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} $

$\frac{x^4}{(x-1)(x-2)}=f(x)+\frac{A}{(x-1)}+\frac{B}{(x-2)}$

$ \begin{aligned} & \Rightarrow x^4=f(x)(x-2)(x-1)+A(x-2)+B(x-1) \end{aligned} $

at $x=1$

$ 1=-A \Rightarrow A=-1 $

at $x=2$

$ 16=B(2-1) \Rightarrow B=16 $

Put the values of $A$ and $B$ and put

$ \begin{aligned} & x=-2 \\ & 16=f(-2)(-3)(-4)-(-4)+16(-3) \\ \Rightarrow \quad & 12 f(-2)=60 \\ \Rightarrow & f(-2)=5 \end{aligned} $

Hence, $f(-2)+A+B=5-1+16=20$

$ \begin{aligned} & \frac{x^4+1}{x^2+1}=\frac{x^4-1+2}{x^2+1} \\ & \quad=\frac{\left(x^4-1\right)}{x^2+1}+\frac{2}{x^2+1} \\ & \quad=\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+1}+\frac{2}{x^2+1} \\ & \quad=x^2-1+\frac{2}{x^2+1} \end{aligned} $

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

$ =\frac{x^3}{3}-x+2 \tan ^{-1} x+c $

Comparing with

$ A x^3+B x^2+C x+D \tan ^{-1} x+E $

We get, $A=1 / 3, B=0, C=-1, D=2$ Therefore,

$ A+B+C+D=\frac{1}{3}+0-1+2=4 / 3 $

$ \text { } \begin{aligned} \frac{x^2-x+2}{x^2+x+2}=1-\frac{2 x}{x^2+x+2} \\ =1-\frac{2 x+1-1}{x^2+x+2} \\ =1-\frac{2 x+1}{x^2+x+2}-\frac{1}{x^2+x+2} \\ =1-\frac{2 x+1}{x^2+x+2}-\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{7}{4}} \end{aligned} $

$ \begin{aligned} & \therefore \int \frac{x^2-x+2}{x^2+x+2} d x \\ & =\int\left(1-\frac{2 x+1}{x^2+x+2}-\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{7}{4}}\right) d x \\ & =\int 1 \cdot d x-\int \frac{2 x+1}{x^2+x+2} d x -\int \frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{7}{4}} \cdot d x \end{aligned} $

let $x^2+x+2=u \Rightarrow(2 x+1) d x=d u$

$ \begin{aligned} & =x-\int \frac{1}{u} \cdot d u-\frac{1}{\sqrt{7 / 4}} \tan ^{-1}\left(\frac{x+1 / 2}{\sqrt{7 / 4}}\right) \\ & =x-\ln \left(x^2+x+2\right)-\frac{2}{\sqrt{7}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{7}}\right) \end{aligned} $

Comparing with

$ \begin{aligned} & x-\log (f(x))+\frac{2}{\sqrt{7}} \tan ^{-1}(g(x)) \\ & f(x)=x^2+x+2 \text { and } g(x)=\frac{2 x+1}{\sqrt{7}} \end{aligned} $

then, $f(-1)=(-1)^2-1+2=2$

and $g(-1)=\frac{2(-1)+1}{\sqrt{7}}=-1 / \sqrt{7}$

Hence, $f(-1)+\sqrt{7} \cdot g(-1)$

$ =2+\sqrt{7}(-1 / \sqrt{7})=1 $

In $\sec \left(x-\frac{\pi}{3}\right) \sec \left(x+\frac{\pi}{6}\right)$

Let $A=x-\frac{\pi}{3}$ and $B=x+\frac{\pi}{6}$

then $B-A=x / 2 \Rightarrow B=A+\pi / 2$

So, $\sec B=\sec \left(\frac{\pi}{2}+A\right)=-\operatorname{cosec} A$

Then, $\sec A \cdot \sec B=\sec A(-\operatorname{cosec} A)$

$ =-\sec A \cdot \operatorname{cosec} A $

Now, $-\sec A \cdot \operatorname{cosec} A=\frac{-1}{\sin A \cdot \cos A}$

$ \begin{gathered} =\frac{-2}{\sin 2 A}=-2 \operatorname{cosec} 2 A \\ \therefore \int \sec \left(x-\frac{\pi}{3}\right) \sec \left(x+\frac{\pi}{6}\right) d x \\ =-2 \int \operatorname{cosec}\left(2 x-\frac{2 \pi}{3}\right) d x \end{gathered} $

Let $2 x-\frac{2 \pi}{3}=u$

$ \begin{aligned} \Rightarrow 2 \cdot d x & =d u=-\int \operatorname{cosec} u \cdot d u \\ & =-\ln |\operatorname{cosec}(u)-\cot (u)|+c \end{aligned} $

and $\operatorname{cosec}(u)-\cot (u)=\tan (u / 2)$

$ \begin{aligned} & =-\ln \left|\tan \left(\frac{u}{2}\right)\right|+C=-\ln \left|\tan \left(x-\frac{\pi}{3}\right)\right|+c \\ & =\ln \left|\cot \left(x-\frac{\pi}{3}\right)\right|+C=\ln \left|\frac{\cos \left(x-\frac{\pi}{3}\right)}{\sin (x-\pi / 3)}\right|+c \\ & \begin{aligned} \because \sin \left(x-\frac{\pi}{3}\right) & =\sin \left(\frac{\pi}{2}+\left(x-\frac{\pi}{6}\right)\right) \\ = & \cos \left(x-\frac{\pi}{6}\right) \\ = & \ln \left|\frac{\cos (x-\pi / 3)}{\cos (x-\pi / 6)}\right|+c \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \int \frac{a \cos x+3 \sin x}{5 \cos x+2 \sin x} \cdot d x \\ & =\frac{26}{29} x-\frac{k}{29} \log |5 \cos x+2 \sin x|+C \\ & \begin{aligned} \because \frac{d}{d x}(5 \cos x+2 \sin x) & =-5 \sin x+2 \cos x \\ \text { let } a \cos x+3 \sin x & =A(-5 \sin x+2 \cos x) +B(5 \cos x+2 \sin x) \end{aligned} \end{aligned} $

By comparing

$ \begin{aligned} & -5 A+2 B=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & 2 A+5 B=a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

[Eq. (ii) $] \times 2$ - [Eq. (i) $] \times 5$

$ 29 A=2 a-15 \Rightarrow A=\frac{2 a-15}{29} $

So, $I=A \log |5 \cos x+2 \sin x|+B x+C$

Comparing with the original form

We get $A=-\frac{k}{29}, B=\frac{26}{29}$

So, $\quad A=\frac{2 a-15}{29}=\frac{-k}{29}$

$\Rightarrow k=15-2 a \Rightarrow|a+k|=|15-a|$

From Eq. (ii), we get $2\left(\frac{2 a-15}{29}\right)+3\left(\frac{26}{29}\right)=a$

$ \Rightarrow 4 a-30+130=29 a \Rightarrow a=4 $

Hence, $|a+k|=|15-a|=11$

$ \begin{aligned} & \text { } \because 1-\sin ^4 x=\left(1-\sin ^2 x\right)\left(1+\sin ^2 x\right) \\ & =\cos ^2 x\left(1+\sin ^2 x\right) \\ & \begin{aligned} \therefore \int \frac{1}{1-\sin ^4 x} d x & =\int \frac{1}{\cos ^2 x \cdot\left(1+\sin ^2 x\right)} \cdot d x \\ & =\int \frac{\sec ^2 x}{1+\sin ^2 x} d x \end{aligned} \end{aligned} $

Let, $\tan x=t$

$ \Rightarrow \frac{1}{1+\sin ^2 x}=\frac{1+t^2}{1+2 t^2} \quad\left\{\because \sin x=\frac{1}{\sqrt{1+r^2}}\right\} $

and $\sec ^2 x \cdot d x=d t$

$ \begin{aligned} &\begin{aligned} & =\int \frac{1+t^2}{1+2 t^2} \cdot d t \\ \because \frac{1+t^2}{1+2 t^2} & =\frac{1}{2}\left(1+\frac{1}{1+2 t^2}\right) \\ \therefore \quad I & =\frac{1}{2} \int\left(1+\frac{1}{1+2 t^2}\right) d t \\ & =\frac{1}{2}\left(t+\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} t)\right)+C \\ & =\frac{1}{2} \tan x+\frac{1}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+C \end{aligned}\\ &\begin{aligned} & A \tan x+B \tan ^{-1}(\sqrt{2} \tan x)+C \\ & \text { We get, } A=\frac{1}{2}, B=\frac{1}{2 \sqrt{2}} \\ & \therefore A^2-B^2=\frac{1}{4}-\frac{1}{8}=\frac{1}{8} \end{aligned} \end{aligned} $

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

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

Compare coefficient, we get

$ \begin{array}{l|r|l} \hline \begin{array}{l} \text { Coefficient } \\ \text { of } \boldsymbol{x}^{\mathbf{1}} \end{array} & \begin{array}{l} \text { Coefficient } \\ \text { of } \boldsymbol{x}^{\mathbf{2}} \end{array} & \text { constant } \\ \hline 0=A+B+C & 1=3 A+B & 0=2 A-2 B-C \\ \ldots \text { (i) } & \ldots \text { (ii) } & A-B=\frac{C}{2} \ldots \text { (ii) } \\ \hline \end{array} $

$ \begin{aligned} &\text { From Eqs. (i) and (iii), we get }\\ &\begin{aligned} A+B & =-C \\ A-B & =C / 2 \\ \hline 2 A & =\frac{-C}{2} \\ A & =\frac{-C}{4} \\ B & =\frac{-3 C}{4} \end{aligned} \end{aligned} $

From by Eq. (i),

$ \begin{aligned} & -\frac{3 C}{4}-\frac{3 C}{4}=1 \\ & C=\frac{-4}{6}=\frac{-2}{3} \\ & \therefore A+B-C \\ & \Rightarrow \frac{-C}{4}-\frac{3 C}{4}-C=\frac{-8 C}{4}=-2 C=\frac{4}{3} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } I=\int \frac{5 \tan x}{\tan x-2} d x \\ & \Rightarrow \int \frac{5 \sin x}{\sin x-2 \cos x} d x \\ & \text { Let } 5 \sin x=A(\sin x-2 \cos x) \quad+B(\cos x+2 \sin x) \end{aligned} \end{aligned} $

$ \begin{aligned} &5=A+2 B \text { and } 0=-2 A+B \\ \therefore \quad B & =2 A \\ & A=1 \\ B & =2 \end{aligned} $

$ \begin{array}{r} \Rightarrow \quad I=\int \frac{(\sin x-2 \cos x)}{(\sin x-2 \cos x)} d x +\frac{2(\cos x+2 \sin x)}{\sin x-2 \cos x} d x \end{array} $

$ \begin{aligned} & =x+2 \log |\sin x-2 \cos x|+C \\ \therefore \quad a & =1, b=2 \\ \Rightarrow a+b & =3 \end{aligned} $

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

$ \begin{aligned} & \text { Put } \quad x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta \\ & \Rightarrow \quad \quad \theta=\tan ^{-1} x \\ & \begin{aligned} I & =\int \tan \theta \cdot \cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \sec ^2 \theta d \theta \\ & =\int \cos ^{-1}(\cos 2 \theta) \cdot \tan \theta \sec ^2 \theta d \theta \\ & =\int_1^{2 \theta \tan \theta \sec ^2 \theta d \theta} \end{aligned} \end{aligned} $

Using by part

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

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

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

Let $\sqrt{x}=t$

$ \begin{aligned} \Rightarrow \frac{1}{2 \sqrt{x}} d x & =d t \\ I & =\int \frac{2 d t}{(1+t) \sqrt{1-t^2}} \end{aligned} $

Put $t=\sin \theta \Rightarrow d t=\cos \theta d \theta$

$ \begin{aligned} & \therefore \int \frac{2 \cos \theta d \theta}{(1+\sin \theta) \cdot \cos \theta} \\ & \quad \quad =2 \int \frac{d \theta}{1+\sin \theta}=2 \int \frac{1-\sin \theta}{1-\sin ^2 \theta} d \theta \\ & \quad= 2 \int\left(\sec ^2 \theta-\sec \theta \tan \theta\right) d \theta \\ & \quad= 2(\tan \theta-\sec \theta)+C \\ & \quad= 2\left(\frac{\sin \theta-1}{\cos \theta}\right)+C \\ & \quad= 2\left(\frac{t-1}{\sqrt{1-t^2}}\right)+C=2\left(\frac{\sqrt{x}-1}{\sqrt{1-x}}\right)+C \end{aligned} $

$ \begin{aligned} & =-2\left(\frac{1-\sqrt{x}}{\sqrt{1-x}}\right)+C \\ & =-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C \\ \{\because & \left.-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \times \sqrt{\frac{1-\sqrt{x}}{1-\sqrt{x}}} \Rightarrow \frac{-2(1-\sqrt{x})}{\sqrt{1-x}}\right\} \end{aligned} $

$ \begin{aligned} &\text { Let } I=\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x\\ &\begin{aligned} & \text { put } x=a \tan ^2 \theta \\ & \begin{aligned} \Rightarrow d x & =2 a \tan \theta \cdot \sec ^2 \theta d \theta \\ \therefore I & =\int \sin ^{-1} \sqrt{\frac{a \tan ^2 \theta}{a \sec ^2 \theta}} 2 a \tan \theta \sec ^2 \theta d \theta \\ & =2 a \int \theta_1 \cdot \tan \theta \sec ^2 \theta \cdot d \theta \\ \Rightarrow I & =2 a\left[\theta \cdot \int \tan \theta \sec ^2 \theta d \theta\right. \left.-\int\left(1 \cdot \int \tan \theta \sec ^2 \theta d \theta\right) d \theta\right] \end{aligned} \end{aligned} \end{aligned} $

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

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } I=\int \frac{x}{x \tan x+1} d x \\ & \quad=\int \frac{x \cos x}{x \sin x+\cos x} d x \\ & \text { Put } x \sin x+\cos x=t \\ & \Rightarrow(x \cos x+\sin x-\sin x) d x=d t \\ & \Rightarrow \quad d t=x \cos d x \\ & \quad I=\int \frac{d t}{t} \\ & \Rightarrow \quad I=\ln |x \sin x+\cos x|+C \\ & \begin{array}{rlrl} \therefore \quad & f(x) =x \sin x+\cos x \\ \Rightarrow \quad f\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}+1\right) \frac{1}{\sqrt{2}} \\ & =\frac{\pi+4}{4 \sqrt{2}} \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \quad \frac{2 x^4-3 x^2+4}{\left(x^2+1\right)\left(x^2+2\right)}=\frac{2 x^4-3 x^2+4}{x^4+3 x^2+2} \\ & \quad=\frac{2 x^4+6 x^2+4}{x^4+3 x^2+2}-\frac{9 x^2}{x^4+3 x^2+2} \\ & \quad=2-\frac{9 x^2}{x^4+3 x^2+2} \\ & \therefore \frac{9 x^2}{x^4+3 x^2+2}=\frac{A}{x^2+1}+\frac{B}{x^2+2} \\ & \Rightarrow \quad 9 x^2=A\left(x^2+2\right)+B\left(x^2+1\right) \\ & \Rightarrow \quad 9=A+B \text { and } 2 A+B=0 \\ & \Rightarrow \quad B=-2 A \\ & \therefore \quad A=-9, B=18 \\ & \therefore \frac{2 x^4-3 x^2+4}{\left(x^2+1\right)\left(x^2+2\right)}=2+\frac{9}{x^2+1}-\frac{18}{x^2+2} \\ & \quad a=2, p=0, q=9, m=0, n=-18 \\ & \therefore \quad n / q=-2 \\ & p+m-a=0+0-2=-2 \end{aligned} \end{aligned} $

Let $I=\int 1 \cdot(\log 2 x)^3 d x$

Using by part

$ \begin{aligned} = & x(\log 2 x)^3-\int\left(3(\log 2 x)^2 \cdot \frac{1}{x} \cdot x\right) d x \\ = & x(\log 2 x)^3-3 \int(\log 2 x)^2 d x \\ = & (x \log 2 x)^3-3 \cdot x(\log 2 x)^2 +3 \int\left(x \cdot 2 \cdot \log 2 x \cdot \frac{1}{x}\right) d x \\ = & x\left[(\log 2 x)^3-3(\log 2 x)^2\right]+6 \int(\log 2 x) d x \\ = & x\left[(\log 2 x)^3-3(\log 2 x)^2\right]+6 x \log 2 x-6 \int d x \\ = & x\left[(\log 2 x)^3-3(\log 2 x)^2+6 \log 2 x-6\right]+C \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{x+1}{(x-2 \sqrt{1-x}} d x \\ & \quad \sqrt{1-x}=t \\ & \Rightarrow \quad 1-x=t^2 \\ & \Rightarrow \quad x=1-t^2 \\ & \text { and } \quad \frac{-1}{2 \sqrt{1-x}} d x=d t \\ & \Rightarrow \quad \frac{d x}{\sqrt{1-x}}=-2 d t \\ & \therefore \quad I=\int \frac{1-t^2+1}{\left(1-t^2-2\right)}(-2) d t=-2 \int \frac{t^2-2}{t^2+1} d t \\ & \quad=-2 \int\left(\frac{t^2+1}{t^2+1}-\frac{3}{t^2+1}\right) d x \\ & \quad=-2\left(t-3 \tan ^{-1} t\right)+C \\ & \quad=-2 \sqrt{1-x}+6 \tan ^{-1} \sqrt{1-x}+C \\ & \quad=6 \tan ^{-1} \sqrt{1-x}-2 \sqrt{1-x}+C \end{aligned} $

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

$ \begin{aligned} & =\frac{\cos x}{x}\left(\frac{-1}{x \sin x+\cos x}\right)-\int \frac{x \sin x+\cos x}{x^2}. \frac{1}{(x \sin x+\cos x)} d x\\ & =\frac{-\cos x}{x(x \sin x+\cos x)} d x \\ & \therefore f(x)=\frac{1}{x}+C \\ & \therefore \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{x(x \sin x+\cos x)}+\frac{1}{x} \\ & \quad=\frac{2}{\pi} \end{aligned} $

$ \begin{aligned} &\text { } I=\int \sin ^3 x \cos ^2 x d x\\ &\begin{aligned} & \text { Let } \quad \cos x=t \\ & \Rightarrow \quad-\sin x d x=d t \\ & \therefore \quad I=\int\left(1-t^2\right) t^2(-d t) \\ & =\int\left(t^4-t^2\right) d t=\frac{t^5}{5}-\frac{t^3}{3}+C \\ & =\frac{\cos ^5 x}{5}-\frac{\cos ^3 x}{3}+C \\ & =\frac{\cos x\left(1-\sin ^2 x\right)^2}{5}-\frac{\cos x\left(1-\sin ^2 x\right)}{3}+C \\ & =\frac{\sin ^4 x \cos x}{5}-\frac{2 \sin ^2 x \cos x}{5}+\frac{\cos x}{5} -\frac{\cos x}{3}+\frac{\cos x \sin ^2 x}{3}+C \\ & =\frac{\sin ^4 x \cdot \cos x}{5}-\frac{\sin ^2 x \cos x}{15}-\frac{2 \cos x}{15}+C \end{aligned} \end{aligned} $

Given, $\frac{3 x^3-7 x+1}{(x-2)^5}$

$ \begin{aligned} & =\frac{A}{(x-2)}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^3}+\frac{D}{(x-2)^4}+\frac{E}{(x-2)^5} \\ & =\frac{A(x-2)^4+B(x-2)^3+C(x-2)^2+D(x-2)+E}{(x-2)^5} \\ & =A x^4+(-8 A+B) x^3+(24 A-6 B+C) x^2 +(-32 A+12 B-4 C+D) x+16 A-8 B+4 C-2 D+E \end{aligned} $

Comparing the coefficient, we get

$ \begin{aligned} &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=0 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-8 A+B=3 \\ & \Rightarrow \quad B=3+8 A=3 \\ & 24 A-6 B+C=0 \\ & \Rightarrow \quad C=6 B-24 A=18-0=18 \\ & -32 A+12 B-4 C+D=-7 \\ & \Rightarrow \quad D=-7+32 A-12 B+4 C=-7+0-36+72=29 \\ & 16 A-8 B+4 C-2 D+E=1 \\ & \Rightarrow \quad E=1-16 A+8 B-4 C+2 D \\ & =1-0+24-72+58=11 \\ & \therefore \quad A(B+C+D+E)=0(3+18+29+11)=0 \end{aligned} $

$ \text { } \begin{aligned} & \begin{aligned} & (\sqrt{\tan x}+\sqrt{\cot x}) d x \\ & =\int\left(\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}\right) d x \\ & =\int \frac{\tan x+1}{\sqrt{\tan x}} d x \end{aligned} \end{aligned} $

Let $u=\sqrt{\tan x}$, then $u^2=\tan x$ and

$ \begin{aligned} 2 u d u & =\sec ^2 x d x \\ & =\left(1+\tan ^2 x\right) d x \\ & =\left(1+u^4\right) d x \\ & =\int \frac{u^2+1}{u} \cdot \frac{2 u}{1+u^4} d u \\ & =\int \frac{2\left(u^2+1\right)}{1+u^4} d u=\int \frac{2\left(1+\frac{1}{u^2}\right)}{\frac{1}{u^2}+u^2} d u \\ & =\int \frac{2\left(1+\frac{1}{u^2}\right)}{\left(u-\frac{1}{u}\right)^2+2} d u \end{aligned} $

$ \begin{aligned} &\text { Again, Let } v=u-\frac{1}{u}\\ &\begin{aligned} \Rightarrow \quad d v & =\left(1+\frac{1}{u^2}\right) d u \\ & =\int \frac{2}{2\left(\frac{v^2}{2}+1\right)} d v=\int \frac{1}{\left(\frac{v^2}{2}+1\right)} d v \\ & =\int \frac{1}{\left(\frac{v}{\sqrt{2}}\right)^2+1} d v \\ & =\tan ^{-1}\left(\frac{v}{\sqrt{2}}\right) \cdot \sqrt{2}+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { where } c=\text { constant }\\ &\begin{aligned} & =\sqrt{2} \tan ^{-1}\left(\frac{u-\frac{1}{u}}{\sqrt{2}}\right)+C \\ & =\sqrt{2} \tan ^{-1}\left(\frac{\sqrt{\tan x}-\frac{1}{\sqrt{\tan x}}}{\sqrt{2}}\right)+C \\ & =\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+C \end{aligned} \end{aligned} $

Let $I=\int \frac{\sqrt{x-2}}{2 x+4} d x$

and $u=\sqrt{x-2}$

$\Rightarrow u^2=x-2$ and $x=u^2+2$

Differentiate $x$ w.r.t $u$, we get

$ \frac{d x}{d u}=2 u \Rightarrow d x=2 u d u $

$ \begin{aligned} &\begin{aligned} I & =\int \frac{\sqrt{x-2}}{2 x+4} d x=\int \frac{2 u \cdot u}{2\left(u^2+2\right)+4} d u \\ & =\int \frac{2 u^2}{2 u^2+8} d u=\int \frac{u^2}{u^2+4} d u \\ & =\int \frac{u^2+4-4}{u^2+4} d u=\int\left(1-\frac{4}{u^2+4}\right) d u \\ & =u-4 \cdot \frac{1}{2} \cdot \tan ^{-1}\left(\frac{u}{2}\right)+C \end{aligned}\\ &\text { where, } C=\text { constant }\\ &\begin{aligned} & =u-2 \tan ^{-1}\left(\frac{u}{2}\right)+C \\ & =\sqrt{x-2}-2 \tan ^{-1}\left(\frac{\sqrt{x-2}}{2}\right)+C \end{aligned} \end{aligned} $

Given,

$ \begin{aligned} I & =\int x^{49}\left[\tan ^{-1}\left(x^{50}\right)+\frac{x^{50}}{1+x^{100}}\right] d x \\ & =\frac{x^n}{k} f(x)+c \end{aligned} $

Let $u=x^{50}$, then $d u=50 x^{49} d x$

$ \text { } \begin{aligned} so, I & =\int x^{49}\left[\tan ^{-1} x^{50}+\frac{x^{50}}{1+x^{100}}\right] d x \\ & =\int x^{49}\left[\tan ^{-1}(u)+\frac{u}{1+u^2}\right] \cdot \frac{d u}{50 x^{49}} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{1}{50} \int\left(\tan ^{-1}(u)+\frac{u}{1+u^2}\right) d u \\ & =\frac{1}{50}\left[\int \tan ^{-1}(u) d u+\int \frac{u}{1+u^2} d u\right] \\ & =\frac{1}{50}\left[u \tan ^{-1}(u)-\int \frac{u}{1+u^2} d u+\int \frac{u}{1+u^2} d u\right] \end{aligned}\\ &\text { [using product rule] } \end{aligned} $

$ \begin{aligned} & =\frac{1}{50}\left[u \tan ^{-1}(u)\right]+c \\ & =\frac{1}{50} \cdot x^{50} \cdot \tan ^{-1}\left(x^{50}\right)+c \end{aligned} $

Comparing to original equation, we get

$ n=50, k=50, f(x)=\tan ^{-1}\left(x^{50}\right) $

Now, $f(x)-f\left(\sqrt[k]{x^n}\right)$

$ \begin{aligned} & =f(x)-f\left(x^{n / k}\right) \\ & =\tan ^{-1}\left(x^{50}\right)-f\left(x^{50 / 50}\right) \\ & =\tan ^{-1}\left(x^{50}\right)-f(x) \\ & =\tan ^{-1}\left(x^{50}\right)-\tan ^{-1}\left(x^{50}\right)=0 \end{aligned} $

$ \therefore f(x)-f\left(\sqrt[k]{x^n}\right)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, $k-n=50-50=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

$ f(x)-f\left(\sqrt[k]{x^n}\right)=k-n $