Indefinite Integration

$ \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} $

$ \text { } \begin{aligned} I & =\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x \\ & =\int \frac{(\sqrt{1+x})^2+\sqrt{x(1+x)}}{\sqrt{x}+\sqrt{1+x}} d x \\ & =\int \frac{\sqrt{1+x}(\sqrt{1+x}+\sqrt{x})}{\sqrt{x}+\sqrt{1+x}} d x \\ I & =\int \sqrt{1+x} d x=\frac{2}{3}(1+x)^{3 / 2}+C \end{aligned} $

Given, $\int x^2 \cos ^2 x d x=\frac{1}{6} f(x)$ +g(x) \sin 2 x+h(x) \cos 2 x+C $

LHS

$ \begin{aligned} I & =\int x^2 \cos ^2 x d x=\int x^2\left(\frac{1+\cos 2 x}{2}\right) d x \\ I & =\frac{1}{2} \int\left(x^2+x^2 \cos 2 x\right) d x \\ & =\frac{1}{2} \int x^2 d x+\frac{1}{2} \int x^2 \cos 2 x d x \\ & =\frac{1}{2} \cdot \frac{x^3}{3}+\frac{1}{2} I_1 \end{aligned} $

Where $I_1=\int x^2 \cos ^2 x d x$

We have,

$ \begin{aligned} & \quad I=\int \frac{e^{\sin x}(\sin 2 x-8 \cos x)}{2(\sin x-3)^2} d x \\ \Rightarrow \quad & \quad I=\int \frac{e^{\sin x}(2 \sin x-8) \cos x}{2(\sin x-3)^2} d x \end{aligned} $

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

$ \begin{array}{ll} \Rightarrow & \quad I=\int \frac{e^t(t-4)}{(t-3)^2} d t \\ & \quad I=\int e^t\left(\frac{1}{t-3}-\frac{1}{(t-3)^2}\right) d t \\ & \quad\left[\because \int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+C\right] \\ \Rightarrow & \quad I=e^t \cdot \frac{1}{t-3}+C \\ \Rightarrow & \quad I=e^{\sin x} \frac{1}{\sin x-3}+C \end{array} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \int\left(3 t^2 \sin \frac{1}{t}-t \cos \frac{1}{t}\right) d t=f(t) \sin \frac{1}{t}+C \\ & \int \sin \frac{1}{t} \cdot 3 t^2 d t-\int t \cos \frac{1}{t} d t=f(t) \sin \frac{1}{t}+C \\ & \begin{aligned} \sin \frac{1}{t} \cdot t^3-\int \cos \left(\frac{1}{t}\right)\left(-\frac{1}{t^2}\right) t^3 d t & -\int t \cos \frac{1}{t} d t \end{aligned} \end{aligned} \end{aligned} $

$ \begin{array}{lc} \Rightarrow & t^3 \sin \frac{1}{t}+\int t \cos \frac{1}{t} d t -\int t \cos \frac{1}{t} d t=f(t) \sin \frac{1}{t}+C \\ \Rightarrow & t^3 \sin \frac{1}{t}=f(t) \sin \frac{1}{t}+C \\ \therefore & f(t)=t^3 \\ & f(2)=2^3=8 \end{array} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } I=\int(\log x)^3 \cdot x^4 d x \\ & I=\int t^3 e^{4 t} \cdot e^t d t \quad \text { Put } \log x=t \\ & x=e^t \\ & d x=e^t d t \\ & I=\int t^3 e^{5 t} d t \\ & \Rightarrow \quad I=e^{5 t}\left[\frac{t^3}{5}-\frac{3 t^2}{25}+\frac{6 t}{125}-\frac{6}{625}\right]+C \\ & \Rightarrow I=x^5 \\ & \begin{aligned} {\left[\begin{array}{rl} \frac{1}{5}(\log x)^3 & -\frac{3}{25}(\log x)^2 \\ & +\frac{6}{125}(\log x)-\frac{6}{625} \end{array}\right]+C } \end{aligned} \end{aligned} \end{aligned} $

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

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

$ \begin{aligned} & 1-2 \sin x \cos x=t^2 \\ & 2 \sin x \cos x=1-t^2 \\ & =\int \frac{2 d t}{\left(2-t^2\right)\left(1+t^2\right)}=-2 \int \frac{d t}{\left(t^2+1\right)\left(t^2-2\right)} \\ & =-\frac{2}{3} \int\left(\frac{1}{t^2-2}-\frac{1}{t^2+1}\right) d t \\ & =-\frac{2}{3}\left[\frac{1}{2 \sqrt{2}} \ln \left(\frac{t-\sqrt{2}}{t+\sqrt{2}}\right)-\tan ^{-1} t\right] \\ & =-\frac{1}{3 \sqrt{2}} \ln \left(\frac{\sin x-\cos x-\sqrt{2}}{\sin x-\cos x+\sqrt{2}}\right) +\frac{2}{3} \tan ^{-1}(\sin x-\cos x)+C \end{aligned} $

$ \begin{aligned} & \begin{array}{r} =\frac{1}{3 \sqrt{2}} \ln \left|\frac{\sin x-\cos +\sqrt{2}}{\sin x-\cos x-\sqrt{2}}\right| +\frac{2}{3} \tan ^{-1}(\sin x-\cos x)+C \end{array} \\ & \begin{aligned} \because A=\frac{1}{3 \sqrt{2}}, B=\frac{2}{3} \text { and } t=\sin x-\cos x \end{aligned} \\ & \because\left(\frac{B}{A}, t\right)=(2 \sqrt{2}, \sin x-\cos x) \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} $

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