Indefinite Integration

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

$ \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 $

$ \begin{aligned} & \text { } \operatorname{Let} I=\int \frac{x}{\sqrt{x^2-2 x+5}} d x \\ & =\int \frac{x-1+1}{\sqrt{x^2-2 x+5}} d x \\ & =\int \frac{x-1}{\sqrt{x^2-2 x+5}}+\int \frac{1}{\sqrt{x^2+2 x+5}} d x \end{aligned} $

Let $u=x^2-2 x+5$, then $d u=(2 x-2) d x$

$ \begin{aligned} & =2(x-1) d x \\ & =\frac{1}{2} \int \frac{d u}{\sqrt{u}}+\int \frac{1}{\sqrt{(x-1)^2+4}} d x \\ & =\frac{1}{2}\left(2 u^{\frac{1}{2}}\right)+\sin h^{-1} \frac{(x-1)}{2}+C \\ & \quad\left[\because \int \frac{1}{\sqrt{x^2+a^2}}=\sin h^{-1} \frac{x}{a}\right] \end{aligned} $

Where, $C=$ constant

$ \begin{gathered} =\sqrt{u}+\sinh ^{-1}\left(\frac{(x-1)}{2}\right)+C \\ =\sqrt{x^2-2 x+5}+\sinh ^{-1}\left(\frac{x-1}{2}\right)+C \end{gathered} $

$ \begin{aligned} &\text { For } 0 < x < 1 \text {, }\\ &\begin{aligned} & I=\int\left[\tan ^{-1}\left(1-x+x^2\right)+\tan ^{-1}(1-x)\right] d x \\ & =\int \tan ^{-1}\left[\frac{\left(1-x+x^2\right)+(1-x)}{1-\left(1-x+x^2\right)(1-x)}\right] d x \\ & =\int \tan ^{-1}\left(\frac{2-2 x+x^2}{x\left(2-2 x+x^2\right)}\right) d x \\ & =\int \tan ^{-1}\left(\frac{1}{x}\right) d x \end{aligned} \end{aligned} $

$ \begin{aligned} & =\int \cot ^{-1}(x) d x=\int\left(\frac{\pi}{2}-\tan ^{-1} x\right) d x \\ & =\int \frac{\pi}{2} d x-\int \tan ^{-1} x d x \\ & =\frac{\pi}{2} x-\left[x \tan ^{-1} x-\frac{1}{2} \ln \left(1+x^2\right)+C\right] \end{aligned} $

[Using product rule]

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

where $C=$ constant of integration

$ \begin{aligned} & =x\left(\frac{\pi}{2}-\tan ^{-1} x\right)+\frac{1}{2} \ln \left(1+x^2\right)+C \\ & =x \cot ^{-1} x+\frac{1}{2} \ln \left(1+x^2\right)+C \\ & =x \cot ^{-1} x+\log \sqrt{1+x^2}+C \end{aligned} $

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

Comparing both sides, we get

$ \begin{aligned} & & A+B & =0 \\ \Rightarrow & & A & =-B \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

and $C-B=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

and $2 A-C=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

put the value of $A$ from Eq. (i) to Eq. (iii)

$ \begin{array}{ll} \Rightarrow & -2 B-C=1 \\ \Rightarrow & C=-2 B-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{array} $

By Eqs. (ii) and (iv),

We get

$ (-2 B-1)-B=3 \Rightarrow B=-4 / 3 $

and $A=-(-4 / 3)=4 / 3$

$ \begin{aligned}Also,\,\, C & =-2 B-1=-2(-4 / 3)-1 \\ & =\frac{8}{3}-1=5 / 3 \end{aligned} $

$ \begin{aligned}thus,\,\, 5(A-B) & =5\left(\frac{4}{3}-(-4 / 3)\right) \\ & =5\left(\frac{8}{3}\right)=\frac{40}{3} \end{aligned} $

and $8 C=8 \times \frac{5}{3}=\frac{40}{3}$

Therefore, $5(A-B)=8 C$

Let $I=\int \frac{\sec ^2 x-7}{\sin ^7 x} d x$

$ \begin{aligned} & \Rightarrow \quad I=\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x \\ & \Rightarrow \quad I=\int \frac{\sin ^5 x\left(1-7 \cos ^2 x\right) d x}{\sin ^{12} x \cos ^2 x} \end{aligned} $

Put $\sin ^6 x \cos x=t$

$ \begin{aligned} & \left(6 \sin ^5 x \cos ^2 x-\sin ^7 x\right) d x=d t \\ \Rightarrow & \sin ^5 x\left(6 \cos ^2 x-\sin ^2 x\right) d x=d t \\ \Rightarrow & \sin ^5 x\left(6 \cos ^2 x-1+\cos ^2 x\right) d x=d t \\ \Rightarrow & \sin ^5 x\left(1-7 \cos ^2 x\right) d x=-d t \\ \therefore & I=\int \frac{-d t}{t^2} \\ \Rightarrow & I=\frac{1}{t}+C \\ \Rightarrow & I=\frac{1}{\sin ^6 x \cos x}+C \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \int\left(x^6+x^4+x^2\right) \sqrt{2 x^4+3 x^2+6} d x \\ & =\int x\left(x^5+x^3+x\right) \sqrt{2 x^4+3 x^2+6} d x \\ & =\int\left(x^5+x^3+x\right) \sqrt{2 x^6+3 x^4+6 x^2} d x \\ & \text { Put } 2 x^6+3 x^4+6 x^2=t \\ & \Rightarrow 12\left(x^5+x^3+x\right) d x=d t \\ & =\frac{1}{12} \int \sqrt{t} \cdot d t=\frac{1}{12} \frac{t^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}+C \\ & \text { = } \frac{t^{\frac{3}{2}}}{18}+C=\frac{1}{18}\left(2 x^6+3 x^4+6 x^2\right)^{\frac{3}{2}}+C \\ & \text { So, } f(x)=\frac{1}{18}\left(2 x^6+3 x^4+6 x^2\right)^{3 / 2} \\ & \Rightarrow f(3)=\frac{1}{18}(9(195))^{\frac{3}{2}}=\frac{3}{2}(195)^{\frac{3}{2}} \end{aligned} \end{aligned} $

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

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

$ \begin{aligned} & \text { } \int \frac{d x}{2 \cos x+3 \sin x+4} \\ & \text { Let } t=\tan \frac{x}{2} \Rightarrow \sin x=\frac{2 t}{1+t^2} \\ & \cos x=\frac{1-t^2}{1+t^2} \\ & \text { and } d x=\frac{2 d t}{1+t^2} \end{aligned} $

$ \begin{aligned} & =\int \frac{2 d t}{\left(1+t^2\right)\left[\frac{2\left(1-t^2\right)}{\left(1+t^2\right)}+\frac{6 t}{\left(1+t^2\right)}+4\right]} \\ & =\int \frac{2 d t}{2-2 t^2+6 t+4+4 t^2} \\ & =\int \frac{d t}{t^2+3 t+3}=\int \frac{d t}{\left(t+\frac{3}{2}\right)^2+\frac{3}{4}} \\ \text { put } u & =t+\frac{3}{2} \Rightarrow d u=d t \\ & =\int \frac{1}{u^2+\left(\frac{\sqrt{3}}{2}\right)^2} d u \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{1}{\left(\frac{\sqrt{3}}{2}\right)} \tan ^{-1}\left(\frac{u}{\left(\frac{\sqrt{3}}{2}\right)}\right)+C \\ & =\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 u}{\sqrt{3}}\right)+C \\ & =\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+3}{\sqrt{3}}\right)+C \\ & =\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan \frac{x}{2}+3}{\sqrt{3}}\right)+C \end{aligned}\\ &\text { Thus, } f(x)=\tan ^{-1}\left(\frac{2 \tan \left(\frac{x}{2}\right)+3}{\sqrt{3}}\right)\\ &\begin{aligned} \therefore f\left(\frac{2 \pi}{3}\right) & =\tan ^{-1}\left(\frac{2 \tan \left(\frac{\pi}{3}\right)+3}{\sqrt{3}}\right) \\ & =\tan ^{-1}\left(\frac{2 \sqrt{3}+3}{\sqrt{3}}\right) \\ & =\tan ^{-1}(2+\sqrt{3})=\frac{5 \pi}{12} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} & \because \int \frac{1}{\left((x+4)^3(x+1)^5\right)^{\frac{1}{4}}} d x \\ &=A \cdot\left(\frac{x+4}{x+1}\right)^n+C \\ &=\int \frac{1}{(x+1)^2\left(\frac{x+4}{x+1}\right)^{3 / 2}} d x \\ & \text { Let } t=\frac{x+4}{x+1} \\ & \Rightarrow d t=\frac{-3}{(x+1)^2} d x=-\int \frac{1}{3} t^{-3 / 4} d t \\ &=-\frac{1}{3}\left[\frac{\frac{-3}{4}+1}{\frac{-3}{4}+1}\right]+C \\ &=-\frac{1}{3}\left[4 t^{1 / 4}\right]+C \\ &=-\frac{4}{3}\left(\frac{x+4}{x+1}\right)^{1 / 4}+C \end{aligned} \end{aligned} $

Comparing with $A=\frac{(x+4)^n}{x+1}+C$

$ A=-\frac{4}{3}, n=\frac{1}{4} $

Therefore, $n+\frac{1}{A}=\frac{1}{4}+\frac{1}{\left(\frac{-4}{3}\right)}$

$ \begin{aligned} & =\frac{1}{4}-\frac{3}{4} \\ & =\frac{-2}{4}=\frac{-1}{2} \end{aligned} $

Let $I=\int \frac{x+1}{x^3-1} d x$

$ \begin{aligned} & =\int \frac{2}{3(x-1)} d x+\int \frac{-2 x-1}{3\left(x^2+x+1\right)} d x \\ & =\frac{2}{3} \ln |x-1|-\frac{1}{3} \ln \left|x^2+x+1\right|+C \end{aligned} $

where $C$ is constant of integration.

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

$ \text { } \begin{aligned} Let \,\,I & =\int \frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x \\ & =\int \frac{(x+4)\left(x^3-4 x^2+4\right)}{\left(x^3-4 x^2+2\right.} d x \\ & =\int(x+4) d x=\frac{x^2}{2}+4 x+C_1 \\ & =\frac{x^2+8 x+2 C_1}{2} \\ & =\frac{x^2+8 x+C}{2} \end{aligned} $

where $C_1$ is constant of integration $C=2 C_1$.

Let $I=\int \frac{\sec ^2 x}{(\sec x+\tan x)^{\frac{5}{2}}} d x$

Put $\sec x+\tan x=t$

$\therefore \sec x-\tan x=\frac{1}{t}$

$\therefore \sec x=\frac{1}{2}\left(t+\frac{1}{t}\right)$ and $\tan x=\frac{1}{2}\left(t-\frac{1}{t}\right)$

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

$ \begin{aligned} & \Rightarrow \quad\left(\sec x \tan x+\sec ^2 x\right) d x=d t \\ & \Rightarrow \quad d x=\frac{d t}{\sec x \tan x+\sec ^2 x} \\ & I=\int \frac{\sec ^2 x}{\frac{5}{t^2}} \times \frac{d t}{\sec x \tan x+\sec ^2 x} \\ & =\int \frac{\sec x}{t^{\frac{5}{2}}} \times \frac{d t}{(\sec x+\tan x)} \\ & =\int\left(\frac{1}{2}\left(t+\frac{1}{t}\right)\right) \cdot \frac{1}{t^{\frac{5}{2}}} \times \frac{1}{t} d t \\ & =\frac{1}{2} \int\left(t+\frac{1}{t}\right) t^{\frac{-7}{2}} d t \end{aligned} $

$ \begin{aligned} & =\frac{1}{2}\left[\int t^{\frac{-5}{2}} d t+\int t^{\frac{-9}{2}} d t\right] \\ & =\frac{1}{2}\left[\frac{t^{\frac{-3}{2}}}{\frac{-3}{2}}+\frac{t^{\frac{-7}{2}}}{\frac{-7}{2}}\right]+C \\ & =\frac{1}{2}\left[\frac{-2}{3} \times \frac{1}{t^{\frac{3}{2}}}-\frac{2}{7} \frac{1}{\frac{7}{2}}\right]+C \\ & =\frac{-1}{3}(\sec x-\tan x)^{\frac{3}{2}}-\frac{1}{7}(\sec x-\tan x)^{\frac{7}{2}} +C \end{aligned} $

$ \begin{aligned}Let\,\,& I=\int \frac{1}{\cos x}\left[\frac{1}{\sin x}-\frac{1}{\sin x+3 \cos x}\right] d x \\ & =\int \frac{2}{\sin 2 x} d x-\int \frac{\sec ^2 x}{\tan x+3} d x \\ & =2 \int \operatorname{cosec} 2 x d x-\int \frac{d z}{z+3}, \text { where } \\ & z=\tan x \\ & =\frac{2 \log _e|\operatorname{cosec} 2 x-\cot 2 x|}{2}-\ln |z+3|+C \\ & =\ln |\operatorname{cosec} 2 x-\cot 2 x|-\ln |\tan x+3|+C \end{aligned} $

$ \begin{aligned} & =\ln \left|\frac{\frac{1}{\sin 2 x}-\frac{\cos 2 x}{\sin 2 x}}{\frac{\sin x}{\cos x}+3}\right|+C \\ & =\ln \left|\frac{\frac{1-\cos 2 x}{\sin 2 x}}{\frac{\sin x+3 \cos x}{\cos x}}\right|+C \\ & =\ln \left|\frac{2 \sin ^2 x}{2 \sin x \cdot \cos x} \times \frac{\cos x}{\sin x+3 \cos x}\right|+C \\ & =\ln \left|\frac{\sin x}{\sin x+3 \cos x}\right|+C \end{aligned} $

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

                                                              (I)             (II)

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

$ \begin{aligned} &\text { }\\ &\begin{aligned} & \frac{a x+5}{\left(x^2+b\right)(x+3)}=\frac{x+21}{12\left(x^2+b\right)}+\frac{c}{12(x+3)} \\ & \Rightarrow 12(a x+5)=(x+21)(x+3)+c\left(x^2+b\right) \\ & \Rightarrow 12 a x+60=x^2+3 x+21 x+63 +c x^2+b c \\ & \Rightarrow 12 a x+60=(1+c) x^2+24 x+63+b c \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Comparing coefficient }\\ &\begin{array}{rlrlrl} \Rightarrow 1+c & =0 & 12 a & =24 & 63+b c & =60 \\ c & =-1 & a & =2 & 63-b & =60 \end{array} \end{aligned} $

$ \begin{array}{ll} \Rightarrow \quad b =3 \\ b^2 =9=a^3-c \\ \therefore \quad b^2 =a^3-c \end{array} $

$ \begin{aligned} & \text { } \int\left(\sum_{r=0}^{\infty} \frac{x^r 2^r}{r!}\right) d x \\ & =\int\left(1+\frac{(2 x)^1}{1!}+\frac{(2 x)^2}{2!}+\frac{(2 x)^3}{3!}+\ldots\right) d x \\ & =\int e^{2 x} d x=\frac{e^{2 x}}{2}+C \end{aligned} $

$ \begin{aligned} & \text { } \int \frac{1}{12 \cos x+5 \sin x} d x \\ & R=\sqrt{12^2+5^2}=\sqrt{169}=13 \\ & \tan \alpha=\frac{5}{12} \\ & =\int \frac{1}{13\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)} d x \\ & =\frac{1}{13} \int \frac{1}{\cos (x-\alpha)} d x \\ & =\frac{1}{13} \log \left|\tan \left(\frac{\pi}{4}+\frac{x}{2}-\frac{1}{2} \tan ^{-1} \frac{5}{12}\right)\right|+C \end{aligned} $

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

Put $u=\sin x$

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

$ \text { } \begin{aligned} & \int \frac{13 \cos 2 x-9 \sin 2 x}{3 \cos 2 x-4 \sin 2 x} d x \\ = & \int\left(3-\frac{1(-6 \sin 2 x-8 \cos 2 x)}{2(3 \cos 2 x-4 \sin 2 x)}\right) d x \\ = & 3 x-\frac{1}{2} \log |3 \cos 2 x-4 \sin 2 x|+C \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \int \sqrt{x^2+x+1} d x \\ & =\int \sqrt{x^2+x+\frac{1}{4}-\frac{1}{4}+1} d x \\ & =\int \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} d x \\ & =\int \sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d x \\ & =\frac{x+\frac{1}{2}}{2} \sqrt{x^2+x+1}+\frac{3}{8} \ln \left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right|+C \\ & =\frac{2 x+1}{4} \sqrt{x^2+x+1} +\frac{3}{8} \sin ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \end{aligned} \end{aligned} $

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

Put $x=1$

$ 2=2 B \Rightarrow B=1 $

In Eq. (i) put $x=0$

$ \begin{aligned} & 1=-A+B+D \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow \quad & A=D \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

In Eq. (i) put $x=-1$

$ \begin{aligned} 0 & =-4 A+2-4 C+4 D \\ 4 C & =2 \Rightarrow C=\frac{1}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

In Eq. (i) put $x=2$

$ 3=5 A+5+1+D \Rightarrow 6 A=-3 $

$ \begin{gathered} A=-\frac{1}{2}=D \\ \sqrt{3 A^2+4 D^2+5 C^2+B^2} \end{gathered} $

$ \begin{aligned} & =\sqrt{\frac{3}{4}+\frac{4}{4}+\frac{5}{4}+1} \\ & =\sqrt{\frac{16}{4}}=2 \end{aligned} $

$ \begin{aligned} I= & \int \frac{1}{9 \cos ^2 x-24 \sin x \cos x+16 \sin ^2 x} d x =\\ \Rightarrow & I=\int \frac{1}{(3 \cos x-4 \sin x)^2} d x \\ \Rightarrow & I=\int \frac{\sec ^2 x}{(3-4 \tan x)^2} d x \\ & 3-4 \tan x=t \\ & -4 \sec ^2 x d x=d t \\ \Rightarrow & I=\frac{1}{-4} \int t^{-2} d t \\ & =\frac{1}{4} \cdot \frac{1}{t}+C=\frac{\cos x}{4(3 \cos x-4 \sin x)}+C \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \int \frac{1}{\cot \frac{x}{2} \cot \frac{x}{3} \cot \frac{x}{6}} d x \\ & =\int \tan \frac{x}{2} \tan \frac{x}{3} \tan \frac{x}{6} d x \\ & \because \frac{x}{6}=\frac{x}{2}-\frac{x}{3} \\ & \Rightarrow \tan \frac{x}{6}=\frac{\tan \frac{x}{2}-\tan \frac{x}{3}}{1+\tan \frac{x}{2} \tan \frac{x}{3}} \\ & \begin{array}{r} \int\left(\tan \frac{x}{2}-\tan \frac{x}{3}-\tan \frac{x}{6}\right) d x \\ =-2 \log \left|\cos \frac{x}{2}\right|+3 \log \left|\cos \frac{x}{3}\right| \\ +6 \log \left|\sec \frac{x}{6}\right|+k \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} & \therefore A=-2, B=3, C=6 \\ & \Rightarrow A+B+C=7 . \end{aligned} $

$I=\int \frac{\sin x+\cos x}{\sin x-\cos x} d x$ put $\sin x-\cos x=t$

$ \begin{aligned} \Rightarrow \quad & -(\cos x+\sin x) d x=d t \\ I & =-\int \frac{1}{t} d t \\ & =-\log |\sin x-\cos x|+C \\ & =-\log |\cos x-\sin x|+C \end{aligned} $

$ \begin{aligned} & \text { } I=\int \frac{x^4-1}{x^2 \sqrt{x^4+x^2+1}} d x \\ & \Rightarrow \quad I=\int \frac{x^2-\frac{1}{x^2}}{\sqrt{x^4+x^2+1}} d x \\ & \quad=\int \frac{x-\frac{1}{x^3}}{\sqrt{x^2+\frac{1}{x^2}+1}} d x \\ & \text { Put } \quad x^2+\frac{1}{x^2}+1=t^2 \\ & \Rightarrow \quad 2 x-\frac{2}{x^3} d x=2 t d t \\ & I=\int d t=t+C=\frac{\sqrt{x^4+x^2+1}}{x}+C \end{aligned} $

$I=\int \frac{(3 x-2) \tan \left(\sqrt{9 x^2-12 x+1}\right)}{\sqrt{9 x^2-12 x+1}} d x$

Put $9 x^2-12 x+1=t^2$

$ \begin{aligned} \Rightarrow \quad & (18 x-12) d x=2 t d t \\ \Rightarrow \quad & (3 x-2) d x=\frac{1}{3} t d t \\ \Rightarrow \quad I & =\frac{1}{3} \int \frac{t \tan t}{t} d t=\frac{1}{3} \int \tan t d t \\ & \quad=\frac{1}{3} \log |\sec t|+C \\ & \quad=\frac{1}{3} \log \left|\sec \sqrt{9 x^2-12 x+1}\right|+C \end{aligned} $

Given $\int e^{\sin x}(1+\sec x \tan x) d x$

$ \begin{array}{r} =e^{\sin x} f(x)+c \\ \int e^{\sin x}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos ^3 x}\right) \cos x d x \end{array} $

Put $\sin x=t$

$ \begin{aligned} & \cos x d x=d t \\ & \cos x=\sqrt{1-t^2} \end{aligned} $

$ \int e^t\left[\frac{1}{\sqrt{1-t^2}}+\frac{t}{\left(\left(1-t^2\right)^{3 / 2}\right)}\right] d t $

[Using result $\int e^x\left[f(x)+f^{\prime}(x)\right] d x$

$ \begin{gathered} \left.=e^x f(x)+C\right] \\ \int e^t\left[\frac{1}{\sqrt{1-t^2}}+\left(\frac{1}{\sqrt{1-t^2}}\right)^1\right] d t=e^t \cdot \frac{1}{\sqrt{1-t^2}} \\ =e^{\sin x} \cdot \frac{1}{\cos x}=e^{\sin x} \cdot \sec x+C \end{gathered} $

Comparing with $e^{\sin x} f(x)+C$

$ \begin{aligned} \therefore \quad f(x) & =\sec x=1 \\ x & =0,2 \pi \end{aligned} $

$\therefore$ Number of solution is 2 .

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \int \frac{d x}{(x-1)^{3 / 2}(x-3)^{1 / 2}} \\ & \text { Let } I=\int \frac{d x}{(x-1)^2\left(\frac{x-3}{x-1}\right)^{\frac{1}{2}}} \\ & \text { Put } \frac{x-3}{x-1}=t^2 \\ & \frac{x-1-x+3}{(x-1)^2} d x=2 t d t \\ & \frac{1}{(x-1)^2} d x=t d t \\ & \quad I=\int \frac{t d t}{t}=\int d t=t=\sqrt{\frac{x-3}{x-1}}+C \\ & \because f(x)=\frac{x-3}{x-1} \\ & f(-1)-f(0)=\frac{-4}{-2}-3=2-3=-1 \end{aligned} \end{aligned} $

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

Put $2-x^2=t^2$

$ \begin{array}{ll} \therefore & -x d x=t d t \text { and } x^2=2-t^2 \\ \Rightarrow & I=\int \frac{-t d t}{\left(t^2-1\right) t} \\ \Rightarrow & I=\int \frac{d t}{1-t^2}=\frac{1}{2} \ln \left(\frac{1+t}{1-t}\right) \\ & =\frac{1}{2} \ln \left(\frac{1+\sqrt{2-x^2}}{1-\sqrt{2-x^2}}\right)+C \end{array} $