Indefinite Integration

$ \text { } \begin{aligned} \frac{3 x+1}{(x-1)^2\left(x^2+1\right)}= & \frac{A}{x-1} +\frac{B}{(x-1)^2}+\frac{C x+D}{x^2+1} \\ 3 x+1= & A(x-1)\left(x^2+1\right)+B\left(x^2+1\right) +(C x+D)(x-1)^2 \\ = & x^3(A+C)+x^2(-A+B+D-2 C) +x(A+C-2 D)-A+B+D \end{aligned} $

$ \begin{aligned} &\begin{aligned} & A+C=0 \\ & -A+B+D-2 C=0 \\ & A+C-2 D=3 \\ & -A+B+D=1 \\ & -A-C+B+D-C=0 \\ & B+D-C=0 \end{aligned}\\ &\text { So, } 2(A-C+B+D)=2 A\\ &\begin{array}{rlr} 2 D=-3 \Rightarrow D=\frac{-3}{2} & \\ B-A=\frac{5}{2} & \\ -A+B+D-2 C & =0 \\ \frac{5}{2}-\frac{3}{2}+2 A & =0\,\,\,\,\,\,\, [\because C=-A]\\ \Rightarrow \quad 2 A & =-1 \end{array} \end{aligned} $

$ \begin{aligned} &\text { } y=(f(x))^{g(x)}\\ &\begin{aligned} & \Rightarrow \frac{d y}{d x}=\{f(x)\}^{g(x)}\left(\frac{g(x) \cdot f^{\prime}(x)}{f(x)}+g^{\prime}(x) \ln f(x)\right) \\ & \Rightarrow \frac{d y}{d x}=y\left(f^{\prime}(x) \frac{g(x)}{f(x)}+g^{\prime}(x) \ln f(x)\right) \\ & \Rightarrow H(x)=\frac{g(x)}{f(x)} \text { and } G(x)=\ln f(x) \\ & \begin{aligned} \Rightarrow \int \frac{G(x) H(x) f^{\prime}(x)}{g(x)} d x=\int \frac{\ln f(x)}{g(x)} \times \frac{g(x)}{f(x)} \times \frac{f^{\prime}(x)}{d x} \\ \quad=\int \ln f(x) d(\ln f(x)) \\ \quad=\frac{[\ln f(x)]^2}{2}+C \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } I_2-I_1=\int \frac{e^{3 x}-e^x}{e^{4 x}+e^{2 x}+1} d x \\ & =\int \frac{e^x\left(e^{2 x}-1\right)}{e^{4 x}+e^{2 x}+1} d x \\ & \Rightarrow I_2-I_1 \int \frac{t^2-1}{t^4+t^2+1} d t \quad\left(\text { put } t=e^x\right) \\ & \Rightarrow I_2-I_1=\int \frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2-1} d t \end{aligned} $

$ \begin{aligned} & =\frac{1}{2 \times 1} \ln \left|\frac{t+\frac{1}{t}-1}{t+\frac{1}{t}+1}\right|+C \\ & =\frac{1}{2} \ln \left|\frac{e^x+e^{-x}-1}{e^x+e^{-x}+1}\right|+C \end{aligned} $

Given,

$ I=\int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x $

Put $x=\cos ^2(2 \theta)$

$ \begin{aligned} & d x=-\sin (4 \theta) \times 2 d \theta \\ & I=\int \frac{1}{\cos ^2(2 \theta)} \times \frac{\sin \theta}{\cos \theta} \times-4 \sin 2 \theta \cos 2 \theta d \theta \\ & I=\int \frac{-8 \sin ^2 \theta \cdot \cos \theta \cdot \cos 2 \theta d \theta}{\cos \theta \cdot \cos ^2(2 \theta)} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =-\int \frac{18 \sin ^2 \theta}{\cos 2 \theta} d \theta \\ & =+4 \int \frac{\cos 2 \theta-1}{\cos 2 \theta} d \theta \\ & =4 \theta-4 \int \sec 2 \theta d \theta \\ & =4 \theta-\frac{4 \ln |\sec 2 \theta+\tan 2 \theta|}{2}+C \\ & =2 \ln \left|\frac{1}{\sec 2 \theta+\tan 2 \theta}\right|+4 \theta+C \\ & =2 \ln \left|\frac{1}{\sec 2 \theta+\tan 2 \theta}\right| \\ & \quad+2 \times\left(\frac{\pi}{2}-\sin ^{-1} \sqrt{x}\right)+C \\ & =2 \ln \left|\frac{\sqrt{x}}{\sqrt{1-x}+1}\right|-2 \sin ^{-1} \sqrt{x}+C \end{aligned}\\ &\text { So, } f(x)=\operatorname{sech}^{-1} \sqrt{x} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } 3 x+2=\alpha \cdot \frac{d}{d x}\left(4 x^2+4 x+5\right)+\beta \\ & 3 x+2=(8 x+4) \alpha+\beta \\ & \Rightarrow 3=8 \alpha \text { and } 4 \alpha+\beta=2 \\ & \Rightarrow \alpha=\frac{3}{8} \text { and } \beta=2-\frac{3}{2}=\frac{1}{2} \\ & \text { Let } I=\int \frac{\frac{3}{8}(8 x+4)+\frac{1}{2}}{4 x^2+4 x+5} d x \\ & =\frac{3}{8} \log \left(4 x^2+4 x+5\right)+\frac{1}{2} \int \frac{d x}{(2 x+1)^2+2^2} \\ & =\frac{3}{8} \log \left(4 x^2+4 x+5\right)+\frac{1}{2} \times \frac{1}{4} \times \tan ^{-1}\left(\frac{2 x+1}{2}\right)+C \end{aligned} \end{aligned} $

$ \Rightarrow A+B=\frac{3}{8}+\frac{1}{8}=\frac{1}{2} $

Let $I=\int x^3 \sin 3 x d x$

$ \begin{aligned} & I=x^3 \int \sin 3 x d x-\int\left(\frac{d}{d x}\left(x^3\right) \int \sin 3 x d x\right) d x \\ & =\frac{-x^3 \cos 3 x}{3}-\int \frac{3 x^2 \times(-\cos 3 x) d x}{3} \\ & =\frac{-x^3 \cos 3 x}{3}+\int x^2 \cos 3 x d x+c_1 \\ & \text { Let } I_1=\int x^2 \cos 3 x d x \\ & I_1=x^2 \int \cos 3 x d x-\int\left(\frac{d}{d x}\left(x^2\right) \int \cos 3 x d x\right) d x \\ & I_1=\frac{x^2 \sin 3 x}{3}-\int \frac{2 x \times \sin 3 x}{3} d x+c_2 \\ & \text { Let } I_3=\int x \sin 3 x d x \end{aligned} $

$ \begin{aligned} & \Rightarrow I_3=x \int \sin 3 x d x-\int\left(\frac{d}{d x}(x) \int \sin 3 x d x\right) d x \\ & \Rightarrow I_3=\frac{-x \cos 3 x}{x}-\int \frac{-\cos 3 x}{3} d x \\ & \Rightarrow I_3=\frac{-x \cos 3 x}{3}+\frac{\sin 3 x}{9}+c_3 \\ & \text { So, } I=\frac{-x^3 \cos 3 x}{3}+\frac{x^2 \sin 3 x}{3} \\ & \quad-\frac{2}{3}\left(\frac{-x \cos 3 x}{3}+\frac{\sin 3 x}{9}\right)+C \\ & \Rightarrow I=\frac{1}{27}\left[-9 x^3 \cos 3 x+9 x^2 \sin 3 x\right.+6 x \cos 3 x-2 \sin 3 x]+C \\ & \Rightarrow \quad f(x)=6 x-9 x^3 \\ & \quad g(x)=9 x^2-2 \\ & \Rightarrow \quad f(1)=-3 \quad, \quad g(1)=7 \\ & \Rightarrow f(1)+g(1)=4 \end{aligned} $

$ \begin{aligned} I_1+I_2 & =\int\left(\sin ^6 x+\cos ^6 x\right) d x \\ & =\int 1-\frac{3}{4}\left(\frac{1-\cos 4 x}{2}\right) d x \\ & =\int\left(1-\frac{3}{8}+\frac{3}{8} \cos 4 x\right) d x \\ & =\frac{5}{8} \int d x+\frac{3}{8} \int \cos 4 x d x \\ & =\frac{5 x}{8}+\frac{3 \sin 4 x}{32}+C \\ & =\frac{1}{32}(20 x+3 \sin 4 x)+C \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{x+\cos x}{1-\sin x} d x \\ & \begin{aligned} \Rightarrow I & =\int \frac{(x+\cos x)(1+\sin x)}{\cos ^2 x} d x \\ \Rightarrow I & =\int x\left(\sec ^2 x+\tan x \cdot \sec x\right) d x +\int(\sec x+\tan x) d x \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow I & =x \int \sec ^2 x d x-\int \tan x d x+x \sec x-\int \sec x d x+\int(\sec x+\tan x) d x+C_1 \\ \Rightarrow I & =x(\sec x+\tan x)+C \\ \Rightarrow I & =x\left(\frac{1+\sin x}{\cos x}\right)+C \\ \Rightarrow I & =x \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)+C \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{d x}{(x+2) \sqrt{x^2+x+2}} \\ & \text { Put } z=\frac{1}{x+2} \\ & \Rightarrow \quad \frac{d z}{d x}=-\frac{1}{\left(x+2^2\right.}=-z^2 \\ & \Rightarrow \quad x+2=\frac{1}{z} \\ & \Rightarrow \quad x=\frac{1-2 z}{z} \end{aligned} $

$ \begin{aligned} & \Rightarrow I=\int \frac{-\frac{d z}{z^2}}{\left(\frac{1}{z}\right) \sqrt{\left(\frac{1-2 z}{z}\right)^2+\frac{1-2 z}{z}+2}} \\ & \Rightarrow I=-\int \frac{d z}{\sqrt{(1-2 z)^2+z(1-2 z)+2 z^2}} \\ & =-\int \frac{d z}{\sqrt{4 z^2-4 z+1+z-2 z^2+2 z^2}} \\ & =-\int \frac{d z}{\sqrt{4 z^2-3 z+1}} \\ & =-\frac{1}{2} \int \frac{d z}{\sqrt{z^2-\frac{3 z}{4}+\frac{1}{4}}} \\ & =-\frac{1}{2} \int \frac{d z}{\sqrt{z^2-\frac{3 z}{4}+\frac{9}{64}+\frac{1}{4}-\frac{9}{64}}} \\ & =-\frac{1}{2} \int \frac{d z}{\sqrt{\left(z-\frac{3}{8}\right)^2+\frac{7}{64}}} \\ & =\frac{-1}{2} \ln \left(z-\frac{3}{8}+\sqrt{\left(z-\frac{3}{8}\right)^2+\frac{7}{64}}\right)+C \end{aligned} $

$ \begin{aligned} & =\frac{-1}{2} \ln \left(\frac{1}{x+2}-\frac{3}{8}+\sqrt{\left(\frac{1}{x+2}-3\right.} \frac{x^2}{8}+\frac{7}{64}\right)+C \\ & =\frac{-1}{2} \sinh ^{-1}\left(\frac{2-3 x}{\sqrt{7} \times(x+2)}\right)+C \end{aligned} $

$ \begin{aligned} & \text { } \frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{x^2+3} \\ & x^2+1=(A x+B)\left(x^2+3\right)+(C x+D)\left(x^2+2\right) \end{aligned} $

Coefficient of $x^3=0$, coefficient of $x=0$

$ \Rightarrow A+C=0,3 A+2 C=0 \Rightarrow A=C=0 $

Coefficient of $x^2=1$, constant term $=1$

$ B+D=1, \quad 1=3 B+2 D $

$\eqalign{ & 3B\,\, + \,\,2D\, = \,\,1 \cr & 2B\,\, + \,\,2D\, = \,\,2 \cr & \underline { - \,\,\,\,\, - \,\,\,\,\,\, - \,\,\,\,\,\,\,} \cr & B\, = \, - 1,\,\,D\, = \,2 \cr} $

$ \therefore A+B+C+D=1 $

Let’s call the integral $ I $.

$ I = \int \frac{3^x (x \log 3 - 1)}{x^2} dx $

Let’s make a substitution. Set $ t = \frac{3^x}{x} $.

Now, calculate the derivative of $ t $ with respect to $ x $:

$ \frac{dt}{dx} = \frac{3^x \log 3 \cdot x - 3^x}{x^2} = \frac{3^x (x \log 3 - 1)}{x^2} $

So, $ dt = \frac{3^x (x \log 3 - 1)}{x^2} dx $.

This shows that the part inside our integral, $ \frac{3^x (x \log 3 - 1)}{x^2} dx $, is simply $ dt $.

So the integral becomes $ I = \int dt = t + C $.

Recall $ t = \frac{3^x}{x} $, so the answer is $ I = \frac{3^x}{x} + C $.

$ \begin{aligned} & \text { } \because \frac{5 \pi}{4} < x < \frac{7 \pi}{4} \\ & \begin{aligned} \int \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x & \\ = & \int \sqrt{\frac{1-\frac{2 \tan x}{1+\tan ^2 x}}{1+\frac{2 \tan ^2}{1+\tan ^2 x}}} d x \\ = & \int \frac{\sqrt{(\tan x-1)^2}}{\sqrt{(\tan x+1)^1}} d x \\ = & -\int \frac{1-\tan x}{1+\tan x} d x \\ = & -\int \tan \left(\frac{\pi}{4}-x\right) d x+C \\ = & \frac{-\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|}{-1}+C \\ = & \log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } I=\int x \tan ^{-1} \sqrt{\frac{1+x^2}{1-x^2}} d x \\ & x^2=\cos \theta \Rightarrow 2 x d x=-\sin \theta d \theta \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow I & \left.=\int \tan ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} \cdot\left(\frac{-\sin \theta}{2}\right) d \theta\right) \\ & =\frac{1}{2} \int \tan ^{-1}\left(\cot \frac{\theta}{2}\right)(-\sin \theta) d \theta \\ & =\frac{1}{2} \int\left(\frac{\pi}{2}-\frac{\theta}{2}\right)(-\sin \theta) d \theta \\ & =\frac{1}{2} \int\left(\frac{\theta}{2} \sin \theta-\frac{\pi}{2} \sin \theta\right) d \theta \\ & =\frac{\theta}{4}(-\cos \theta)+\frac{\sin \theta}{4}+\frac{\pi}{4} \cos \theta+C \\ & =\frac{-x^2}{4} \cos ^{-1} x^2+\frac{\sqrt{1-x^4}}{4}+\frac{\pi}{4} x^2+C \\ & =\frac{x^2}{4}\left(\pi-\cos ^{-1} x^2\right)+\frac{1}{4} \sqrt{1-x^4}+C \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \begin{aligned} & I=\int \frac{d x}{(2 \cos x+\sin x)^2} \\ & I=\int \frac{\sec ^2 x}{(2+\tan x)^2} \\ & \text { Put } \tan x+2=t \\ & \sec ^2 x d x=d t \\ & I=\int \frac{d t}{t^2} \\ &=\frac{-1}{t}+c=\frac{-1}{\tan x+2}+C \\ &=\frac{-\cos x}{\sin x+2 \cos x}+C \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \frac{x+3}{(x+1)\left(x^2+2\right)}=\frac{a}{x+1}+\frac{b x+c}{x^2+2} \\ & \Rightarrow x+3=a\left(x^2+2\right)+(b x+c)(x+1) \\ & \text { put } x=-1 \\ & \qquad \quad 2=a(3) \Rightarrow a=\frac{2}{3} \end{aligned} $

put $x=0$

$ \begin{aligned} & 3=2 a+c \\ & c=3-\frac{4}{3}=\frac{5}{3} \end{aligned} $

put $x=1$

$ \begin{gathered} 4=3 a+(b+c) 2 \\ 4=2+\left(b+\frac{5}{3}\right) 2 \\ 1=b+\frac{5}{3} \Rightarrow b=\frac{-2}{3} \\ \therefore a-b+c=\frac{2}{3}+\frac{2}{3}+\frac{5}{3}=\frac{9}{3}=3 \end{gathered} $

$ \begin{aligned} &\text { } I=\int e^{-x}\left(x^3-2 x^2+3 x-4\right) d x\\ &\begin{aligned} & \text { Let }-x=t \\ & \begin{aligned} -d x= & d t \\ I= & \int e^t\left(-t^3-2 t^2-3 t-4\right)(-d t) \\ = & \int e^t\left(t^3+2 t^2+3 t+4\right) d t \\ = & \int e^t\left(t^3+3 t^2-t^2-2 t+5 t+5-1\right) d t \\ = & \int e^t\left(t^3+3 t^2\right) d t+\int e^t\left(-t^2-2 t\right) d t +\int e^t(5 t+5) d t-\int e^t d t \\ = & e^t t^3+e^t\left(-t^2\right)+e^t(5 t)-e^t+C \\ = & e^t\left(t^3-t^2+5 t-1\right)+C \\ = & e^{-x}\left(-x^3-x^2-5 x-1\right)+C \\ = & -e^{-x}\left(x^3+x^2+5 x+1\right)+C \end{aligned} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} & \int\left(1+\tan ^2 x\right)(1+2 x \tan x) d x \\ = & \int\left(\sec ^2 x+2 x \tan x \sec ^2 x\right) d x \end{aligned} $

$ = \int {{{\sec }^2}x\,dx\, + \,\int {\mathop {2x}\limits_I \,\tan \,x\,{{\mathop {\sec }\limits_{II} }^2}x\,dx} } $

$ \begin{aligned} & =\tan x+2 x\left(\frac{\tan ^2 x}{2}\right)-\int \tan ^2 x \cdot d x+C \\ & =\tan x+x \tan ^2 x-\int\left(\sec ^2 x-1\right) d x+C \\ & =\tan x+x \tan ^2 x-\tan x+x+C \\ & =x\left(\tan ^2 x+1\right)+C=x \sec ^2 x+C \end{aligned} $

$ \begin{aligned} &\text { } I=\int x^2 \frac{\tan ^{-1} x}{\left(1+x^2\right)^2} d x\\ &\text { Put } x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta\\ &\begin{aligned} I & =\int \tan ^2 \theta \cdot \frac{\theta \cdot \sec ^2 \theta d \theta}{\sec ^4 \theta} \\ & =\int \theta \cdot \frac{\tan ^2 \theta}{\sec ^2 \theta} d \theta=\int \theta\left(\sin ^2 \theta\right) d \theta \\ & =\int \theta\left(\frac{1-\cos 2 \theta}{2}\right) d \theta \\ & =\frac{1}{2} \int \theta(1-\cos 2 \theta) d \theta \\ & =\frac{1}{2} \int \theta d \theta-\int \theta \cos 2 \theta d \theta \\ & =\frac{1}{2}\left(\frac{\theta^2}{2}-\frac{\theta \sin 2 \theta}{2}-\frac{\cos 2 \theta}{4}\right)+C \\ & =\frac{\theta^2}{4}-\frac{\theta}{4} \sin 2 \theta-\frac{1}{8} \cos 2 \theta+C \\ & =\frac{\left(\tan ^{-1} x\right)^2}{4}-\frac{\tan ^{-1} x}{4} \cdot\left(\frac{2 x}{1+x^2}\right)-\frac{1}{8} \frac{1-x^2}{1+x^2}+C \\ & =\frac{\left(\tan ^{-1} x\right)^2}{4}-\frac{4 x \tan ^{-1} x+1-x^2}{8\left(1+x^2\right)}+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \begin{aligned} I & =\int \frac{\log x}{(1+x)^3} d x \\ & =\int_1^{\log x} \cdot \frac{1}{(1+x)^3} d x \\ \Rightarrow \quad I & =\log x \int \frac{1}{(1+x)^3} d x -\int\left(\frac{1}{x} \cdot \int \frac{1}{(1+x)^3} d x\right) d x \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{\log x}{-2(1+x)^2}+\frac{1}{2} \int \frac{1}{x} \cdot \frac{1}{(1+x)^2} d x . \\ & =\frac{1}{2}\left(\frac{-\log x}{(1+x)^2}\right)+\frac{1}{2} \int\left(\frac{1}{x}-\frac{1}{1+x}-\frac{1}{(1+x)^2}\right) d x \\ & =\frac{1}{2}\left(\frac{-\log x}{(1+x)^2}+\log x-\log (1+x)\right.\left.+\frac{1}{x+1}\right)+C \end{aligned} $

$ =\frac{1}{2}\left(\frac{1}{1+x}-\frac{\log x}{(1+x)^2}+\log \left(\frac{x}{1+x}\right)\right)+C $

We are given:

$ \frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+1} $

First, get a common denominator to combine the right side:

$ \frac{A}{x+2}+\frac{B x+C}{x^2+1} = \frac{A(x^2+1)+(B x+C)(x+2)}{(x+2)(x^2+1)} $

Set the numerators equal to each other:

$ x^2-3 = A(x^2+1) + (B x + C)(x+2) $

Expand the right side:

$A(x^2 + 1) = A x^2 + A$
$(B x + C)(x + 2) = B x^2 + 2B x + C x + 2C = B x^2 + (2B + C)x + 2C$

Combine all terms on the right:

$A x^2 + B x^2 = (A+B)x^2$
$(2B + C)x$
$A + 2C$

So we get:

$x^2-3 = (A+B)x^2 + (2B+C)x + (A+2C)$

Now, match the coefficients for $x^2$, $x$, and the constant term:

For $x^2$: $A + B = 1$
For $x$: $2B + C = 0$
For constant: $A + 2C = -3$

Write the equations:

1. $A + B = 1$
2. $2B + C = 0$
3. $A + 2C = -3$

Now solve these equations step by step:

From equation 2: $2B + C = 0$ → $C = -2B$

Substitute $C = -2B$ into equation 3: $A + 2(-2B) = -3$
$A - 4B = -3$

Now use $A + B = 1$. Substitute $A = 1 - B$ into the last equation:
$1 - B - 4B = -3$
$1 - 5B = -3$
$-5B = -4$
$B = \frac{4}{5}$

Now $C = -2B = -2 \times \frac{4}{5} = \frac{-8}{5}$
$A = 1 - B = 1 - \frac{4}{5} = \frac{1}{5}$

Now find $3 A + 2 B - C$:

$3\left(\frac{1}{5}\right) + 2\left(\frac{4}{5}\right) - \left(\frac{-8}{5}\right)$
$= \frac{3}{5} + \frac{8}{5} + \frac{8}{5}$
$= \frac{3 + 8 + 8}{5}$
$= \frac{19}{5}$

Let $I=\int\left(\frac{1}{x^2}+\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\right) d x$

again let $I_1=\int \frac{1}{x^2} d x$

And $I_2=\int \frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x} d x$

$\therefore I_1=\int x^{-2} d x=\frac{x^{-2+11}}{-2+1}+c_1=-\frac{1}{x}+c_1$

And $I_2=\int\left(\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}+\frac{\cos ^3 x}{\sin ^2 x \cos ^2 x}\right) d x$

$=\int(\tan x \sec x+\cot x \operatorname{cosec} x) d x$

$=\int \tan x \sec x d x+\int \cot x \operatorname{cosec} x d x$

$=\sec x+(-\operatorname{cosec} x)+c_2$

$\therefore I=I_1+I_2=-\frac{1}{x}+\sec x-\operatorname{cosec} x+C$

$\Rightarrow I=-\frac{1}{x}+\frac{1}{\cos x}-\frac{1}{\sin x}+C$

$=\frac{-(\cos x \sin x)+x \sin x-x \cos x}{x \cos x \sin x}+C$

$=\frac{x(\sin x-\cos x)-\cos x \sin x}{x \cos x \sin x}+C$

$ \begin{aligned} & \text {Given, } I_n=\int \frac{1}{\left(x^2+1\right)^n} d x \\ & \begin{aligned} = & \frac{x}{\left(x^2+1\right)^n}-\int\left(\frac{d}{d x}\left(\frac{1}{\left(x^2+1\right)^n}\right) \int d x\right) d x \\ = & \frac{x}{\left(x^2+1\right)^n}+2 n \int \frac{x^2}{\left(x^2+1\right)^{n+1}} d x \\ \Rightarrow & I_n=\frac{x}{\left(x^2+1\right)^n} +2 n \int\left(\frac{1}{\left(x^2+1\right)^n}-\frac{1}{\left(x^2+1\right)^{n+1}}\right) d x \\ \Rightarrow & I_n=\frac{x}{\left(x^2+1\right)^n}+2 n\left(I_n-I_{n+1}\right)+C \\ \Rightarrow & 2 n I_{n+1}=\frac{x}{\left(x^2+1\right)^n}+(2 n-1) I_n+C \\ \Rightarrow & 2 n I_{n+1}-(2 n-1) I_n=\frac{x}{\left(x^2+1\right)^n}+C \end{aligned} \end{aligned} $

Let $I=\int \frac{x^3 d x}{x^4+3 x^2+2}$

put $x^2=t \Rightarrow 2 x d x=d t \Rightarrow x d x=\frac{d t}{2}$

$ \begin{aligned} \therefore I & =\int \frac{t \cdot \frac{d t}{2}}{t^2+3 t+2} \\ I & =\frac{1}{2} \int \frac{t d t}{t^2+3 t+2}=\frac{1}{2} \int \frac{t d t}{(t+1)(t+2)} \end{aligned} $

Now, $\frac{t}{(t+1)(t+2)}=\frac{A}{t+1}+\frac{B}{t+2}$

$ \Rightarrow \quad t=A(t+2)+B(t+1) $

When, $t+2=0 \Rightarrow t=-2$

$ \therefore \quad-2=B(-1) \Rightarrow B=2 $

when, $t+1=0 \Rightarrow t=-1$

$ \begin{aligned} & \therefore \quad-1=A(1)+B(0) \Rightarrow A=-1 \\ & \therefore \frac{t}{(t+1) t+2)}=\frac{-1}{t+1}+\frac{2}{t+2} \\ & \therefore I=\frac{1}{2} \int\left(\frac{-1}{t+1}+\frac{2}{t+2}\right) d t \end{aligned} $

$ \begin{aligned} I & =\frac{1}{2}[-\log (t+1)+2 \log (t+2)]+C \\ & \left.=-\frac{1}{2} \log \left(x^2+1\right)+\log \left(x^2+2\right)\right]+C \\ & =\log \left(x^2+2\right)+\log \left(x^2+1\right)^{-1 / 2}+C \\ & =\log \left(\frac{x^2+2}{\sqrt{x^2+1}}\right)+C \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{d x}{\left(x^2+9\right) \sqrt{x^2+16}} \\ & \text { put } x=4 \tan \theta \Rightarrow d x=4 \sec ^2 \theta d \theta \\ & \begin{aligned} \therefore I & =\int \frac{4 \sec ^2 \theta d \theta}{\left(16 \tan ^2 \theta+9\right) \sqrt{16 \tan ^2 \theta+16}} \\ & =\int \frac{4 \sec ^2 \theta d \theta}{\left(16 \tan ^2 \theta+9\right) 4 \sec \theta} \\ & =\int \frac{\sec \theta d \theta}{16 \tan ^2 \theta+9} \end{aligned} \end{aligned} $

$ \begin{aligned} & \quad=\int \frac{\cos \theta d \theta}{16 \sin ^2 \theta+9 \cos ^2 \theta}=\int \frac{\cos \theta d \theta}{7 \sin ^2 \theta+9} \\ & \text { Again put } \sin \theta=t \\ & \Rightarrow \cos \theta d \theta=d t \\ & \therefore I=\int \frac{d t}{7 t^2+9} \\ & I=\int \frac{d t}{7\left(t^2+\frac{9}{7}\right)}=\frac{1}{7} \int \frac{d t}{t^2+\left(\frac{3}{\sqrt{7}}\right)^2} \\ & \quad=\frac{1}{7} \cdot \frac{1}{\frac{3}{\sqrt{7}}} \tan ^{-1}\left(\frac{t}{\frac{3}{\sqrt{7}}}\right)+C \\ & \quad=\frac{1}{3 \sqrt{7}} \tan ^{-1}\left(\frac{\sqrt{7} t}{3}\right)+C \\ & \quad=\frac{1}{3 \sqrt{7}} \tan ^{-1}\left(\frac{\sqrt{7} \sin \theta}{3}\right)+C \\ & \quad=\frac{1}{3 \sqrt{7}} \tan ^{-1}\left(\frac{\sqrt{7} x}{3 \sqrt{x^2+16}}\right)+C \end{aligned} $

$ \begin{aligned} & \left(\because x=4 \tan \theta \Rightarrow \sin \theta=\frac{x}{\sqrt{x^2+16}}\right) \\ & \therefore \quad K=\frac{\sqrt{7}}{3} \end{aligned} $

$I=\int \frac{2 \sin x-3 \cos x}{4 \cos x-3 \sin x} d x$

Let $t=4 \cos x-3 \sin x$

$ \Rightarrow \frac{d t}{d x}=-4 \sin x-3 \cos x $

Now, $2 \sin x-3 \cos x$

$ \begin{aligned} & \Rightarrow A(-4 \sin x-3 \cos x)+B(4 \cos x-3 \sin x) \\ & \Rightarrow(-4 A-3 B) \sin x+(-3 A+4 B) \cos x \end{aligned} $

Comparing the coefficients, we get

$ -4 A-3 B=2 \text { and }-3 A+4 B=-3 $

Solving these two equations, we get

$ A=\frac{1}{25} \text { and } B=\frac{-18}{25} $

$ \begin{aligned} & \text { So, } I=\int \frac{2 \sin x-3 \cos x}{4 \cos x-3 \sin x} d x \\ & \begin{array}{r} \Rightarrow \frac{1}{25} \int \frac{-4 \sin x-3 \cos x}{4 \cos x-3 \sin x} d x +\left(\frac{-18}{25}\right) \int \frac{4 \cos x-3 \sin x}{4 \cos x-3 \sin x} d x \end{array} \\ & \Rightarrow \frac{1}{25} \int \frac{d t}{t}-\frac{18}{25} \int d x \Rightarrow \frac{1}{25} \log |t|-\frac{18}{25} x+C \\ & \Rightarrow \frac{1}{25} \log |4 \cos x-3 \sin x|-\frac{18}{25} x+C \\ & =\frac{1}{25}[\log |4 \cos x-3 \sin x|-18 x]+C \end{aligned} $

$ \text { } \begin{aligned} \int e^{4 x}(\sin 3 x-\cos 3 x) d x \\ =\int e^{4 x} \sin 3 x d x-\int e^{4 x} \cos 3 x d x \end{aligned} $

Using the standard integrals

$ \begin{aligned} & \int e^{a x} \sin b x d x=\frac{e^{a x}}{a^2+b^2}(a \sin b x-b \cos b x) \\ & \int e^{a x} \cos b x d x=\frac{e^{a x}}{a^2+b^2}(a \cos b x+b \sin b x) \end{aligned} $

Here, $a=4, b=3$

$ \begin{aligned} & \text { So, } \int e^{4 x}(\sin 3 x-\cos 3 x) d x \\ & \qquad \begin{aligned} = & \frac{e^{4 x}}{4^2+3^2}(4 \sin 3 x-3 \cos 3 x) -\frac{e^{4 x}}{4^2+3^2}(4 \cos 3 x+3 \sin 3 x)+C \\ \Rightarrow \quad & \frac{e^{4 x}}{25}[4 \sin 3 x-3 \cos 3 x -4 \cos 3 x-3 \sin 3 x]+C \\ \Rightarrow \quad & \frac{e^{4 x}}{25}[\sin 3 x-7 \cos 3 x]+C \end{aligned} \end{aligned} $

$I=\int\left(\frac{1-\log x}{1+(\log x)^2}\right)^2 d x$

Let $y=\log x$. Then, $x=e^y \Rightarrow d x=e^y \cdot d y$

So, $I=\int\left(\frac{1-y}{1+y^2}\right)^2 e^y d y$

$ \begin{aligned} = & \int\left(\frac{1+y^2-2 y}{\left(1+y^2\right)^2}\right) e^y d y \\ \Rightarrow \quad & \int\left(\frac{1}{1+y^2}-\frac{2 y}{\left(1+y^2\right)^2}\right) e^y d y \end{aligned} $

Since, $\int\left(f(y)+f^{\prime}(y)\right) e^y d y=f(y) e^y+C$

Here, $f(y)=\frac{1}{1+y^2}$ and $f^{\prime}(y)=\frac{-2 y}{\left(1+y^2\right)^2}$

$ \begin{aligned} \therefore \quad I & =f(y) e^y+C=\frac{1}{1+y^2} e^y+C \\ & =\frac{1}{1+(\log x)^2} \cdot e^{\log x}+C \\ & =\frac{x}{1+(\log x)^2}+C \end{aligned} $

$ \begin{aligned} &\text { Given, } \int(x+2) \sqrt{x^2-x+2} d x\\ &\Rightarrow \quad \frac{1}{3} f(x)+\frac{5}{8} g(x)+\frac{35}{16} h(x)+C \end{aligned} $

Let $x+2=A(2 x-1)+B$

Comparing coefficients, we get

$ \begin{aligned} & \quad 2 A=1 \\ & \Rightarrow \quad A=\frac{1}{2},-A+B=2 \\ & \Rightarrow \quad B=2+A=2+\frac{1}{2}=\frac{5}{2} \\ & \text { So, } x+2=\frac{1}{2}(2 x-1)+\frac{5}{2} \\ & \int(x+2) \sqrt{x^2-x+2} d x \\ & =\int\left(\frac{1}{2}(2 x-1)+\frac{5}{2}\right) \sqrt{x^2-x+2} d x \\ & =\frac{1}{2} \int(2 x-1) \sqrt{x^2-x+2} d x +\frac{5}{2} \int \sqrt{x^2-x+2} d x \end{aligned} $

$ \begin{aligned} & \text { Let } u=x^2-x+2 \text {, then } d u=(2 x-1) d x \\ & =\frac{1}{2} \int \sqrt{u} d u+\frac{5}{2} \int \sqrt{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2} d x \\ & =\frac{1}{2} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}}+\frac{5}{2}\left[\frac{2 x-1}{4} \sqrt{x^2-x+2}\right.\left.+\frac{7}{8} \ln \left|x-\frac{1}{2}+\sqrt{x^2-x+2}\right|\right]+c \\ & =\frac{1}{3}\left(x^2-x+2\right)^{\frac{3}{2}}+\frac{5(2 x-1)}{8} \sqrt{x^2-x+2} +\frac{35}{16} \ln \left|x-\frac{1}{2}+\sqrt{x^2-x+2}\right|+c \end{aligned} $

Comparing with original equation, we get

$ \begin{aligned} & f(x)=\left(x^2-x+2\right)^{\frac{3}{2}} \\ & g(x)=(2 x-1) \sqrt{x^2-x+2} \\ & h(x)=\ln \left|x-\frac{1}{2}+\sqrt{x^2-x+2}\right| \end{aligned} $

So, $f(-1)=\left[(-1)^2-(-1)+2\right]^{\frac{3}{2}}$

$ \begin{aligned} &=[1+1+2]^{\frac{3}{2}}=4^{\frac{3}{2}}=8 \\ & g(-1)=(2(-1)-1) \sqrt{(-1)^2-(-1)+2} \\ &=-3 \sqrt{1+1+2}=-3 \sqrt{4} \\ &=-3 \times 2=-6 \\ & h\left(\frac{1}{2}\right)=\ln \left|\frac{1}{2}-\frac{1}{2}+\sqrt{\left(\frac{1}{2}\right)^2-\frac{1}{2}+2}\right| \\ & \Rightarrow \ln \mid 0 \left.+\sqrt{\frac{1}{4}-\frac{1}{2}+2} \right\rvert\, \\ & \Rightarrow \ln \left|\sqrt{\frac{1-2+8}{4}}\right| \Rightarrow \ln \left|\sqrt{\frac{7}{4}}\right|=\ln \left(\frac{\sqrt{7}}{2}\right) \end{aligned} $

$ \begin{aligned} &\text { So, } f(-1)+g(-1)+h\left(\frac{1}{2}\right)=8-6+\ln \left(\frac{\sqrt{7}}{2}\right)\\ &= 2+\ln \left(\frac{\sqrt{7}}{2}\right) \end{aligned} $

Let $ u=\tan \left(\frac{x}{2}\right) \Rightarrow d u=\frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) d x $

$ \Rightarrow $ Using the substitution $ \sin x=\frac{2 u}{u^{2}+1} $

and $ \cos x=\frac{1-u^{2}}{u^{2}+1} $ and $ d x=2 \frac{d u}{u^{2}+1} $

$ \Rightarrow \int \frac{2 d u}{\left(1-u^{2}\right)\left[3\left(\frac{u^{2}+1}{1-u^{2}}-\frac{2 u}{u^{2}-1}\right)+2\right]} $

then we get;

$ \int \frac{2}{u^{2}+6 u+5} d u=2 \int \frac{1}{(u+3)^{2}-4} d u $

Substitute $ s=u+3 $ and $ d s=d u $

$ =2 \int \frac{1}{s^{2}-4} d s=2 \int \frac{-1}{4\left(1-\frac{s^{2}}{4}\right)} $

$ \begin{array}{l} =\frac{-1}{2} \int \frac{1}{1-\frac{s^{2}}{4}} d s \Rightarrow p=s / 2 \\\\ d p=\frac{d s}{2}=-\int \frac{1}{1-p^{2}} d p=-\tan h^{-1}(p)+C \end{array} $

$ \Rightarrow-\tan h^{-1}\left(\frac{u+3}{2}\right)+C $

$ \Rightarrow-\tan h^{-1}\left[\left(\frac{1}{2}\left(\tan \left(\frac{x}{2}\right)+3\right)\right)\right]+C $

$ =\frac{1}{2}\left[\log \sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right] -\log \left[\sin \left(\frac{x}{2}\right)+5 \cos \left(\frac{x}{2}\right)\right]+C $

$ =\frac{1}{2} \log \left[\frac{\sin x / 2+\cos x / 2}{\sin x /+5 \cos x / 2}\right]+C $

$ =\frac{1}{2} \log \left[\frac{\tan x / 2+1}{\tan x / 2+5}\right]+C $

$ I=\int \frac{d x}{4+3 \cot x}=\int \frac{\sin x}{4 \sin x+3 \cos x} d x $

$ \sin x=1(4 \sin x+3 \cos x) +m(4 \cos x-3 \sin x) $

$ 1=4 l-3 m $

$ 0=3 l+4 m \quad 4 l-3\left(-\frac{3}{4} l\right)=1 $

$ 4 m=-3 l \quad 4 l+\frac{9 l}{4}=1 $

$ m=-\frac{3}{4} l \quad \frac{25}{4} l=1 $

$ l=\frac{4}{25} $

$ m=-\frac{3}{4} \times \frac{4}{25}=-\frac{3}{25} $

$ \begin{aligned} I & =\frac{4}{25} \int d x+\left(-\frac{3}{25}\right) \int \frac{4 \cos x-3 \sin x}{4 \sin x+3 \cos x} d x \\\\ & =\frac{4 x}{25}-\frac{3}{25} \ln (4 \sin x+3 \cos x)+C \end{aligned} $

$ \int \frac{d x}{(x+1) \sqrt{x^{2}+4}} $

Put, $ x+1=\frac{1}{t} \Rightarrow d x=-\frac{1}{t^{2}} d t $

$ =\int \frac{-d t}{\sqrt{1-2 t+5 t^{2}}} $

$ \begin{array}{l} =\int \frac{-\frac{1}{t^{2}} d t}{\frac{1}{t} \sqrt{\left(\frac{1}{t}-1\right)^{2}+4}} \\\\ =\int \frac{-\frac{1}{t^{2}} d t}{\frac{1}{t} \sqrt{\frac{1}{t^{2}}+1-\frac{2}{t}+4}}\end{array} $

$ =-\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{t^{2}-\frac{2 t}{5}+\frac{1}{5}}} $

$ =-\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{t^{2}-\frac{2 t}{5}+\frac{1}{5}+\frac{1}{25}-\frac{1}{25}}} $

$ =-\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{\left(t-\frac{1}{5}\right)^{2}+\frac{4}{25}}} $

$ =-\frac{1}{\sqrt{5}} \int \frac{d t}{\sqrt{\left(t-\frac{1}{5}\right)^{2}+\left(\frac{2}{5}\right)^{2}}} $

$ =-\frac{1}{\sqrt{5}} \sinh ^{-1}\left(\frac{t-(1 / 5)}{2 / 5}\right) $

$ =-\frac{1}{\sqrt{5}} \sinh ^{-1}\left(\frac{5 t-1}{2}\right) $

$ =-\frac{1}{\sqrt{5}} \sinh ^{-1}\left(\frac{\frac{5}{x+1}-1}{2}\right) $

$ =-\frac{1}{\sqrt{5}} \sinh ^{-1}\left(\frac{5-x-1}{2(x+1)}\right) $

$ =-\frac{1}{\sqrt{5}} \sinh ^{-1}\left(\frac{4-x}{2(x+1)}\right)+c $

$ \int e^{x}\left(x^{3}+x^{2}-x+4\right) d x $

$ \begin{aligned} = & e^{x}\left[x^{3}+x^{2}-x+4-3 x^{2}-2 x+1\right. +6 x+2-6]+C \\\\ = & e^{x}\left[x^{3}-2 x^{2}+3 x+1\right]+C \\\\ \therefore & f(x)=x^{3}-2 x^{2}+3 x+1 \\\\ & f(1)=1-2+3+1=5-2=3 \end{aligned} $

$ \begin{array}{l} \text { We know that } \\ x^{4}+x^{2}+1=\left(x^{2}+x+1\right)\left(x^{2}-x+1\right) \\ \{\therefore a=1\} \\ \frac{1}{x^{4}+x^{2}+1}=\frac{A x+B}{x^{2}+x+1}+\frac{C x+D}{x^{2}-x+1} \\ \frac{1}{\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)} \\ =\frac{(A x+B)\left(x^{2}-x+1\right)+(C x+D)\left(x^{2}+x+1\right)}{\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)} \\ 1=(A x+B)\left(x^{2}-x+1\right)+(C x+D)\left(x^{2}+x+1\right) \\ 1=A x^{3}-A x^{2}+A x+B x^{2}-B x+B \\ \quad+C x^{3}+C x^{2}+C x+D x^{2}+D x+D \end{array} $

On comparing, we get

Coefficient of $ x^{3}=0 A+C=0 $

Coefficient of $ x^{2}=0 $

$ -A+B+C+D=0 $

Coefficient of $ x=0 $

$ \begin{array}{l} A-B+C=0 \\ B+D=1 \end{array} $

From Eqs. (i) and (iv), we get

$ A-C=1 $

and from Eqs. (i) and (v) we get,

$ A=\frac{1}{2}, C=-\frac{1}{2} $

So, $ A+B-C+D=1=2 a $

$ \int \frac{d x}{x^{4}+8 x^{2}+9}=\int \frac{d x}{\left(x^{2}+4-\sqrt{7}\right)\left(x^{2}+4+\right.} $

Use partial fraction,

$ \Rightarrow \int \frac{d x}{\left(x^{2}+4-\sqrt{7}\right)\left(x^{2}+4+\sqrt{7}\right)}=\frac{1}{2 \sqrt{7}} $

$ \int \frac{\left(x^{2}+4+\sqrt{7}\right)-\left(x^{2}+4-\sqrt{7}\right)}{\left(x^{2}+4-\sqrt{7}\right)\left(x^{2}+4+\sqrt{7}\right)} d x $

$ =\frac{1}{2 \sqrt{7}}\left[\int \frac{d x}{x^{2}+4-\sqrt{7}}-\int \frac{d x}{x^{2}+4+\sqrt{7}}\right] $

$ =\frac{1}{2 \sqrt{7}}\left[\frac{1}{\sqrt{4-\sqrt{7}}} \tan ^{-1}\left(\frac{x}{\sqrt{4-\sqrt{7}}}\right)\right. $

$ \left.-\frac{1}{\sqrt{4+\sqrt{7}}} \tan ^{-1}\left(\frac{x}{\sqrt{4+\sqrt{7}}}\right)\right]+C $

There is no comparison hence, no options are correct.

To find $ f(1) - f(-1) $, we start by evaluating the given integral:

$ \int \left(1 + x - \frac{1}{x}\right) e^{x + \frac{1}{x}} \, dx $

This can be expanded as:

$ \int e^{x + \frac{1}{x}} \, dx + \int x \left(1 - \frac{1}{x^{2}}\right) e^{x + \frac{1}{x}} \, dx $

Integrating by parts, we have:

$ = x e^{x + \frac{1}{x}} - \int x \left(1 - \frac{1}{x^{2}}\right) e^{x + \frac{1}{x}} \, dx + \int x \left(1 - \frac{1}{x}\right) e^{x + \frac{1}{x}} \, dx $

This simplifies to:

$ = x e^{x + \frac{1}{x}} + C $

Thus, the function $ f(x) $ is given by:

$ f(x) = x e^{x + \frac{1}{x}} $

To find $ f(1) $ and $ f(-1) $:

$ f(1) = 1 \cdot e^{1 + \frac{1}{1}} = e^{2} $

$ f(-1) = -1 \cdot e^{-1 + \frac{1}{-1}} = -1 \cdot e^{-2} = -e^{-2} $

Therefore, the difference is:

$ f(1) - f(-1) = e^{2} - (-e^{-2}) = e^{2} + e^{-2} $

To evaluate the integral $ \int \frac{1}{x^{m} \sqrt[m]{x^{m}+1}} \, dx $, we can use the substitution method.

First, rewrite the integral as:

$ \int \frac{1}{x^{m+1} \sqrt[m]{1+\frac{1}{x^{m}}}} \, dx $

Now, let:

$ 1 + \frac{1}{x^m} = t $

Then, differentiate both sides with respect to $ x $:

$ -\frac{m}{x^{m+1}} \, dx = dt $

This implies:

$ \frac{dx}{x^{m+1}} = -\frac{1}{m} \, dt $

Substituting in the integral gives:

$ \int -\frac{1}{m} t^{-\frac{1}{m}} \, dt $

Integrating with respect to $ t $:

$ -\frac{1}{m} \cdot \frac{t^{-\frac{1}{m} + 1}}{-\frac{1}{m} + 1} + C $

Simplify the expression:

$ -\frac{1}{m-1} \left( t^{\frac{m-1}{m}} \right) + C $

Substitute back the expression for $ t $:

$ -\frac{1}{m-1} \left( \frac{1+x^m}{x^m} \right)^{\frac{m-1}{m}} + C $

Hence, the solution is:

$ -\frac{1}{m-1} \left( \frac{\sqrt[m]{x^m+1}}{x} \right)^{m-1} + C $

$ \begin{array}{l} \int \sqrt{(\operatorname{cosec} x+1)} d x=k \tan ^{-1}(f(x))+c \\\\ \Rightarrow \int \sqrt{\frac{1+\sin x}{\sin x}} d x \\\\ \int \sqrt{\frac{(1+\sin x)(1-\sin x)}{\sin x(1-\sin x)}} d x \\\\ \int \sqrt{\frac{1-\sin ^{2} x}{\sin x(1-\sin x)}} d x \\\\ \int \frac{\cos x d x}{\sqrt{\sin x(1-\sin x)}} \end{array} $

Let $ \sin x=t $

$ \cos x d x=d t $

$ \Rightarrow \quad \int \frac{d t}{\sqrt{t-t^{2}}} $

$ \int \frac{d t}{\sqrt{\frac{1}{4}-\left(\frac{1}{2}-t\right)^{2}}} $

Let $ \frac{1}{2}-t=u \Rightarrow-d t=d u $

then, $ 2 \int \frac{-d u}{\sqrt{1-(2 \mu)^{2}}} $

$ \begin{array}{l} =-2 \frac{1}{2}\left(\sin ^{-1}(4 t)\right)+C \\\\ =-\sin ^{-1}(1-2 t)+C \\\\ =-\sin ^{-1}(1-2 \sin x)+C \\\\ =-\tan ^{-1}\left[\frac{1-2 \sin x}{2 \sqrt{\sin ^{2} x-\sin x}}\right]+C \end{array} $

On comparing, we get

$ \begin{array}{l} k=-1 \text { and } f(x)=\frac{1-2 \sin x}{2 \sqrt{\sin ^{2} x-\sin x}} \\\\ k f\left(\frac{\pi}{6}\right)=-1 \times\left[\frac{1-2 \times \frac{1}{2}}{2 \sqrt{\frac{1}{4}-\frac{1}{2}}}\right]=0 \end{array} $

$ \therefore $ No option is matching,

Given,

$ \begin{aligned} I & =\int \sqrt{4 \cos ^{2} x-5 \sin ^{2} x} \cos x d x \\\\ & =\int \sqrt{4-9 \sin ^{2} x} \cos x d x \end{aligned} $

Let $ \sin x=t \Rightarrow \cos x d x=d t $

$ \begin{array}{l} I=\int \sqrt{4-9 t^{2}} d t=3 \int \sqrt{\left(\frac{2}{3}\right)^{2}-t^{2} d t} \\\\ \frac{1}{2} \sin x \sqrt{4-9 \sin ^{2} x}+\frac{2}{2} \sin ^{-1}\left(\frac{3 \sin x}{2}\right)+C \end{array} $

Given,

$ I=\int\left(\frac{4 \tan ^{4} x+3 \tan ^{2} x-1}{\tan ^{2} x+4}\right) d x $

$ =\int\left(\frac{4 \tan ^{4} x+4 \tan ^{3} x-\tan ^{2} x-1}{\tan ^{2} x+4}\right) d x $

$ =\int \frac{\left(4 \tan ^{2} x-1\right)\left(\tan ^{2} x+1\right)}{\left(\tan ^{2} x+4\right)} d x $

Also, let $ \tan x=t \Rightarrow \sec ^{2} x d x=d t $

$ \begin{array}{l} \Rightarrow\left(1+\tan ^{2} x\right) d x=d t \\\\ \therefore \quad I=\int \frac{\left(4 t^{2}-1\right)}{t^{2}+4} d t \\\\ I=4 \int\left[\frac{\left(t^{2}+4\right)}{\left(t^{2}+4\right)}-\frac{17}{4} \times \frac{1}{\left(t^{2}+4\right)}\right] d t \\\\ I=4 \int\left(1-\frac{17}{4\left(t^{2}+4\right)}\right) d t \\\\ I=4\left[t-\frac{17}{4} \times \frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C\right] \\\\ I=4 \tan x-\frac{17}{2} \tan ^{-1}\left(\frac{\tan x}{2}\right)+C \end{array} $

Given, $ \int \frac{\left(\sin ^{4} x+2 \cos ^{2} x-1\right) \cos x}{(1+\sin x)^{6}} d x $

Let $ \sin x=t \Rightarrow \cos x d x=d t $

then, $ \int \frac{\left[t^{4}+2\left(1-t^{2}\right)-1\right]}{(1+t)^{6}} d t $

$ \begin{array}{ll} \because \quad & t^{4}+2-2 t^{2}-1 \\\\ \Rightarrow \quad & t^{4}-2 t^{2}+1=\left(t^{2}-1\right)^{2} \\\\ & \int \frac{\left(t^{2}-1\right)^{2}}{(1+t)^{6}} d t \\\\ & \int \frac{(t-1)^{2}(t+1)^{2}}{(1+t)^{6}} d t \end{array} $

$ \int \frac{(t-1)^{2}}{(1+t)^{4}} d t $

Let $ 1+t=\mu, d t=d \mu $

$ \begin{array}{l} \int \frac{(\mu-1-1)^{2}}{\mu^{4}} d \mu \\\\ \int\left(\frac{\mu^{2}}{\mu^{4}}+\frac{4}{\mu^{4}}-\frac{4 \mu}{\mu^{4}}\right) d \mu \end{array} $

$ \int \frac{1}{\mu^{2}} d \mu+4 \int \frac{1}{\mu^{4}} d \mu-4 \int \frac{1}{\mu^{3}} d \mu $

$ =\frac{\mu^{-2+1}}{-2+1}+\frac{4 \mu^{-4}}{-4+1}-\frac{4 \mu^{-3+1}}{-3+1}+C $

$ =\frac{-1}{\mu}-\frac{4}{3 \mu^{3}}+\frac{2}{\mu^{2}}+C $

$ =\frac{-3 \mu^{2}-4+6 \mu}{3 \mu^{3}}+C $

Put $ x=1+\sin x $

$ (\because t=\sin x) $

$ =\frac{-3(1+\sin x)^{2}-4+6(1+\sin x)}{3(1+\sin x)^{3}}+C $

$ =\frac{-3-3 \sin ^{2} x-6 \sin x-4+6+6 \sin x}{3(1+\sin x)^{3}}+C $

$ =\frac{-3 \sin ^{2} x-1}{3(1+\sin x)^{3}}+C=\frac{-\left(3 \sin ^{2} x+1\right)}{3(\sin x+1)^{2}}+C $

$ \begin{aligned} & \text { Given, } \frac{3 x^4-2 x^2+1}{(x-2)^4} \\\\ & =A+\frac{B}{x-2}+\frac{C}{(x-2)^2}+\frac{D}{(x-2)^3}+\frac{E}{(x-2)^4} \\\\ & \Rightarrow 3 x^4-2 x^2+1=A(x-2)^4 \\\\ & \quad+B(x-2)^3+C(x-2)^2+D(x-2)+E \\\\ & \Rightarrow 3 x^4-2 x^2+1=A x^4+(-8 A+B) x^3 \\\\ & \quad+(24 A-6 B+C) x^2 \\\\ & \quad+(-32 A+12 B-4 C+D) x \\\\ &\quad+(16 A-8 B+4 C-2 D+E) \end{aligned} $

On comparing coefficients, we get

$ \begin{array}{ll} & A=3,-8 A+B=0 \\\\ \Rightarrow \quad & B=24 \\\\ & 24 A-6 B+C=-2 \\\\ \Rightarrow \quad & C=70,-32 A+12 B-4 C+D=0 \\\\ \Rightarrow \quad & D=88 \\\\ & 16 A-8 B+4 C-2 D+E=1 \\\\ \Rightarrow \quad & E=41 \\\\ \therefore \quad & 2 A+3 B-C-D+E \end{array} $

$\begin{aligned} & =2(3)+3(24)-70-88+41 \\\\ & =(6+72+41)-(70+88) \\\\ & =119-158=-39\end{aligned}$

Let

$ \begin{aligned} & I=\int e^{-2 x}\left(\tan 2 x-2 \sec ^2 2 x \tan 2 x\right) d x \\\\ & \text { Put }-2 x=t \\\\ & \Rightarrow \quad d x=-\frac{d t}{2} \end{aligned} $

$ \begin{aligned} \therefore I= & \int e^t\left(\tan (-t)-2 \sec ^2(-t) \tan (-t)\right)\left(\begin{array}{r} -\frac{dt}{2}\end{array}\right) \\\\ = & -\frac{1}{2} \int e^t\left(2 \sec ^2 t \tan t-\tan t\right) d t \\\\ = & -\frac{1}{2} \int e^t\left[\left(\sec ^2 t-\tan t\right)\right. \\\\ & \left.\quad+\left(2 \sec ^2 t \tan t-\sec ^2 t\right)\right] d t \end{aligned} $

We know that,

$ \begin{aligned} & \quad \int e^t f(t)+f^{\prime}(t) d t=e^t f(t)+C \\\\ & \text { Here, } \quad f(t)=\sec ^2 t-\tan t \\\\ & \text { and } \quad f^{\prime}(t)=2 \sec ^2 t \tan t-\sec ^2 t \\\\ & \therefore \quad I=-\frac{1}{2} e^t\left[\sec ^2 t-\tan t\right]+C \\\\ & =-\frac{1}{2} e^{-2 x} \cdot\left[\sec ^2(-2 x)-\tan (-2 x)\right]+C \end{aligned} $

$=-\frac{e^{-2 x}}{2}\left[\sec ^2 2 x+\tan 2 x\right]+c$

$ \begin{aligned} &\text {Given, } \int x^3 \sin 3 x d x\\\\ &\begin{aligned} & \quad=f(x) \cos 3 x+g(x) \sin 3 x+C \\\\ & \text { Let, } I=\int x^3 \sin 3 x d x \\\\ & =x^3 \int \sin 3 x d x-\int\left(3 x^2 x \int \sin 3 x d x\right) d x \\\\ & =x^3\left(\frac{-\cos 3 x}{3}\right)-\int\left(3 x^2 \times\left(\frac{-\cos 3 x}{3}\right)\right) d x \\\\ & =-\frac{x^3}{3} \cos 3 x+\int x^2 \cos 3 x d x \\\\ & =-\frac{x^3}{3} \cos 3 x+\left[x^2\left(\frac{\sin 3 x}{3}\right)-\int\left(2 x \times \frac{\sin 3 x}{3}\right) d x]\right. \\\\ & =-\frac{x^3}{3} \cos 3 x+\frac{x^2}{3} \sin 3 x-\frac{2}{3} \int x \sin 3 x d x \\\\ & =-\frac{x^3}{3} \cos 3 x+\frac{x^2}{3} \sin 3 x-\frac{2}{3} \\\\ & \quad\left[\frac{-x}{3} \cos 3 x+\frac{\sin 3 x}{9}\right]+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\left(-\frac{x^3}{3}+\frac{2 x}{9}\right) \cos 3 x+\left(\frac{x^2}{3}-\frac{2}{27}\right) \sin 3 x+6 \\\\ & \therefore f(x)=-\frac{x^3}{3}+\frac{2 x}{9} \text { and } g(x)=\frac{x^2}{3}-\frac{2}{27} \end{aligned}\\\\ &\text { Now, } 27(f(x)+x g(x))\\\\ &\begin{aligned} & =27\left[\left(-\frac{x^3}{3}+\frac{2 x}{9}\right)+x\left(\frac{x^2}{3}-\frac{2}{27}\right)\right] \\\\ & =27\left[\frac{2 x}{9}-\frac{2}{27} x\right]=6 x-2 x=4 x \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{d x}{9 \cos ^2 2 x+16 \sin ^2 2 x} \\\\ & =\int \frac{\sec ^2 2 x d x}{9+16 \tan ^2 2 x}=\frac{1}{9} \int \frac{\sec ^2 2 x d x}{1+\left(\frac{4}{3} \tan 2 x\right)^2} \\\\ & =\frac{1}{24} \int \frac{\frac{4}{3} \cdot\left(2 \sec ^2 2 x\right)}{1+\left(\frac{4}{3} \tan 2 x\right)^2} \end{aligned} $

$ \begin{aligned} & =\frac{1}{24} \tan ^{-1}\left(\frac{4}{3} \tan 2 x\right)+C \\\\ & {\left[\because \int \frac{f^{\prime}(x) d x}{1+[f(x)]^2}=\tan ^{-1}(f(x))+C\right]} \end{aligned} $

Let,

$ I=\int \frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x} d x $

Put $3 x=t \Rightarrow 3 d x=d t$

$ \therefore \quad I=\int \frac{2 \cos t-3 \sin t}{\cos t+2 \sin t}\left(\frac{d t}{3}\right) $

Let, $2 \cos t-3 \sin t=A(\cos t+2 \sin t)$

$ +B\left[\frac{d}{d t}(\cos t+2 \sin t)\right] $

$ \begin{aligned} & \Rightarrow 2 \cos t-3 \sin t=A(\cos t+2 \sin t) \\\\ & \quad+B(-\sin t+2 \cos t) \\\\ & \Rightarrow 2=A+2 B \text { and }-3=2 A-B \end{aligned} $

On solving these equations, we get

$ A=-\frac{4}{3} \text { and } B=\frac{7}{5} $

$ \therefore I=\frac{1}{3}\left[\int \frac{-\frac{4}{5}(\cos t+2 \sin t)}{+\frac{7}{5}(-\sin t+2 \cos t)}(\cos t+2 \sin t] d t\right] $

$ \begin{aligned} & =\frac{1}{3}\left[-\frac{4}{5} t+\frac{7}{5} \log |\cos t+2 \sin t|\right]+C \\\\ & =\frac{7}{15} \log |\cos 3 x+2 \sin 3 x|-\frac{4}{5} x+C \\\\ & {[\because t=3 x]} \end{aligned} $

We have,

$ \begin{aligned} & \frac{x^4}{\left(x^2+1\right)(x-2)}=f(x)+\frac{A x+B}{x^2+1}+\frac{C}{x-2} \\\\ & \frac{x^4}{\left(x^2+1\right)(x-2)} \\\\ & \quad f(x)\left[\left(x^2+1\right)(x-2)\right] \\\\ & =\frac{+(A x+B)(x-2)+C\left(x^2+1\right)}{\left(x^2+1\right)(x-2)} \\\\ & \Rightarrow \frac{x^4}{\left(x^2+1\right)(x-2)} \end{aligned} $

$ \begin{aligned} & f(x)\left[x^3-2 x^2+x-2\right] \\\\ = & \frac{+\left[A x^2+(B-2 A) x-2 B\right]+C x^2+C}{\left(x^2+1\right)(x-2)} \end{aligned} $

Here, $f(x)$ will be linear function only as highest power in LHS is 4

So, let $f(x)=p x+q$

$ \frac{x^4}{\left(x^2+1\right)(x-2)} $

$\begin{aligned} &(p x+q)\left[x^3-2 x^2+x-2\right] \\\\ &= \frac{+\left[A x^2+(B-2 A) x-2 B+c x^2+C\right]}{\left(x^2+1\right)(x-2)} \\\\ & \Rightarrow \frac{x^4}{\left(x^2+1\right)(x-2)} \\\\ & p x^4+(q-2 p) x^3+(p-2 q+A+C) x^2 \\\\ &= \frac{+(q-2 p+B-2 A) x+(C-2 B-2 q)}{\left(x^2+1\right)(x-2)}\end{aligned}$

On comparing both sides, we get

$ \begin{aligned} & p=1, q-2 p=0 \Rightarrow q=2 \\\\ & p-2 q+A+C=0 \Rightarrow A+C=3 \\\\ & q-2 p+B-2 A=0 \Rightarrow B=2 A \\\\ & C-2 B-2 q=0 \Rightarrow C-2 B=4 \end{aligned} $

From Eqs. (i), (ii) and (iii), we get,

$ A=\frac{-1}{5}, C=\frac{16}{5}, B=\frac{-2}{5}, f(x)=x+2 $

So, $f(14)+2 A-B=14+2+0=16=5 C$

We have,

$ \int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) d x=f(x)+C $

Let

$ \begin{array}{l} I=\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) d x \\\\ =\int\left[\left|\sin \frac{x}{2}-\cos \frac{x}{2}\right|+\left|\sin \frac{x}{2}+\cos \frac{x}{2}\right|\right] d x \\\\ =\int\left[\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)-\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)\right] d x \\\\ \qquad\left[\because \frac{5 \pi}{2} < x < \frac{7 \pi}{2} \Rightarrow \frac{5 \pi}{4} < \frac{x}{2} < \frac{7 \pi}{4}\right] \\\\ =\int\left(\cos \frac{x}{2}-\sin \frac{x}{2}-\sin \frac{x}{2}-\cos \frac{x}{2}\right) d x \\\\ =-2 \int \sin \frac{x}{2} d x=4 \cos \frac{x}{2}+C \end{array} $

So, $ f(x)=4 \cos \frac{x}{2} \Rightarrow f^{\prime}(x)=-2 \sin \frac{x}{2} $

$ f^{\prime}\left(\frac{8 \pi}{3}\right)=-2 \sin \frac{4 \pi}{3}=-2 \times \frac{-\sqrt{3}}{2}=\sqrt{3} $

Let $\begin{aligned} & =\int \frac{\left(1-4 \sin ^2 x\right) \cos x}{\cos (3 x+2)} d x \\\\ & =\int \frac{\left(1-4+4 \cos ^2 x\right) \cos x}{\cos (3 x+2)} d x\end{aligned}$

$ =\int \frac{4 \cos ^2 x-3 \cos x}{\cos (3 x+2)} d x=\int \frac{\cos 3 x}{\cos (3 x+2)} d x $

Let $3 x+2=t \Rightarrow x d x=\frac{d t}{3}$

$ \begin{aligned} & I=\int \frac{\cos (t-2)}{\cos t} \frac{d t}{3} \\\\ & =\frac{1}{3} \int \frac{\cos 2 \cos t+\sin 2 \sin t}{\cos (t)} d t \\\\ & =\frac{1}{3} \int(\cos 2+\sin 2 \tan t) d t \\\\ & =\frac{1}{3}[(\cos 2) \cdot t+\sin 2 \ln (\sec t)+K] \\\\ & =\frac{1}{3}[\cos 2 \cdot(3 x+2)+\sin 2 \cdot \ln [\sec (3 x+2)]+K \end{aligned} $

$\begin{aligned}= & (\cos 2) x+\frac{2}{3} \cos 2 +\frac{1}{3} \sin 2 \cdot \ln [\sec (3 x+2)]+\frac{K}{3} \end{aligned}$

$=(\cos 2) x+\frac{1}{3} \sin 2 \ln [\sec (3 x+2)]+C$

$\begin{aligned} & \text { Let } I=\int \frac{x^5+x}{x^8+1} d x \\\\ & =\int \frac{x^4\left(x+\frac{1}{x^3}\right)}{x^4\left(x^4+\frac{1}{x^4}\right)} d x=\int \frac{\left(x+\frac{1}{x^3}\right) d x}{\left(x^2\right)^2+\left(\frac{1}{x^2}\right)^2-2+2} \\\\ & =\int \frac{\left(x+\frac{1}{x^3}\right) d x}{\left(x^2-\frac{1}{x^2}\right)^2+(\sqrt{2})^2}\end{aligned}$

Let $x^2-\frac{1}{x^2}=t \Rightarrow 2\left(x+\frac{1}{x^3}\right) d x=d t$

So, $I=\int \frac{d t / 2}{t^2+(\sqrt{2})^2}=\frac{1}{2} \int \frac{d t}{t^2+\sqrt{2}}$ $=\frac{1}{2}\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)\right]=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left[\frac{x^4-1}{\sqrt{2} x^2}\right]+c$

[put the value of $t=x^2-\frac{1}{x^2}$ ]