{\cos }^2}x$,the expression simplifies to:
$I = \int {({{\sin }^4}x - {{\cos }^4}x)} dx$
Further,${{\sin }^4}x - {{\cos }^4}x = ({{\sin }^2}x - {{\cos }^2}x)({{\sin }^2}x + {{\cos }^2}x) = -\cos 2x \cdot 1 = -\cos 2x$
Thus,$I = \int {-\cos 2x} dx = -\frac{{\sin 2x}}{2} + c$

(B) Given integral: $I = \int \frac{x^{5 m-1}+2 x^{4 m-1}}{\left(x^{2 m}+x^{m}+1\right)^{3}} d x$
Divide numerator and denominator by $x^{6m}$:
$I = \int \frac{x^{5 m-1}+2 x^{4 m-1}}{x^{6m}\left(x^{-2m}+x^{-m}+1\right)^{3}} d x$
$I = \int \frac{x^{-m-1}+2 x^{-2 m-1}}{\left(1+x^{-m}+x^{-2 m}\right)^{3}} d x$
Let $t = 1+x^{-m}+x^{-2 m}$.
Then $dt = (-m x^{-m-1} - 2m x^{-2m-1}) dx = -m(x^{-m-1} + 2x^{-2m-1}) dx$.
So,$(x^{-m-1} + 2x^{-2m-1}) dx = -\frac{dt}{m}$.
Substituting into the integral:
$I = \int \frac{-dt/m}{t^3} = -\frac{1}{m} \int t^{-3} dt = -\frac{1}{m} \left( \frac{t^{-2}}{-2} \right) + C = \frac{1}{2mt^2} + C$.
Substituting $t$ back:
$I = \frac{1}{2m(1+x^{-m}+x^{-2m})^2} + C = \frac{1}{2m(\frac{x^{2m}+x^m+1}{x^{2m}})^2} + C = \frac{x^{4m}}{2m(x^{2m}+x^m+1)^2} + C$.
Thus,$f(x) = \frac{x^{4m}}{2m(x^{2m}+x^m+1)^2}$.

(A) We are given that $\int \frac{dx}{x + x^7} = p(x)$.
Consider the integral $I = \int \frac{x^6}{x + x^7} dx$.
We can rewrite the integrand as:
$I = \int \frac{x^6}{x(1 + x^6)} dx = \int \frac{x^5}{1 + x^6} dx$.
Alternatively,we can manipulate the original expression:
$I = \int \frac{x^6}{x(1 + x^6)} dx = \int \frac{(1 + x^6) - 1}{x(1 + x^6)} dx$.
Splitting the integral,we get:
$I = \int \frac{1 + x^6}{x(1 + x^6)} dx - \int \frac{1}{x(1 + x^6)} dx$.
$I = \int \frac{1}{x} dx - \int \frac{1}{x + x^7} dx$.
Since $\int \frac{1}{x} dx = \ln |x| + c_1$ and $\int \frac{1}{x + x^7} dx = p(x) + c_2$,we have:
$I = \ln |x| - p(x) + c$.

(A) Let $I = \int \frac{\cos 8x + 1}{\cot 2x - \tan 2x} dx.$
First,simplify the denominator:
$\cot 2x - \tan 2x = \frac{\cos 2x}{\sin 2x} - \frac{\sin 2x}{\cos 2x} = \frac{\cos^2 2x - \sin^2 2x}{\sin 2x \cos 2x} = \frac{\cos 4x}{\frac{1}{2} \sin 4x} = 2 \cot 4x.$
Using the identity $\cos 8x + 1 = 2 \cos^2 4x$,the integral becomes:
$I = \int \frac{2 \cos^2 4x}{2 \cot 4x} dx = \int \frac{\cos^2 4x}{\frac{\cos 4x}{\sin 4x}} dx = \int \cos 4x \sin 4x dx.$
Multiply and divide by $2$:
$I = \frac{1}{2} \int 2 \sin 4x \cos 4x dx = \frac{1}{2} \int \sin 8x dx.$
Integrating $\sin 8x$ gives:
$I = \frac{1}{2} \left( -\frac{\cos 8x}{8} \right) + k = -\frac{1}{16} \cos 8x + k.$
Comparing this with $A \cos 8x + k$,we get $A = -\frac{1}{16}$.

(A) Given $f(x) = \int \left( \frac{x^2 + \sin^2 x}{1 + x^2} \right) \sec^2 x \, dx$.
We can rewrite the integrand as:
$f(x) = \int \frac{x^2 \sec^2 x + \sin^2 x \sec^2 x}{1 + x^2} \, dx$
Since $\sin^2 x \sec^2 x = \tan^2 x$,we have:
$f(x) = \int \frac{x^2 \sec^2 x + \tan^2 x}{1 + x^2} \, dx$
Using $\sec^2 x = 1 + \tan^2 x$,we get:
$f(x) = \int \frac{x^2(1 + \tan^2 x) + \tan^2 x}{1 + x^2} \, dx$
$f(x) = \int \frac{x^2 + x^2 \tan^2 x + \tan^2 x}{1 + x^2} \, dx = \int \frac{x^2 + \tan^2 x(1 + x^2)}{1 + x^2} \, dx$
$f(x) = \int \frac{x^2}{1 + x^2} \, dx + \int \tan^2 x \, dx$
$f(x) = \int \frac{x^2 + 1 - 1}{1 + x^2} \, dx + \int (\sec^2 x - 1) \, dx$
$f(x) = \int (1 - \frac{1}{1 + x^2}) \, dx + \int \sec^2 x \, dx - \int 1 \, dx$
$f(x) = x - \tan^{-1} x + \tan x - x + C = \tan x - \tan^{-1} x + C$
Given $f(0) = 0$,we have $0 = \tan 0 - \tan^{-1} 0 + C$,so $C = 0$.
Thus,$f(x) = \tan x - \tan^{-1} x$.
Therefore,$f(1) = \tan 1 - \tan^{-1}(1) = \tan 1 - \frac{\pi}{4}$.

(A) We have the integral $I = \int \frac{\sqrt{1-x^{2}}}{x^{4}} dx.$
Rewrite the integrand as $I = \int \frac{x \sqrt{\frac{1}{x^{2}}-1}}{x^{4}} dx = \int \frac{1}{x^{3}} \sqrt{\frac{1}{x^{2}}-1} dx.$
Let $t = \frac{1}{x^{2}} - 1.$ Then $dt = -\frac{2}{x^{3}} dx,$ which implies $\frac{dx}{x^{3}} = -\frac{1}{2} dt.$
Substituting these into the integral,we get $I = -\frac{1}{2} \int \sqrt{t} dt = -\frac{1}{2} \cdot \frac{t^{3/2}}{3/2} + C = -\frac{1}{3} t^{3/2} + C.$
Substituting back $t = \frac{1-x^2}{x^2},$ we get $I = -\frac{1}{3} \left(\frac{1-x^2}{x^2}\right)^{3/2} + C = -\frac{1}{3} \frac{(1-x^2)^{3/2}}{x^3} + C.$
Since $(1-x^2)^{3/2} = (\sqrt{1-x^2})^3,$ we have $I = -\frac{1}{3x^3} (\sqrt{1-x^2})^3 + C.$
Comparing this with $A(x) (\sqrt{1-x^2})^m + C,$ we identify $m = 3$ and $A(x) = -\frac{1}{3x^3}.$
Thus,$(A(x))^m = \left(-\frac{1}{3x^3}\right)^3 = -\frac{1}{27x^9}.$ Note: The provided options seem to have a sign discrepancy or typo; based on the calculation,the result is $-\frac{1}{27x^9}$.

(B) Given integral $I = \int \frac{dx}{x^{3}(1+x^{6})^{2 / 3}}$.
We can rewrite the integral by factoring out $x^6$ from the parenthesis:
$I = \int \frac{dx}{x^{3} \cdot (x^6)^{2/3} (1 + x^{-6})^{2/3}} = \int \frac{dx}{x^{3} \cdot x^4 (1 + x^{-6})^{2/3}} = \int \frac{dx}{x^7 (1 + x^{-6})^{2/3}}$.
Let $t = 1 + x^{-6}$. Then $dt = -6x^{-7} dx$,which implies $\frac{dx}{x^7} = -\frac{dt}{6}$.
Substituting these into the integral:
$I = \int -\frac{1}{6} t^{-2/3} dt = -\frac{1}{6} \cdot \frac{t^{1/3}}{1/3} + C = -\frac{1}{2} t^{1/3} + C$.
Substituting back $t = 1 + x^{-6} = \frac{x^6+1}{x^6}$:
$I = -\frac{1}{2} \left( \frac{1+x^6}{x^6} \right)^{1/3} + C = -\frac{1}{2} \cdot \frac{(1+x^6)^{1/3}}{x^2} + C$.
Comparing this with the given form $xf(x)(1+x^6)^{1/3} + C$:
$xf(x)(1+x^6)^{1/3} = -\frac{1}{2x^2} (1+x^6)^{1/3}$.
Dividing both sides by $x(1+x^6)^{1/3}$:
$f(x) = \frac{-1}{2x^3}$.

(C) Let $I = \int \frac{dx}{(x^2 - 2x + 10)^2} = \int \frac{dx}{((x - 1)^2 + 9)^2}$.
Substitute $x - 1 = 3 \tan \theta$,so $dx = 3 \sec^2 \theta d\theta$.
The integral becomes $\int \frac{3 \sec^2 \theta d\theta}{(9 \tan^2 \theta + 9)^2} = \int \frac{3 \sec^2 \theta d\theta}{81 \sec^4 \theta} = \frac{1}{27} \int \cos^2 \theta d\theta$.
Using $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get $\frac{1}{54} \int (1 + \cos 2\theta) d\theta = \frac{1}{54} (\theta + \frac{\sin 2\theta}{2}) + C$.
Since $\tan \theta = \frac{x - 1}{3}$,then $\theta = \tan^{-1} \left( \frac{x - 1}{3} \right)$.
Also,$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \left( \frac{x - 1}{\sqrt{(x - 1)^2 + 9}} \right) \left( \frac{3}{\sqrt{(x - 1)^2 + 9}} \right) = \frac{6(x - 1)}{x^2 - 2x + 10}$.
Substituting these back,$I = \frac{1}{54} \left( \tan^{-1} \left( \frac{x - 1}{3} \right) + \frac{3(x - 1)}{x^2 - 2x + 10} \right) + C$.
Comparing with the given form,$A = \frac{1}{54}$ and $f(x) = 3(x - 1)$.

(C) We are given the integral $I = \int \frac{\tan x + \tan \alpha}{\tan x - \tan \alpha} dx$.
Using the identity $\tan x \pm \tan \alpha = \frac{\sin(x \pm \alpha)}{\cos x \cos \alpha}$,we have:
$I = \int \frac{\sin(x + \alpha) / (\cos x \cos \alpha)}{\sin(x - \alpha) / (\cos x \cos \alpha)} dx = \int \frac{\sin(x + \alpha)}{\sin(x - \alpha)} dx$.
Let $t = x - \alpha$,then $x = t + \alpha$ and $dx = dt$. Substituting these into the integral:
$I = \int \frac{\sin(t + 2\alpha)}{\sin t} dt$.
Using the expansion $\sin(t + 2\alpha) = \sin t \cos 2\alpha + \cos t \sin 2\alpha$:
$I = \int \frac{\sin t \cos 2\alpha + \cos t \sin 2\alpha}{\sin t} dt$
$I = \int \cos 2\alpha dt + \int \cot t \sin 2\alpha dt$
$I = t \cos 2\alpha + \sin 2\alpha \ln |\sin t| + C$.
Substituting $t = x - \alpha$ back:
$I = (x - \alpha) \cos 2\alpha + \ln |\sin (x - \alpha)| \sin 2\alpha + C$.
Comparing this with $A(x) \cos 2\alpha + B(x) \sin 2\alpha + C$,we get $A(x) = x - \alpha$ and $B(x) = \ln |\sin (x - \alpha)|$.
Thus,the correct option is $(C)$.

(A) Let $I = \int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}}$.
Substitute $u = \sin x$,then $du = \cos x \, dx$.
$I = \int \frac{du}{u^3 (1+u^6)^{2/3}}$.
Factor out $u^6$ from the parenthesis: $I = \int \frac{du}{u^3 (u^6(\frac{1}{u^6}+1))^{2/3}} = \int \frac{du}{u^3 \cdot u^4 (\frac{1}{u^6}+1)^{2/3}} = \int \frac{du}{u^7 (u^{-6}+1)^{2/3}}$.
Let $t = u^{-6}+1$,then $dt = -6u^{-7} \, du$,so $u^{-7} \, du = -\frac{1}{6} \, dt$.
$I = -\frac{1}{6} \int t^{-2/3} \, dt = -\frac{1}{6} \cdot \frac{t^{1/3}}{1/3} + c = -\frac{1}{2} t^{1/3} + c$.
Substituting back $t = u^{-6}+1 = \frac{1+u^6}{u^6}$,we get $I = -\frac{1}{2} \left(\frac{1+u^6}{u^6}\right)^{1/3} + c = -\frac{1}{2} \frac{(1+\sin^6 x)^{1/3}}{\sin^2 x} + c$.
Comparing with $f(x)(1+\sin^6 x)^{1/\lambda} + c$,we identify $\lambda = 3$ and $f(x) = -\frac{1}{2 \sin^2 x}$.
Then $\lambda f\left(\frac{\pi}{3}\right) = 3 \cdot \left(-\frac{1}{2 \sin^2(\pi/3)}\right) = 3 \cdot \left(-\frac{1}{2 \cdot (3/4)}\right) = 3 \cdot \left(-\frac{2}{3}\right) = -2$.

(A) Let $I = \int \frac{1}{1+\tan x} d x$.
We can write $\tan x = \frac{\sin x}{\cos x}$,so $I = \int \frac{\cos x}{\cos x + \sin x} d x$.
Multiply the numerator and denominator by $2$ and adjust the numerator:
$I = \frac{1}{2} \int \frac{2\cos x}{\cos x + \sin x} d x = \frac{1}{2} \int \frac{(\cos x + \sin x) + (\cos x - \sin x)}{\cos x + \sin x} d x$.
Split the integral:
$I = \frac{1}{2} \int 1 d x + \frac{1}{2} \int \frac{\cos x - \sin x}{\cos x + \sin x} d x$.
For the second integral,let $u = \cos x + \sin x$,then $du = (-\sin x + \cos x) d x$.
Thus,$I = \frac{x}{2} + \frac{1}{2} \int \frac{1}{u} du = \frac{x}{2} + \frac{1}{2} \log |u| + C$.
Substituting back $u = \cos x + \sin x$,we get:
$I = \frac{x}{2} + \frac{1}{2} \log |\cos x + \sin x| + C$.

Let $I = \int \frac{1}{1+\cot x} dx$.
$= \int \frac{1}{1+\frac{\cos x}{\sin x}} dx$
$= \int \frac{\sin x}{\sin x+\cos x} dx$
$= \frac{1}{2} \int \frac{2 \sin x}{\sin x+\cos x} dx$
$= \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin x+\cos x} dx$
$= \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} dx$
$= \frac{x}{2} + \frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} dx$.
Let $\sin x+\cos x = t$,then $(\cos x-\sin x) dx = dt$,which implies $(\sin x-\cos x) dx = -dt$.
Therefore,$I = \frac{x}{2} + \frac{1}{2} \int \frac{-dt}{t}$
$= \frac{x}{2} - \frac{1}{2} \ln |\sin x+\cos x| + C$,where $C$ is an arbitrary constant.

Let $I = \int \frac{1}{1-\tan x} dx$.
$= \int \frac{1}{1-\frac{\sin x}{\cos x}} dx = \int \frac{\cos x}{\cos x-\sin x} dx$.
Multiply and divide by $2$:
$= \frac{1}{2} \int \frac{2 \cos x}{\cos x-\sin x} dx$.
Rewrite the numerator:
$= \frac{1}{2} \int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{\cos x-\sin x} dx$.
$= \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} dx$.
$= \frac{x}{2} + \frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} dx$.
Let $t = \cos x - \sin x$,then $dt = -(\sin x + \cos x) dx$,so $(\cos x + \sin x) dx = -dt$.
$= \frac{x}{2} - \frac{1}{2} \int \frac{1}{t} dt$.
$= \frac{x}{2} - \frac{1}{2} \ln |\cos x - \sin x| + C$,where $C$ is the constant of integration.

Consider $\sin ^{4} x = (\sin ^{2} x)^{2}$.
Using the identity $\sin ^{2} x = \frac{1 - \cos 2x}{2}$,we have:
$\sin ^{4} x = \left(\frac{1 - \cos 2x}{2}\right)^{2} = \frac{1}{4}(1 - 2\cos 2x + \cos ^{2} 2x)$.
Using the identity $\cos ^{2} 2x = \frac{1 + \cos 4x}{2}$,we get:
$\sin ^{4} x = \frac{1}{4} \left(1 - 2\cos 2x + \frac{1 + \cos 4x}{2}\right) = \frac{1}{4} \left(\frac{3}{2} - 2\cos 2x + \frac{1}{2}\cos 4x\right) = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.
Now,integrate with respect to $x$:
$\int \sin ^{4} x \, dx = \int \left(\frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\right) \, dx$.
$= \frac{3}{8}x - \frac{1}{2} \left(\frac{\sin 2x}{2}\right) + \frac{1}{8} \left(\frac{\sin 4x}{4}\right) + C$.
$= \frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$,where $C$ is the constant of integration.

(N/A) We want to evaluate $I = \int \tan^{3} 2x \sec 2x \, dx$.
First,rewrite the integrand:
$\tan^{3} 2x \sec 2x = \tan^{2} 2x \cdot \tan 2x \sec 2x = (\sec^{2} 2x - 1) \tan 2x \sec 2x$.
Substitute this into the integral:
$I = \int (\sec^{2} 2x - 1) \tan 2x \sec 2x \, dx = \int \sec^{2} 2x \tan 2x \sec 2x \, dx - \int \tan 2x \sec 2x \, dx$.
For the first part,let $u = \sec 2x$. Then $du = 2 \sec 2x \tan 2x \, dx$,which implies $\sec 2x \tan 2x \, dx = \frac{1}{2} du$.
Thus,$\int \sec^{2} 2x \tan 2x \sec 2x \, dx = \int u^{2} \cdot \frac{1}{2} du = \frac{1}{2} \cdot \frac{u^{3}}{3} = \frac{u^{3}}{6} = \frac{\sec^{3} 2x}{6}$.
For the second part,$\int \tan 2x \sec 2x \, dx = \frac{\sec 2x}{2}$.
Combining these,we get:
$I = \frac{\sec^{3} 2x}{6} - \frac{\sec 2x}{2} + C$,where $C$ is the constant of integration.

(N/A) We have $\frac{1}{\sin x \cos ^{3} x} = \frac{\sin ^{2} x + \cos ^{2} x}{\sin x \cos ^{3} x}$.
$= \frac{\sin ^{2} x}{\sin x \cos ^{3} x} + \frac{\cos ^{2} x}{\sin x \cos ^{3} x} = \frac{\sin x}{\cos ^{3} x} + \frac{1}{\sin x \cos x}$.
$= \tan x \sec ^{2} x + \frac{\sec ^{2} x}{\tan x}$.
Therefore,$\int \frac{1}{\sin x \cos ^{3} x} dx = \int \tan x \sec ^{2} x dx + \int \frac{\sec ^{2} x}{\tan x} dx$.
Let $t = \tan x$,then $dt = \sec ^{2} x dx$.
Substituting these into the integral,we get $\int t dt + \int \frac{1}{t} dt$.
$= \frac{t^{2}}{2} + \log |t| + C$.
Substituting $t = \tan x$ back,we get $\frac{1}{2} \tan ^{2} x + \log |\tan x| + C$,where $C$ is an arbitrary constant.

(N/A) To evaluate the integral $I = \int \frac{1}{\cos (x-a) \cos (x-b)} dx$,we multiply and divide by $\sin(a-b)$:
$I = \frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\cos (x-a) \cos (x-b)} dx$
Since $a-b = (x-b) - (x-a)$,we can rewrite the numerator:
$I = \frac{1}{\sin (a-b)} \int \frac{\sin [(x-b) - (x-a)]}{\cos (x-a) \cos (x-b)} dx$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin (a-b)} \int \frac{\sin (x-b) \cos (x-a) - \cos (x-b) \sin (x-a)}{\cos (x-a) \cos (x-b)} dx$
$I = \frac{1}{\sin (a-b)} \int [\tan (x-b) - \tan (x-a)] dx$
Integrating $\tan(x)$ gives $\ln|\sec(x)|$ or $-\ln|\cos(x)|$:
$I = \frac{1}{\sin (a-b)} [-\ln|\cos (x-b)| + \ln|\cos (x-a)|] + C$
$I = \frac{1}{\sin (a-b)} \ln \left| \frac{\cos (x-a)}{\cos (x-b)} \right| + C$,where $C$ is an arbitrary constant.

(A) To evaluate $\int \frac{x+2}{2 x^{2}+6 x+5} d x$,we express the numerator as $A \frac{d}{d x}(2 x^{2}+6 x+5) + B$.
$x+2 = A(4x+6) + B = 4Ax + (6A+B)$.
Equating coefficients,$4A = 1 \implies A = \frac{1}{4}$ and $6A + B = 2 \implies 6(\frac{1}{4}) + B = 2 \implies B = 2 - \frac{3}{2} = \frac{1}{2}$.
Thus,the integral becomes $\frac{1}{4} \int \frac{4x+6}{2x^2+6x+5} dx + \frac{1}{2} \int \frac{1}{2x^2+6x+5} dx$.
Let $I_1 = \int \frac{4x+6}{2x^2+6x+5} dx = \log |2x^2+6x+5| + C_1$.
For $I_2 = \int \frac{1}{2x^2+6x+5} dx = \frac{1}{2} \int \frac{1}{x^2+3x+\frac{5}{2}} dx = \frac{1}{2} \int \frac{1}{(x+\frac{3}{2})^2 + (\frac{1}{2})^2} dx$.
Using $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a})$,we get $I_2 = \frac{1}{2} \cdot \frac{1}{1/2} \tan^{-1}(\frac{x+3/2}{1/2}) + C_2 = \tan^{-1}(2x+3) + C_2$.
Combining these,the final result is $\frac{1}{4} \log |2 x^{2}+6 x+5|+\frac{1}{2} \tan ^{-1}(2 x+3)+ C$.

(A) This integral is of the form $\int \frac{(p x+q) d x}{\sqrt{a x^{2}+b x+c}}$. Let us express $x+3 = A \frac{d}{d x}(5-4 x-x^{2}) + B = A(-4-2 x) + B$.
Equating the coefficients of $x$ and the constant terms,we get $-2 A = 1$ and $-4 A + B = 3$,which gives $A = -\frac{1}{2}$ and $B = 1$.
Therefore,$\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x = -\frac{1}{2} \int \frac{(-4-2 x) d x}{\sqrt{5-4 x-x^{2}}} + \int \frac{d x}{\sqrt{5-4 x-x^{2}}} = -\frac{1}{2} I_{1} + I_{2} \dots (1)$.
For $I_{1}$,let $5-4 x-x^{2} = t$,then $(-4-2 x) d x = d t$. Thus,$I_{1} = \int \frac{d t}{\sqrt{t}} = 2 \sqrt{t} + C_{1} = 2 \sqrt{5-4 x-x^{2}} + C_{1} \dots (2)$.
For $I_{2}$,$I_{2} = \int \frac{d x}{\sqrt{9-(x+2)^{2}}} = \sin^{-1} \frac{x+2}{3} + C_{2} \dots (3)$.
Substituting $(2)$ and $(3)$ in $(1)$,we get $\int \frac{x+3}{\sqrt{5-4 x-x^{2}}} d x = -\sqrt{5-4 x-x^{2}} + \sin^{-1} \frac{x+2}{3} + C$.

We need to evaluate the integral: $\int \frac{x-1}{\sqrt{x^{2}-1}} dx$
Split the integral into two parts:
$\int \frac{x-1}{\sqrt{x^{2}-1}} dx = \int \frac{x}{\sqrt{x^{2}-1}} dx - \int \frac{1}{\sqrt{x^{2}-1}} dx$ ............ $(1)$
For the first part,let $I_1 = \int \frac{x}{\sqrt{x^{2}-1}} dx$.
Let $x^{2}-1 = t$,then $2x dx = dt$,or $x dx = \frac{1}{2} dt$.
$I_1 = \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot 2t^{1/2} = \sqrt{t} = \sqrt{x^{2}-1}$.
For the second part,we use the standard integral formula:
$\int \frac{1}{\sqrt{x^{2}-a^{2}}} dx = \log |x + \sqrt{x^{2}-a^{2}}| + C$.
Here $a = 1$,so $\int \frac{1}{\sqrt{x^{2}-1}} dx = \log |x + \sqrt{x^{2}-1}|$.
Combining these results in $(1)$:
$\int \frac{x-1}{\sqrt{x^{2}-1}} dx = \sqrt{x^{2}-1} - \log |x + \sqrt{x^{2}-1}| + C$,where $C$ is an arbitrary constant.

(A) To integrate $\int \frac{1}{\sqrt{9x^{2}+6x+5}} dx$,we first complete the square for the quadratic expression inside the square root:
$9x^{2}+6x+5 = (3x)^{2} + 2(3x)(1) + 1^{2} + 4 = (3x+1)^{2} + 2^{2}$.
Now,the integral becomes $\int \frac{1}{\sqrt{(3x+1)^{2} + 2^{2}}} dx$.
Let $t = 3x+1$,then $dt = 3dx$,which implies $dx = \frac{1}{3} dt$.
Substituting these into the integral:
$\frac{1}{3} \int \frac{1}{\sqrt{t^{2} + 2^{2}}} dt$.
Using the standard formula $\int \frac{1}{\sqrt{x^{2}+a^{2}}} dx = \log |x + \sqrt{x^{2}+a^{2}}| + C$:
$= \frac{1}{3} \log |t + \sqrt{t^{2} + 2^{2}}| + C$.
Substituting $t = 3x+1$ back:
$= \frac{1}{3} \log |(3x+1) + \sqrt{(3x+1)^{2} + 4}| + C$.
$= \frac{1}{3} \log |(3x+1) + \sqrt{9x^{2}+6x+5}| + C$,where $C$ is an arbitrary constant.

We need to evaluate the integral $I = \int \frac{1}{\sqrt{(x-1)(x-2)}} dx$.
First,expand the expression inside the square root:
$(x-1)(x-2) = x^2 - 3x + 2$.
Complete the square for the quadratic expression:
$x^2 - 3x + 2 = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4} = (x - \frac{3}{2})^2 - (\frac{1}{2})^2$.
Substitute this back into the integral:
$I = \int \frac{1}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}} dx$.
Let $t = x - \frac{3}{2}$,then $dt = dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{t^2 - a^2}} dt = \log |t + \sqrt{t^2 - a^2}| + C$:
$I = \log |(x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}| + C$.
Simplifying the expression inside the square root back to the original form:
$I = \log |(x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2}| + C$,where $C$ is an arbitrary constant.

To integrate $\int \frac{1}{\sqrt{8+3x-x^{2}}} dx$,we first complete the square for the quadratic expression $8+3x-x^{2}$.
$8+3x-x^{2} = - (x^{2}-3x-8)$
$= - \left(x^{2}-3x + (\frac{3}{2})^{2} - (\frac{3}{2})^{2} - 8\right)$
$= - \left((x-\frac{3}{2})^{2} - \frac{9}{4} - 8\right)$
$= - \left((x-\frac{3}{2})^{2} - \frac{41}{4}\right)$
$= \frac{41}{4} - (x-\frac{3}{2})^{2}$
Now,the integral becomes $\int \frac{1}{\sqrt{\frac{41}{4} - (x-\frac{3}{2})^{2}}} dx$.
Let $t = x-\frac{3}{2}$,then $dt = dx$.
Using the formula $\int \frac{1}{\sqrt{a^{2}-t^{2}}} dt = \sin^{-1}(\frac{t}{a}) + C$,where $a = \frac{\sqrt{41}}{2}$:
$= \sin^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right) + C$
$= \sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right) + C$,where $C$ is an arbitrary constant.

$(x-a)(x-b)$ can be written as $x^{2}-(a+b) x+a b$.
Therefore,
$x^{2}-(a+b) x+a b = \left[x-\left(\frac{a+b}{2}\right)\right]^{2}-\frac{(a-b)^{2}}{4}$.
$\int \frac{1}{\sqrt{(x-a)(x-b)}} d x = \int \frac{1}{\sqrt{\left[x-\left(\frac{a+b}{2}\right)\right]^{2}-\frac{(a-b)^{2}}{4}}} d x$.
Let $x-\left(\frac{a+b}{2}\right)=t$,then $d x=d t$.
$\int \frac{1}{\sqrt{t^{2}-\left(\frac{a-b}{2}\right)^{2}}} d t = \log \left| t + \sqrt{t^{2}-\left(\frac{a-b}{2}\right)^{2}} \right| + C$.
Substituting back $t = x-\left(\frac{a+b}{2}\right)$,we get:
$\log \left| \left(x-\frac{a+b}{2}\right) + \sqrt{(x-a)(x-b)} \right| + C$.

Let $x+2=A \frac{d}{d x}(x^{2}-1)+B$ ...........$(1)$
$\Rightarrow x+2=A(2 x)+B$
Equating the coefficients of $x$ and constant terms on both sides,we obtain:
$2 A=1 \Rightarrow A=\frac{1}{2}$
$B=2$
From $(1),$ we obtain:
$(x+2)=\frac{1}{2}(2 x)+2$
Then,$\int \frac{x+2}{\sqrt{x^{2}-1}} d x=\int \frac{\frac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$
$=\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x+\int \frac{2}{\sqrt{x^{2}-1}} d x$ ..........$(2)$
In $\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x,$ let $x^{2}-1=t \Rightarrow 2 x d x=d t$
$\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{t}} = \frac{1}{2}[2 \sqrt{t}] = \sqrt{t} = \sqrt{x^{2}-1}$
Then,$\int \frac{2}{\sqrt{x^{2}-1}} d x = 2 \log |x + \sqrt{x^{2}-1}|$
From equation $(2),$ we obtain:
$\int \frac{x+2}{\sqrt{x^{2}-1}} d x = \sqrt{x^{2}-1} + 2 \log |x + \sqrt{x^{2}-1}| + C$
where $C$ is an arbitrary constant.

Let $5 x-2=A \frac{d}{d x}\left(1+2 x+3 x^{2}\right)+B$
$\Rightarrow 5 x-2=A(2+6 x)+B$
Equating the coefficients of $x$ and the constant term on both sides,we obtain:
$5=6 A \Rightarrow A=\frac{5}{6}$
$2 A+B=-2 \Rightarrow B=-2-2\left(\frac{5}{6}\right)=-2-\frac{5}{3}=-\frac{11}{3}$
$\therefore 5 x-2=\frac{5}{6}(2+6 x)-\frac{11}{3}$
$\Rightarrow \int \frac{5 x-2}{1+2 x+3 x^{2}} d x=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^{2}} d x$
$=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^{2}} d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^{2}} d x$
Let $I_{1}=\int \frac{2+6 x}{1+2 x+3 x^{2}} d x$ and $I_{2}=\int \frac{1}{1+2 x+3 x^{2}} d x$
$\therefore \int \frac{5 x-2}{1+2 x+3 x^{2}} d x=\frac{5}{6} I_{1}-\frac{11}{3} I_{2} \quad \dots(1)$
For $I_{1}$,let $1+2 x+3 x^{2}=t$,then $(2+6 x) d x=d t$
$I_{1}=\int \frac{d t}{t}=\log |t|=\log |1+2 x+3 x^{2}| \quad \dots(2)$
For $I_{2}$,$1+2 x+3 x^{2}=3\left(x^{2}+\frac{2}{3} x+\frac{1}{3}\right)=3\left(\left(x+\frac{1}{3}\right)^{2}+\frac{2}{9}\right)=3\left(\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right)$
$I_{2}=\frac{1}{3} \int \frac{d x}{\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}}=\frac{1}{3} \cdot \frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{x+1/3}{\sqrt{2}/3}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right) \quad \dots(3)$
Substituting $(2)$ and $(3)$ into $(1)$:
$\int \frac{5 x-2}{1+2 x+3 x^{2}} d x=\frac{5}{6} \log |1+2 x+3 x^{2}|-\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+C$

(N/A) We have $\frac{6x+7}{\sqrt{(x-5)(x-4)}} = \frac{6x+7}{\sqrt{x^2-9x+20}}$.
Let $6x+7 = A\frac{d}{dx}(x^2-9x+20) + B$.
$6x+7 = A(2x-9) + B$.
Equating the coefficients of $x$ and the constant term,we get $2A = 6 \Rightarrow A = 3$ and $-9A + B = 7 \Rightarrow -27 + B = 7 \Rightarrow B = 34$.
Thus,$\int \frac{6x+7}{\sqrt{x^2-9x+20}} dx = \int \frac{3(2x-9) + 34}{\sqrt{x^2-9x+20}} dx = 3 \int \frac{2x-9}{\sqrt{x^2-9x+20}} dx + 34 \int \frac{1}{\sqrt{x^2-9x+20}} dx$.
Let $I_1 = \int \frac{2x-9}{\sqrt{x^2-9x+20}} dx$. Substituting $t = x^2-9x+20$,$dt = (2x-9)dx$,we get $I_1 = \int t^{-1/2} dt = 2\sqrt{t} = 2\sqrt{x^2-9x+20}$.
Let $I_2 = \int \frac{1}{\sqrt{x^2-9x+20}} dx$. Completing the square,$x^2-9x+20 = (x-\frac{9}{2})^2 - \frac{1}{4} = (x-\frac{9}{2})^2 - (\frac{1}{2})^2$.
Using the formula $\int \frac{1}{\sqrt{x^2-a^2}} dx = \log|x + \sqrt{x^2-a^2}|$,we get $I_2 = \log|x-\frac{9}{2} + \sqrt{x^2-9x+20}|$.
Combining these,the integral is $6\sqrt{x^2-9x+20} + 34\log|x-\frac{9}{2} + \sqrt{x^2-9x+20}| + C$.

(N/A) Let $x+2 = A \frac{d}{dx}(4x-x^2) + B$.
$\Rightarrow x+2 = A(4-2x) + B$.
Equating the coefficients of $x$ and the constant term on both sides,we obtain:
$-2A = 1 \Rightarrow A = -\frac{1}{2}$.
$4A + B = 2 \Rightarrow B = 4$.
Thus,$x+2 = -\frac{1}{2}(4-2x) + 4$.
Therefore,$\int \frac{x+2}{\sqrt{4x-x^2}} dx = \int \frac{-\frac{1}{2}(4-2x) + 4}{\sqrt{4x-x^2}} dx$.
$= -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}} dx + 4 \int \frac{1}{\sqrt{4x-x^2}} dx$.
Let $I_1 = \int \frac{4-2x}{\sqrt{4x-x^2}} dx$ and $I_2 = \int \frac{1}{\sqrt{4x-x^2}} dx$.
So,the integral is $-\frac{1}{2}I_1 + 4I_2$ $(1)$.
For $I_1$,let $4x-x^2 = t$,then $(4-2x)dx = dt$.
$I_1 = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{4x-x^2}$ $(2)$.
For $I_2$,$4x-x^2 = -(x^2-4x) = -(x^2-4x+4-4) = 4-(x-2)^2 = (2)^2-(x-2)^2$.
$I_2 = \int \frac{dx}{\sqrt{(2)^2-(x-2)^2}} = \sin^{-1}\left(\frac{x-2}{2}\right)$ $(3)$.
Substituting $(2)$ and $(3)$ into $(1)$:
$= -\frac{1}{2}(2\sqrt{4x-x^2}) + 4\sin^{-1}\left(\frac{x-2}{2}\right) + C$.
$= -\sqrt{4x-x^2} + 4\sin^{-1}\left(\frac{x-2}{2}\right) + C$,where $C$ is the constant of integration.

We need to evaluate the integral $I = \int \frac{x+2}{\sqrt{x^{2}+2 x+3}} dx$.
First,we express the numerator in terms of the derivative of the quadratic expression inside the square root,which is $\frac{d}{dx}(x^2+2x+3) = 2x+2$.
$\int \frac{x+2}{\sqrt{x^{2}+2 x+3}} dx = \frac{1}{2} \int \frac{2(x+2)}{\sqrt{x^{2}+2 x+3}} dx = \frac{1}{2} \int \frac{2x+4}{\sqrt{x^{2}+2 x+3}} dx$
$= \frac{1}{2} \int \frac{2x+2+2}{\sqrt{x^{2}+2 x+3}} dx = \frac{1}{2} \int \frac{2x+2}{\sqrt{x^{2}+2 x+3}} dx + \int \frac{1}{\sqrt{x^{2}+2 x+3}} dx$
Let $I_1 = \int \frac{2x+2}{\sqrt{x^{2}+2 x+3}} dx$ and $I_2 = \int \frac{1}{\sqrt{x^{2}+2 x+3}} dx$.
For $I_1$,let $t = x^2+2x+3$,then $dt = (2x+2)dx$.
$I_1 = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{x^2+2x+3}$.
For $I_2$,complete the square: $x^2+2x+3 = (x+1)^2 + 2 = (x+1)^2 + (\sqrt{2})^2$.
$I_2 = \int \frac{dx}{\sqrt{(x+1)^2 + (\sqrt{2})^2}} = \log |(x+1) + \sqrt{(x+1)^2 + 2}| = \log |(x+1) + \sqrt{x^2+2x+3}|$.
Combining these,$I = \frac{1}{2}(2\sqrt{x^2+2x+3}) + \log |(x+1) + \sqrt{x^2+2x+3}| + C$.
Final result: $I = \sqrt{x^2+2x+3} + \log |(x+1) + \sqrt{x^2+2x+3}| + C$.

Let $(x+3) = A \frac{d}{dx}(x^{2}-2x-5) + B$
$(x+3) = A(2x-2) + B$
Equating the coefficients of $x$ and the constant term on both sides,we obtain:
$2A = 1 \Rightarrow A = \frac{1}{2}$
$-2A + B = 3 \Rightarrow -2(\frac{1}{2}) + B = 3 \Rightarrow -1 + B = 3 \Rightarrow B = 4$
Therefore,$(x+3) = \frac{1}{2}(2x-2) + 4$
Now,$\int \frac{x+3}{x^{2}-2x-5} dx = \int \frac{\frac{1}{2}(2x-2)+4}{x^{2}-2x-5} dx$
$= \frac{1}{2} \int \frac{2x-2}{x^{2}-2x-5} dx + 4 \int \frac{1}{x^{2}-2x-5} dx$
Let $I_{1} = \int \frac{2x-2}{x^{2}-2x-5} dx$ and $I_{2} = \int \frac{1}{x^{2}-2x-5} dx$
For $I_{1}$,let $x^{2}-2x-5 = t$,then $(2x-2)dx = dt$. So,$I_{1} = \int \frac{dt}{t} = \log |x^{2}-2x-5|$.
For $I_{2}$,$x^{2}-2x-5 = (x-1)^{2} - 6 = (x-1)^{2} - (\sqrt{6})^{2}$.
Using $\int \frac{1}{x^{2}-a^{2}} dx = \frac{1}{2a} \log |\frac{x-a}{x+a}|$,we get $I_{2} = \frac{1}{2\sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|$.
Substituting $I_{1}$ and $I_{2}$ back,the integral is $\frac{1}{2} \log |x^{2}-2x-5| + \frac{4}{2\sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}| + C$
$= \frac{1}{2} \log |x^{2}-2x-5| + \frac{2}{\sqrt{6}} \log |\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}| + C$.

Let $5 x+3=A \frac{d}{d x}(x^{2}+4 x+10)+B$
$\Rightarrow 5 x+3=A(2 x+4)+B$
Equating the coefficients of $x$ and the constant term,we obtain:
$2 A=5 \Rightarrow A=\frac{5}{2}$
$4 A+B=3 \Rightarrow B=3-4(\frac{5}{2})=3-10=-7$
$\therefore 5 x+3=\frac{5}{2}(2 x+4)-7$
$\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{(x+2)^{2}+6}} d x$
For the first integral,let $t=x^{2}+4 x+10$,then $dt=(2 x+4)dx$. Thus,$\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^{2}+4 x+10}$.
For the second integral,using $\int \frac{dx}{\sqrt{x^{2}+a^{2}}}=\log |x+\sqrt{x^{2}+a^{2}}|$,we get $\log |(x+2)+\sqrt{(x+2)^{2}+6}| = \log |(x+2)+\sqrt{x^{2}+4 x+10}|$.
Combining these,the result is $5\sqrt{x^{2}+4 x+10}-7 \log |(x+2)+\sqrt{x^{2}+4 x+10}|+C$.

(A) We need to evaluate the integral $I = \int \sqrt{x^{2}+2 x+5} \, dx$.
First,complete the square for the quadratic expression: $x^{2}+2 x+5 = (x^{2}+2 x+1) + 4 = (x+1)^{2} + 2^{2}$.
Now,the integral becomes $I = \int \sqrt{(x+1)^{2} + 2^{2}} \, dx$.
Using the standard formula $\int \sqrt{t^{2}+a^{2}} \, dt = \frac{1}{2} t \sqrt{t^{2}+a^{2}} + \frac{a^{2}}{2} \log |t + \sqrt{t^{2}+a^{2}}| + C$,where $t = x+1$ and $a = 2$:
$I = \frac{1}{2}(x+1) \sqrt{(x+1)^{2} + 2^{2}} + \frac{2^{2}}{2} \log |(x+1) + \sqrt{(x+1)^{2} + 2^{2}}| + C$.
Simplifying the expression:
$I = \frac{1}{2}(x+1) \sqrt{x^{2}+2 x+5} + 2 \log |x+1 + \sqrt{x^{2}+2 x+5}| + C$.