Indefinite Integration

To solve the integral $\int \frac{1}{x^{m} \sqrt[m]{x^{m}+1}} \, dx$, we start by making the substitution $u = 1 + \frac{1}{x^m}$. This implies:

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

Rearranging gives:

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

Now, rewrite the integral in terms of $u$:

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

This integral can be computed by:

$ \int -\frac{1}{m} u^{-\frac{1}{m}} \, du = -\frac{1}{m} \frac{u^{1 - \frac{1}{m}}}{1 - \frac{1}{m}} + C $

Simplifying the integral gives:

$ = -\frac{1}{m-1} u^{1 - \frac{1}{m}} + C $

Substitute back $u = 1 + \frac{1}{x^m}$:

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

Realizing that:

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

This simplifies to:

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

Finally, note that:

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

Thus, the expression becomes:

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

Let

$ \begin{aligned} I & =\int \frac{\tan x}{\sec ^2 x\left(1+\sec ^6 x\right)^{2 / 3}} d x \\ & =\int \frac{\tan x}{\sec ^2\left(\frac{\cos ^6 x+1}{\cos ^6 x}\right)^{2 / 3}} d x \\ & =\int \frac{\tan x}{\sec ^6 x\left(\cos ^6 x+1\right)^{2 / 3}} d x \\ & =\int \frac{\sin x \cos ^5 x}{\left(1+\cos ^6 x\right)^{2 / 3}} d x \end{aligned} $

Suppose, $1+\cos ^6 x=t$

$ 6 \cos ^5 x(-\sin x) d x=d t $

$ \sin x \cos ^5 x d x=-\frac{1}{6} d t $

So, $I=\int \frac{-1}{6} \frac{d t}{(t)^{2 / 3}}=-\frac{1}{2}(t)^{1 / 3}+C$

On putting the value of $t=1+\cos ^6 x$

$ I=-\frac{1}{2}\left(1+\cos ^6 x\right)^{1 / 3}+C $

Let $I=\int \frac{1 x d x}{(x-1)^{5 / 7}(x+1)^{9 / 7}}$

$ I=\int \frac{1}{(x+1)^2} \cdot \frac{1}{\left(\frac{x-1}{x+1}\right)^{5 / 7}} d x $

On putting $\quad \frac{x-1}{x+1}=t$

$ \begin{aligned} & \frac{1(x+1)-1(x-1)}{(x+1)^2} d x=d t \\ & \Rightarrow \quad \frac{2}{(x+1)^2} d x=d t \\ & I=\frac{1}{2} \int \frac{d t}{(t)^{5 / 7}}=\frac{1}{2} \frac{t^{2 / 7}}{2 / 7}+C \\ & =\frac{7}{4}\left(\frac{x-1}{x+1}\right)^{\frac{2}{7}}+C \end{aligned} $

$ \text { Let } I=\int \frac{1+\sqrt{3} \cot x}{1-\sqrt{3} \cot x} d x $

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

We have, $\int \frac{1}{\operatorname{cosec} x+\cos x} d x$

$ \begin{aligned} & =\frac{1}{2 \sqrt{3}} \log |f(x)|-\int \frac{\cos x-\sin x}{2+\sin 2 x} d x+C \\ & =\int \frac{\sin x+\cos x+\sin x-\cos x}{2+\sin 2 x} d x \\ & =\underbrace{\int \frac{\sin x+\cos x}{2+\sin 2 x} d x}_{\text {solve this }}-\underbrace{\int \frac{\cos x-\sin x}{2+\sin 2 x} d x}_{\text {matched to RHS }} \end{aligned} $

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

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

$ \begin{aligned} & \Rightarrow(\cos x+\sin x) d x=d t \\ & \quad I=\int \frac{d t}{3-t^2}=\frac{1}{2 \sqrt{3}} \ln \left(\frac{\sqrt{3}+t}{\sqrt{3}-t}\right) \\ & \therefore f(x)=\left(\frac{\sqrt{3}+(\sin x-\cos x)}{\sqrt{3}-(\sin x-\cos x)}\right) \end{aligned} $

$ \begin{aligned} f\left(\frac{\pi}{3}\right) & =\frac{\sqrt{3}+\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)}{\sqrt{3}-\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)} \\ & =\frac{3 \sqrt{3}-1}{\sqrt{3}+1} \end{aligned} $

Let $I=\int x^{49} \frac{\tan ^{-1}\left(x^{50}\right)}{1+\left(x^{50}\right)^2} d x$

On putting $x^{50}=t$

$ 50 x^{49} d x=d t $

$ \begin{aligned} &\therefore \quad l=\frac{1}{50} \int \frac{\tan ^{-1} t}{1+t^2} d t\\ &\text { Again, putting }\\ &\begin{aligned} \tan ^{-1} t & =u \\ \Rightarrow \frac{1}{1+t^2} d t & =d u \end{aligned} \end{aligned} $

$ \begin{aligned} \therefore \quad l & =\frac{1}{50} \int u d u=\frac{u^2}{100}+C \\ & =\frac{\left(\tan ^{-1} x^{50}\right)}{100}+C \end{aligned} $

But, $l=k\left(\tan ^{-1} x^{50}\right)^2+C$         [given]

$ \therefore \quad k=\frac{1}{100} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } I=\int(\log x)^3 x^5 d x \\ & I=(\log x)^3 \int x^5 d x-\int\left[\frac{d}{d x}(\log x)^3 \int x^5 d x\right] d x \\ & =(\log x)^3 \times \frac{x^6}{6}-\int\left[\frac{3 \times 1}{x}(\log x)^2 \times \frac{x^6}{6}\right] d x \\ & =\frac{x^6}{6}(\log x)^3-\frac{1}{2} \int\left[(\log x)^2 \times x^5\right] d x \\ & =\frac{x^6}{6}(\log x)^3-\frac{1}{2}\left[(\log x)^2 \int x^5 d x\right. \\ & \quad-\int\left[\frac{d}{d x}(\log x)^2 \int x^5 d x\right] d x \\ & =\frac{x^6}{6}(\log x)^3-\frac{1}{2}\left[(\log x)^2 \times \frac{x^6}{6}\right. \\ & \left.\quad-\int\left(\frac{2 \log x}{x} \times \frac{x^6}{6}\right)\right] d x \\ & =\frac{x^6}{6}(\log x)^3-\frac{1}{2}\left[\frac{x^6}{6}(\log x)^2\right. \\ & \left.-\frac{1}{3} \int \log x \cdot x^5 d x\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{x^6}{6}(\log x)^3-\frac{x^6(\log x)^2}{12} \\ & \quad+\frac{1}{6}\left[\log x \cdot \frac{x^6}{6}-\int \frac{1}{x} \cdot \frac{x^6}{6} d x\right] \\ & =\frac{x^6}{6}(\log x)^3-\frac{x^6}{12}(\log x)^2+\frac{1}{6} \\ & \quad\left[\frac{\log x \cdot x^6}{6}-\frac{1}{6} \cdot \frac{x^6}{6}\right] \\ & =\frac{x^6}{6}(\log x)^3-\frac{x^6}{12}(\log x)^2+\frac{x^6}{36} \cdot \log x \\ & \quad-\frac{x^6}{216}+C \end{aligned} $

$ \begin{aligned} & =\frac{x^6}{216}\left[36(\log x)^3-18(\log x)^2\right. \\ & +6(\log x)-1]+C \end{aligned} $

Compare with given equation, then we have, $A=216, B=36$

$ C=-18, D=6 $

Then, $(A-(B+C+D))$

$ =216-(36+6-18)=192 $

Hence, the value of

$ (A-(B+C+D))=192 $

Let $x^2=y$

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

Hence, we can write,

$ \begin{aligned} \frac{1}{(y+4)(y+1)} & =\frac{A}{(y+4)}+\frac{B}{(y+1)} \\ & =\frac{-1 / 3}{(y+4)}+\frac{1 / 3}{(y+1)} \end{aligned} $

Putting back, $y=x^2$

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

Integrating on both sides w.r.t. $x$, we get

$ \begin{aligned} & \int \frac{1}{\left(x^2+4\right)\left(x^2+1\right)} d x \\ & =\int \frac{-1 / 3}{\left(x^2+4\right)} d x+\int \frac{1}{3\left(x^2+1\right)} d x \\ & =\frac{-1}{3} \int \frac{1}{x^2+2^2} d x+\frac{1}{3} \int \frac{1}{x^2+1^2} d x \\ & =\frac{-1}{3} \times \frac{1}{2} \tan ^{-1} \frac{x}{2}+\frac{1}{3} \times \frac{1}{1} \tan ^{-1} \frac{x}{1}+C \\ & \quad\left[\because \int \frac{1}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C\right] \\ & =\frac{-1}{6} \tan ^{-1} \frac{x}{2}+\frac{1}{3} \tan ^{-1} x+C \\ & =\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C \end{aligned} $

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

$ \begin{aligned} & =\int \frac{d x}{(x-1)^{3 / 4}(x+2)^2(x+2)^{-3 / 4}} \\ & =\int \frac{(x+2)^{3 / 4}}{(x-1)^{3 / 4}} \cdot \frac{1}{(x+2)^2} d x \\ & =\frac{+1}{3} \int\left[\frac{(x-1)}{(x+2)}\right]^{-3 / 4} \cdot\left(\frac{+3}{(x+2)^2}\right) d x \\ & \text { On putting } t=\frac{x-1}{x+2} \\ & \quad \begin{aligned} \Rightarrow d t & =\frac{+3}{(x+2)^2} d x \\ I & =\frac{+1}{3} \int(t)^{-3 / 4} d t \\ & =\frac{+1}{3} \times \frac{+4 t^{1 / 4}}{1}+C \\ & =\frac{4}{3}\left[\frac{x-1}{x+2}\right]^{1 / 4}+C \end{aligned} \end{aligned} $

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

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

$ \begin{aligned} &\begin{aligned} & =\int\left(\sqrt{\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^2}\right) \\ & \left.+\sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}\right) d x \\ & =\int\left[\left|\sin \frac{x}{2}-\cos \frac{x}{2}\right|+\left|\sin \frac{x}{2}+\frac{\cos \frac{x}{2}}{2}\right|\right] d x \end{aligned}\\ &\text { In the given range }\\ &\begin{aligned} & \left|\sin \frac{x}{2}-\cos \frac{x}{2}\right|=\sin \frac{x}{2}-\cos \frac{x}{2} \\ & \text { and }\left|\sin \frac{x}{2}+\cos \frac{x}{2}\right|=-\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) \\ & \therefore I=\int\left(\sin \frac{x}{2}-\cos \frac{x}{2}-\sin \frac{x}{2}-\cos \frac{x}{2}\right) d x \\ & \quad=\int\left(-2 \cos \frac{x}{2}\right) d x \\ & \quad=-4 \sin x / 2+C \end{aligned} \end{aligned} $

From question, we get

$ f(x)=-4 \sin \frac{x}{2} $

$ \begin{aligned}Now, f\left(\frac{\pi}{3}\right)-f(0) & =-4 \sin \left(\frac{\pi}{6}\right)+4 \sin 0 \\ & =-4 \times \frac{1}{2}+0=-2 \end{aligned} $

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

Let $2 \sin ^2 x-3 \sin x+4=t$

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

[putting the value of $t$ ]

On comparing RHS from question, we get

$ f(x)=\log \left|2 \sin ^2 x-3 \sin x+4\right| $

Now, $\left.f\left(\frac{\pi}{2}\right)-f(0)=\log \right\rvert\, 2 \sin ^2(\pi / 2)$

$ \begin{aligned} & -3 \sin (\pi / 2)+4-\log \left|2 \sin ^2(0)-3 \sin (0)+4\right| \\ & =\log |2 \times 1-3 \times 1+4|-\log |0-0+4| \\ & =\log |3|-\log |4| \\ & =\log \left|\frac{3}{4}\right| \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{2 x+3}{\sqrt{3 x^2-2 x+1}} d x \\ & =\int \frac{2 x-\frac{2}{3}+\frac{2}{3}+3}{\left(3 x^2-2 x+1\right)^{1 / 2}} d x \\ & =\int \frac{2 x-\frac{2}{3}}{\left(3 x^2-2 x+1\right)^{1 / 2}} d x+\int \frac{(11 / 3) d x}{\left(3 x^2-2 x+1\right)^{1 / 2}} \end{aligned} $

Now, $I_1=\int \frac{(2 x-2 / 3) d x}{\left(3 x^2-2 x+1\right)^{1 / 2}}$

Let $3 x^2-2 x+1=t \Rightarrow x^2-\frac{2}{3} x+\frac{1}{3}=\frac{t}{3}$

$ \begin{aligned} & \Rightarrow\left(2 x-\frac{2}{3}\right) d x=\frac{1}{3} d t \\ & \therefore I_1=\frac{1}{3} \int \frac{1}{t^{1 / 2}} d t=\frac{1}{3} \frac{t^{1 / 2}}{1 / 2}+C_1 \\ & \quad=\frac{2}{3} t^{1 / 2}+C_1=\frac{2}{3}\left(3 x^2-2 x+1\right)^{1 / 2}+C_1 \end{aligned} $

Now,

$ \begin{aligned} I_2 & =\frac{11}{3 \sqrt{3}} \int \frac{d x}{\left(x^2-\frac{2}{3} x+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2+\frac{1}{3}\right)^{1 / 2}} \\ & =\frac{11}{3 \sqrt{3}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{3}\right)^2+\frac{2}{9}}} \\ & =\frac{11}{3 \sqrt{3}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2}} \end{aligned} $

$ \begin{aligned} & =\frac{11}{3 \sqrt{3}} \sinh ^{-1}\left[\frac{x-\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right]+C_2 \\ & =\frac{11}{3 \sqrt{3}} \sinh ^{-1}\left(\frac{3 x-1}{\sqrt{2}}\right)+C_2 \end{aligned} $

On putting the value of $I_1$ and $I_2$, we get

$ \begin{aligned} I=\frac{2}{3} \sqrt{3 x^2-2 x+1} & \\ & +\frac{11}{3 \sqrt{3}} \sinh ^{-1}\left(\frac{3 x-1}{\sqrt{2}}\right)+C \end{aligned} $

[where, $C=C_1+C_2$ ]

$ \text { } \begin{aligned} Let\,\,I & =\int \frac{d x}{16-7 \sin ^2 x} \\ & =\int \frac{\sec ^2 x d x}{16\left(1+\tan ^2 x\right)-7 \tan ^2 x} \end{aligned} $

$ \begin{aligned} &=\int \frac{\sec ^2 x d x}{16+9 \tan ^2 x}\\ &\text { Let } \quad \tan x=t\\ &\begin{aligned} & \Rightarrow \sec ^2 x d x=d t=\int \frac{d t}{16+9 t^2} \\ & \because \quad \int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a} \\ & =\frac{1}{12} \tan ^{-1}\left(\frac{3 t}{4}\right)+C=\frac{1}{12} \tan ^{-1}\left(\frac{3 \tan x}{4}\right)+c \end{aligned} \end{aligned} $

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

$ \begin{aligned} & \text { Let } \sec x+\tan x=t \\ & \Rightarrow \sec x-\tan x=\frac{1}{t} \end{aligned} $

Now, $\sec x(\sec x+\tan x) d x=d t$

$ \begin{aligned} & \sec x d x=\frac{d t}{t} \text { and } \frac{1}{2}\left(t+\frac{1}{t}\right)=\sec x \\ & \begin{aligned} \therefore I & =\frac{1}{2} \int \frac{\left(t+\frac{1}{t}\right)}{t^2} \frac{d t}{t}=\frac{1}{2} \int\left(\frac{1}{t^2}+\frac{1}{t^4}\right) d t \\ & =\frac{1}{2}\left(\frac{t^{-2+1}}{-2+1}+\frac{t^{-4+1}}{-4+1}\right)+C \\ & =\frac{1}{2}\left(-\frac{1}{t}-\frac{1}{3 t^3}\right)+C=-\frac{1}{6}\left(\frac{3 t^2+1}{t^3}\right)+C \\ & =-\frac{1}{6}\left(\frac{1+3(\sec x+\tan x)^2}{(\sec x+\tan x)^3}\right)+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } I=\int \frac{1}{3 \cos x-4 \sin x+5} d x \\ & \cos x=\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} \text { and } \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} \\ & \therefore I=\int \frac{d x}{3\left(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)-4\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)+5} \\ & =\int \frac{\left(1+\tan ^2 \frac{x}{2}\right) d x}{3-3 \tan ^2 \frac{x}{2}-8 \tan \frac{x}{2}+5+5 \tan ^2 \frac{x}{2}} \\ & =\int \frac{\sec ^2 \frac{x}{2} d x}{2 \tan ^2 \frac{x}{2}-8 \tan \frac{x}{2}+8} \\ & \begin{aligned} \text { Let } \tan \frac{x}{2} & =t \Rightarrow \sec ^2 \frac{x}{2} d x=2 d t \\ \therefore \quad I & =\int \frac{2 d t}{2 t^2-8 t+8}=\int \frac{d t}{t^2-4 t+4} \\ & =\int(t-2)^{-2} d t \\ & =\frac{(t-2)^{-1}}{-1}+C=\frac{1}{2-t}+C \\ & =\frac{1}{2-\tan }+C \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{a x}{(x-2)\left(x^2+1\right)} \\ & \frac{1}{(x-2)\left(x^2+1\right)}=\frac{A}{x-2}+\frac{B x+C}{x^2+1} \\ & \Rightarrow 1=A\left(x^2+1\right)+(B x+C)(x-2) \end{aligned} $

on comparing coefficients of $x^2, x$ and constants and solving them, we have

$ \begin{aligned} & A=\frac{1}{5}, B=-\frac{1}{5} \text { and } C=-\frac{2}{5} \\ & \therefore I=\frac{1}{5} \int \frac{d x}{x-2}-\frac{1}{5} \int \frac{x}{x^2+1} d x-\frac{2}{5} \int \frac{d x}{x^2+1} \\ & =\frac{1}{5} \log |(x-2)|-\frac{1}{5} \log \sqrt{x^2+1}-\frac{2}{5} \tan ^{-1} x+C \\ & =\frac{1}{5}\left[\log \frac{|x-2|}{\sqrt{x^2+1}}-2 \tan ^{-1} x\right]+C \end{aligned} $

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

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

Put $x=0$,

$ \begin{aligned} & 1=B(-1)^2+C(-1)+D \\ & 1=B-C+D \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Put $x=1$,

$ 2=D(2) \Rightarrow D=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Put $x=-1$,

$ \begin{aligned} & 0=4(B-A)+C(-2)(2)+D(2) \\ \Rightarrow & 0=4 B-4 A-4 C+2 D \\ \Rightarrow & 2 B-2 A-2 C+D=0 \\ \Rightarrow & 2 B-2 A-2 C=-1 \\ \Rightarrow & 2(B-C)-2 A=-1 \quad[\because(B-C)=0] \\ \Rightarrow & 0-2 A=-1 \\ \Rightarrow & A=\frac{1}{2} \end{aligned} $

Put $x=2$,

$ \begin{aligned} & 3=(2 A+B)+C(5)+D(5) \\ \Rightarrow & 3=2 \times \frac{1}{2}+B+5 C+5 \\ \Rightarrow & B+5 C=-3 \text { and } B-C=0 \end{aligned} $

Now, $B=-\frac{3}{6}=-\frac{1}{2}$ and $C=-\frac{1}{2}$

Now,

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

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} \int \frac{\sin ^2 x}{\cos ^4 x} d x & =\int \tan ^2 x \sec ^2 x d x \\ & =\frac{\tan ^3 x}{3}+C \end{aligned} \end{aligned} $

So, $(1) \rightarrow(C)$

Also, $\int \frac{\sin ^4 x}{\cos ^2 x} d x=\int \frac{\sin ^2 x\left(1-\cos ^2 x\right)}{\cos ^2 x}dx$

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

$(2) \rightarrow(D)$

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

(3) $\rightarrow$ (B)

$ \begin{aligned} &For \text { (4) } \int \frac{\sin ^3 x}{\cos ^3 x} d x=\int \tan ^3 x d x \\ & =\int \tan x \sec ^2 x d x-\int \tan x d x \\ & =\frac{\tan ^2 x}{2}+\log |\cos x|+C \end{aligned} $

$(4) \rightarrow(\mathrm{A})$

Given,

$ \begin{aligned} & \int \frac{x}{(a+x)^5} d x=\frac{1}{k(a+x)^4}(f(x))+C \\ & \text { We have, LHS }=\int \frac{x}{(a+x)^5} d x \\ & \begin{aligned} \text { or, } I & =\int \frac{a+x-a}{(a+x)^5} d x \\ \quad & =\int \frac{a+x}{(a+x)^5} d x-\int \frac{a}{(a+x)^5} d x \\ & =\int \frac{1}{(a+x)^4} d x-\int \frac{a}{(a+x)^5} d x \\ & =\frac{-1}{3} \frac{1}{(a+x)^3}+\frac{a}{4} \frac{1}{(a+x)^4}+C \\ & =\frac{-4 x-a}{12(a+x)^4}+C \end{aligned} \end{aligned} $

On comparing it with given equation, we get

$ f(x)=-4 x-a \text { and } k=12 $

Then,

$ \begin{aligned} \frac{f(-a)}{a k} & =\frac{4 a-a}{a \times 12} \\ & =\frac{3 a}{12 a}=\frac{1}{4} \end{aligned} $

$ \int x^4(\log x)^3 d x$

$ \begin{aligned} = & (\log x)^3 \frac{x^5}{5}-\int 3(\log x)^2 \cdot \frac{1}{x} \cdot \frac{x^5}{5} d x \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{5} \int x^4(\log x)^2 d x \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{5}\left[(\log x)^2 \frac{x^5}{5}\right.\left.\quad-\int 2(\log x) \cdot \frac{1}{x} \cdot \frac{x^5}{5} d x\right] \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{25}(\log x)^2 x^5 \quad+\frac{6}{25} \int x^4(\log x) d x \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{25}(\log x)^2 x^5+\frac{6}{25} \quad\left[\frac{x^5}{5}(\log x)-\int \frac{1}{x} \cdot \frac{x^5}{5} d x\right] \\ = & \frac{x^5}{5}(\log x)^3-\frac{3}{25}(\log x)^2 x^5 \quad+\frac{6}{125}(\log x) x^5-\frac{6}{125} \frac{x^5}{5}+K \\ = & x^5\left\{\frac{1}{5}(\log x)^3-\frac{3}{25}(\log x)^2\right. \left.\quad+\frac{6}{125}(\log x)-\frac{6}{625}\right\}+C \end{aligned} $

Thus, $A=\frac{1}{5}, B=-\frac{3}{25}, C=\frac{6}{125}$ and

$ D=-\frac{6}{625} $

$\begin{aligned} \therefore A+ & B+C+5 D \\ & =\frac{1}{5}-\frac{3}{25}+\frac{6}{125}-\frac{6}{125}=\frac{2}{25}\end{aligned}$

$\begin{aligned} &(x-1)(x-2)(x-3) \\ & \quad=x^3-6 x^2+11 x-6\end{aligned}$

$\begin{aligned} & \left.x^3-6 x^2+11 x-6\right) x^4 \\ & \frac{x^4-6 x^3+11 x^2-6 x+6}{6 x^3-11 x^2+6 x} \\ & \frac{6 x^3-36 x^2+66 x-36}{(-)(+)(-)} \\ & 25 x^2-60 x+36 \\ & \therefore x^4=(x+6)(x-1)(x-2)(x-3) \\ & \Rightarrow \frac{25 x^2-60 x+36}{(x-1)(x-2)(x-3)} \\ & (x-1)(x-2)(x-3) \\ & x^4(x+6) \\ & p(x)=(x+6) \\ & \text { and } \frac{25 x^2-60 x+1 / 6}{(x-1)(x-2)(x-3)} \\ & =\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\ & \Rightarrow C=\frac{25\left(3^2\right)-60 \times 3+36}{(3-1)(3-2)}=\frac{81}{2} \\ & \therefore p\left(\frac{3}{2}\right)+C=\left(\frac{3}{2}+6\right)+\frac{81}{2}=48\end{aligned}$

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

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

$ \begin{aligned} & \ \int \frac{x^2\left(\sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x \\ & \begin{aligned} =\int \frac{x^2 \sec ^2 x+x^2 \tan x}{(x \tan x+1)^2} d x \\ =\int\left[\frac{-x}{x \tan x+1}+\frac{x\left(x \sec ^2 x-1\right)}{(x \tan x+1)^2}\right. \\ \left.+\frac{2 x^2 \tan x+2 x}{(x \tan x+1)^2}\right] d x \\ I=\int\left[\frac{-x}{x \tan x+1}+x \frac{x \sec ^2 x-1}{(x \tan x+1)^2}\right] d x \\ I_1 \\ +\int \frac{2 x^2 \tan x+2 x}{(x \tan x+1)^2} d x \\ I_2 \end{aligned} \\ & I_1=\int\left[\frac{-x}{x \tan x+1}+x \frac{x I_1+I_2}{\left(x \sec ^2 x-1\right.}\right] d x \\ & \because \int\left(f(x)+x f^{\prime}(x)\right) d x=x f(x)+C \\ & \therefore I_1=\frac{-x^2}{x \tan x+1} \end{aligned} $

$\begin{aligned} & \begin{aligned} & I_2=\int \frac{2 x^2 \tan x+2 x}{(x \tan x+1)^2} d x \\ &=\int \frac{2 x}{x \tan x+1} d x \\ &=\int \frac{2 x \cos x}{x \sin x+\cos x} d x \\ & \text { Let } x \sin x+\cos x=t \\ & \therefore \quad I_2=\int \frac{2 d t}{t}=2 \ln t \\ & I_2=2 \ln |x \sin x+\cos x| \\ & I=\frac{-x^2}{x \tan x+1}+2 \ln |x \sin x+\cos x|+C\end{aligned}\end{aligned}$

$ \begin{aligned} & \\ & I=\int \sin (101 x)(\sin x)^{99} d x \\ & =\int \sin (100 x+x)(\sin x)^{99} d x \\ & =\int \sin (100 x) \cos x(\sin x)^{99} d x+ \\ & \int \cos (100 x) \sin ^{100} x d x \\ & =\int \cos x(\sin x)^{99} \sin 100 x d x+I_1 \\ & I_1=\int \cos (100 x) \sin ^{100} x d x \\ & =\sin ^{100} x \int \cos 100 x d x- \\ & \int\left[\left(\int \cos 100 x d x\right)\left(\frac{d \sin ^{100} x}{d x}\right)\right] d x \\ & =\frac{\sin ^{100} x \sin 100 x}{100}- \\ & \int\left(\frac{\sin 100 x}{100} \cdot 100(\sin x)^{99} \cos x\right) d x \\ & I=\int \cos x(\sin x)^{99} \sin 100 x d x \\ & +\frac{\sin ^{100} x \sin 100 x}{100}-\int \cos x(\sin x)^{99} \\ & \sin 100 x \times d x+C \\ & I=\int \frac{\sin ^{100} x \sin (100 x)}{100}+C \\ & \lambda=100, \mu=100 \\ & \therefore \frac{\lambda}{\mu}=1 \end{aligned} $

$ \begin{aligned} & \ \int e^x\left(\sin ^2 2 x-8 \cos 4 x\right) d x \\ & =\int e^x\left(\frac{1-\cos 4 x}{2}+2 \sin 4 x\right. \\ & -2 \sin 4 x-8 \cos 4 x) d x \\ & =\int e^x\left(\frac{1-\cos 4 x}{2}+2 \sin 4 x\right) d x \\ & -2 \int e^x(\sin 4 x+4 \cos 4 x) d x \\ & =e^x\left(\frac{1-\cos 4 x}{2}\right)-2 e^x \sin 4 x+C \\ & =e^x\left(\frac{1-\cos 4 x}{2}-2 \sin 4 x\right)+C \\ & \Rightarrow f(x)=\frac{1-\cos 4 x}{2}-2 \sin 4 x \\ & \therefore f\left(\frac{\pi}{4}\right)=\frac{1+1}{2}-0=1 \end{aligned} $

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

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

$ \begin{aligned} & =\frac{A x(2 x-3)+B(2 x-3)+C x^2}{x^2(2 x-3)} \\ \Rightarrow x-2 & =x(2 x-3) A+(2 x-3) B+C x^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

At $x=0$,

$ -2=-3 B \Rightarrow B=\frac{2}{3} $

At $x=\frac{3}{2}$,

$ \frac{-1}{2}=C \cdot \frac{9}{4} \Rightarrow C=\frac{-2}{9} $

At $x=2$,

$ \begin{aligned} 0 & =2 A+B+4 C \\ \Rightarrow \quad A & =\frac{1}{2}(-B-4 C) \\ \Rightarrow \quad A & =\frac{1}{2}\left(\frac{-2}{3}+\frac{8}{9}\right)=\frac{1}{2}\left(\frac{-6+8}{9}\right) \\ & =\frac{1}{2} \times \frac{2}{9}=\frac{1}{9} \\ 2(A-C) & =2\left(\frac{1}{9}+\frac{2}{9}\right)=2 \cdot \frac{3}{9}=\frac{2}{3}=B \end{aligned} $

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

On comparing the corresponding coefficients

$ \begin{aligned} A+C & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ A+B+D & =1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ A+B+C & =-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ B+D & =1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{aligned} $

From Eqs. (i)and (iii), we have

$ B=-1 $

From Eq. (iv), we have

$ -1+D=1 \Rightarrow D=2 $

From Eq. (ii), we have

$ \begin{aligned} A+(-1)+2 & =1 \\ A & =0 \end{aligned} $

From Eq. (i), we have

$ \begin{aligned} A+C & =0 \\ 0+C & =0 \Rightarrow C=0 \end{aligned} $

$ \begin{aligned} \therefore A+2 B+C+2 D & \\ & =0+2(-1)+0+2(2) \\ & =-2+4=2 \end{aligned} $

$ \text { } \begin{aligned} & f(x)=\int \frac{2-3 \sin ^2 x}{1+\cos 2 x} d x \\ f(x) & =\int \frac{2-3 \sin ^2 x}{2 \cos ^2 x} d x \\ & =\int\left[\sec ^2 x-\frac{3}{2} \tan ^2 x\right] d x \\ & =\int\left[\sec ^2 x-\frac{3}{2}\left(\sec ^2 x-1\right)\right] d x \\ & =\int\left[\sec ^2 x-\frac{3}{2} \sec ^2 x+\frac{3}{2}\right] d x \\ & =\int\left[\frac{-1}{2} \sec ^2 x+\frac{3}{2}\right] d x \\ f(x) & =\frac{-1}{2} \tan x+\frac{3}{2} x+c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} &\begin{aligned} f\left(\frac{\pi}{4}\right) & =\frac{-1}{2} \tan \frac{\pi}{4}+\frac{3}{2} \cdot \frac{\pi}{4}+c \\ 1 & =-\frac{1}{2}+\frac{3 \pi}{8}+c \\ \Rightarrow \quad c & =\frac{3}{2}-\frac{3 \pi}{8} \Rightarrow c=\frac{3}{8}(4-\pi) \end{aligned}\\ &\text { From Eq. (i), we have }\\ &\begin{aligned} & f(x)=\frac{-1}{2} \tan x+\frac{3}{2} x+\frac{3}{8}(4-\pi) \\ & f(0)=0+0+\frac{3}{8}(4-\pi)=\frac{3}{8}(4-\pi) \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } I=\int\left(\frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}\right)^2 d x \\ & \begin{aligned} I & =\int\left(\frac{\sin x+2 \sin x \cos x}{1+\cos x+2 \cos ^2 x-1}\right)^2 d x \\ I & =\int\left[\frac{\sin x(1+2 \cos x)}{\cos x(1+2 \cos x)}\right]^2 d x \\ & =\int \tan ^2 x d x \,\,\,\,\left[\because \cos x \neq \frac{1}{2}\right]\\ & =\int\left(\sec ^2 x-1\right) d x \\ & =\tan x-x+c \end{aligned} \end{aligned} \end{aligned} $

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

$ =\frac{1}{2}\left[\begin{array}{c} \int \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+(\sqrt{5})^2} d x -\int \frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}\right)^2+1} d x \end{array}\right] $

$ \begin{aligned} & =\frac{1}{2}\left[\frac{1}{\sqrt{5}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{5}}\right)-\tan ^{-1}\left(x+\frac{1}{x}\right)\right] \\ & +k \\ & =\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{5} x}\right)-\frac{1}{2} \tan ^{-1}\left(\frac{x^2+1}{x}\right) \\ & +k \\ & =a \tan ^{-1}\left(\frac{b\left(x^2-1\right)}{x}\right)+c \tan ^{-1}\left(\frac{d\left(x^2+1\right)}{x}\right) \\ & +k \\ & \therefore a=\frac{1}{2 \sqrt{5}}, b=\frac{1}{\sqrt{5}}, c=\frac{-1}{2} \text { and } d=1 \\ & \text { Now, } 5(c+d+a b)=5\left(\frac{-1}{2}+1+\frac{1}{10}\right) \\ & =5\left(\frac{1}{2}+\frac{1}{10}\right) \\ & =\frac{5}{2}+\frac{1}{2}=\frac{6}{2} \\ & =3 \end{aligned} $

Given,

$ \begin{aligned} & \frac{2 x^2-3 x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3} \\ & \Rightarrow 2 x^2-3 x+5=A(x-7)^2 +B(x-7)+C\end{aligned} $

On putting $x=7$ both sides, we get

$ \begin{aligned} & (2 \times 49)-(3 \times 7)+5=0+0+C \\ & \Rightarrow \quad C=82 \\ & \therefore 2 x^2-3 x+5=A(x-7)^2+B(x-7)+82 \\ & \Rightarrow 2 x^2-3 x+5 \\ & \quad=A x^2-(14 A-B) x+7(7 A-B)+82 \end{aligned} $

On comparing both sides, we get

$ \begin{array}{rlrl} & A =2,14 A-B=3 \\ \Rightarrow & 28-B =3 \\ \Rightarrow & B =25 \\ \therefore & 2 A-3 B+C =4-75+82=11 \end{array} $

Given,

$ \begin{aligned} & \frac{3 x^2+a x+3}{(2 x+3)\left(x^2+2\right)}=\frac{3}{2 x+3}+\frac{B x+C}{x^2+2} \\ & \Rightarrow 3 x^2+a x+3 \\ & \quad \quad=3\left(x^2+2\right)+(B x+C)(2 x+3) \\ & \Rightarrow 3 x^2+a x+3 \\ & \quad=3 x^2+6+2 B x^2+3 B x+2 C x+3 C \\ & \Rightarrow 3 x^2+a x+3 \\ & \quad=(3+2 B) x^2+(3 B+2 C) x+3 C+6 \end{aligned} $

On comparing both sides, we get

$ \begin{array}{rlrlrl} 3+2 B=3, & 3 C+6=3, & 3 B+2 C & =a \\ \Rightarrow \quad B=0 & C=-1 & 0-2 & =a \\ & & a & =-2 \end{array} $

$ \therefore a(B+C)=(-2)(0-1)=2 $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \int\left(2^x-\sqrt{1+\sin 2 x}+\frac{1}{x^2}-\frac{1}{x}\right) d x \\ & \frac{3 \pi}{4} < x <\frac{7 \pi}{4} \\ & =\int\left(2^x-\sqrt{(\sin x+\cos x)^2}+\frac{1}{x^2}-\frac{1}{x}\right) d x \\ & =\int\left(2^x-|\sin x+\cos x|+\frac{1}{x^2}-\frac{1}{x}\right) d x \\ & =\int\left(2^x+\sin x+\cos x+\frac{1}{x^2}-\frac{1}{x}\right) d x \\ & =\frac{2^x}{\log _e 2}-\cos x+\sin x-\frac{1}{x}-\log _e x+c \end{aligned} \end{aligned} $

$ \begin{aligned} & \text {Given, } \tan \alpha=\frac{4}{3} \\ & \Rightarrow \cos \alpha=\frac{3}{5} \text { and } \sin \alpha=\frac{4}{5} \end{aligned} $

$ \text { } \begin{aligned}Now, \int & \frac{1}{3 \cos x-4 \sin x} d x \\ \quad= & \frac{1}{5} \int \frac{1}{\frac{3}{5} \cos x-\frac{4}{5} \sin x} d x \\ \quad= & \frac{1}{5} \int \frac{1}{\cos \alpha \cos x-\sin \alpha \sin x} d x \\ \quad= & \frac{1}{5} \int \frac{1}{\cos (\alpha+x)} d x \\ \quad= & \frac{1}{5} \int \sec (\alpha+x) d x \\ \quad= & \frac{1}{5} \ln \tan \left(\frac{\pi}{4}+\frac{\alpha}{2}+\frac{x}{2}\right)+c \end{aligned} $

Given, $x \neq(2 n+1) \frac{\pi}{2}$

Now,

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

On putting $1+\sin x=t, \cos x d x=d t$

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

The given options are wrong in this question. The correct answer is

$ \frac{\sin x}{(1+\sin x)^2}+C $

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

Comparing the coefficients on both sides, we get

$ A+C=0 \Rightarrow A=-C \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\quad B+D=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ 3 A+C=0 $

From Eq. (i), we get $A=C=0$

$ 3 B+D=-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On subtracting Eq. (ii) from Eq. (iii), we get

$ \begin{aligned} 2 B & =-3 \\ \Rightarrow \quad B & =-\frac{3}{2} \\ D & =1-\left(-\frac{3}{2}\right)[\because \text { from Eq. (ii) }] \\ & =\frac{5}{2} \end{aligned} $

$ \begin{aligned} &\text { Given, } g(x) \text { is the anti-derivative of }\\ &\begin{array}{ll} f(x) & \\ & f(x) d x=g(x) \\ \Rightarrow & f(x)=g^{\prime}(x) \\ & \frac{d}{d x}\left[\log _e\left(1+(g(x))^2+c\right]\right. \end{array} \end{aligned} $

$ =\frac{2 g(x) g^{\prime}(x)}{1+(g(x))^2}=\frac{2 g(x) f(x)}{1+(g(x))^2} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \begin{array}{r} f(x) \int\left(\tan ^2 x+\cot ^2 x\right)\left.+4 \frac{\left(\sin ^3 x+\cos ^3 x\right)}{\sin ^2 2 x}\right) d x \\ =\int\left[\sec ^2 x-1+\operatorname{cosec}^2 x-1\right] \left.+\frac{4\left(\sin ^3 x+\cos ^3 x\right)}{4 \sin ^2 x \cos ^2 x}\right] d x \\ =\int\left[\sec ^2 x+\operatorname{cosec}^2 x-2\right]\left.+\frac{\sin x}{\cos ^2 x}+\frac{\cos x}{\sin ^2 x}\right] d x \end{array} \end{aligned} \end{aligned} $

$ \begin{array}{r} =\int\left[\sec ^2 x+\operatorname{cosec}^2 x-2+\sec x \tan x\right. +\cot x \operatorname{cosec} x] d x \\ =\tan x-\cot x-2 x+\sec x -\operatorname{cosec} x+c \end{array} $

$ \begin{gathered} f\left(\frac{\pi}{4}\right)=0 \\ \Rightarrow \tan \frac{\pi}{4}-\cot \frac{\pi}{4}-2 \frac{\pi}{4}+\sec \frac{\pi}{4} \\ -\operatorname{cosec} \frac{\pi}{4}+c=0 \\ 1-1-\frac{\pi}{2}+\sqrt{2}-\sqrt{2}+c=0 \\ c=\frac{\pi}{2} \\ f(x)=\tan x-\cot x-2 x+\sec x \\ -\operatorname{cosec} x+\frac{\pi}{2} \\ \Rightarrow 3\left[f\left(\frac{\pi}{6}\right)+2\right]=3\left[\tan \frac{\pi}{6}-\cot \frac{\pi}{6}-\frac{2 \pi}{6}\right. \\ \left.+\sec \frac{\pi}{6}-\operatorname{cosec} \frac{\pi}{6}+\frac{\pi}{2}+2\right] \\ =3\left[\frac{1}{\sqrt{3}}-\sqrt{3}-\frac{\pi}{3}+\frac{2}{\sqrt{3}}-\frac{2}{1}+\frac{\pi}{2}+2\right] \\ =3\left[\frac{3}{\sqrt{3}}-\sqrt{3}+\frac{\pi}{6}\right] \end{gathered} $

$ =3\left[\sqrt{3}-\sqrt{3}+\frac{\pi}{6}\right]=\frac{\pi}{2} $

Let

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

Let $3 \sin x=t, \cos x d x=\frac{1}{3} d t$

$ \begin{aligned} I= & \frac{1}{3} \int \sqrt{4-t^2} d t \\ = & \frac{1}{6}\left[t \sqrt{4-t^2}+4 \sin ^{-1} \frac{t}{2}\right] \\ = & \frac{1}{6} \times\left[3 \sin x \sqrt{4-9 \sin ^2 x}-\right. \\ & \left.+4 \sin ^{-1}\left(\frac{3 \sin x}{2}\right)\right] \end{aligned} $

$ \begin{aligned} =\frac{1}{2} \sin x \sqrt{4} & -9 \sin ^2 x +\frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right)+c \end{aligned} $

$ \begin{aligned} & \text {Let } \frac{42-13 x}{x^2+x-6}=\frac{M}{x+3}+\frac{N}{x-2} \\ & 42-13 x=M x-2 M+N x+3 N \\ & \Rightarrow \quad M+N=-13 \text { and }-2 M+3 N=42 \\ & -2(-13-N)+3 N=42 \\ & 26+2 N+3 N=42 \\ & 5 N=16 \Rightarrow N=16 / 5 \\ & M=-13-\frac{16}{5}=\frac{-65-16}{5}=-\frac{81}{5} \\ & \frac{42-3 x}{x^2+x-16}=\frac{-81}{5(x+3)}+\frac{16}{5(x-2)} \\ & =\frac{-81}{5 x+15}+\frac{16}{5 x-10} \end{aligned} $

Here,$A=-81, B=16, l=p=5$

$ \begin{aligned} m & =15, q=-10 \\ \frac{A l p}{B m q} & =\frac{-81 \times 5 \times 5}{16 \times 15 \times(-10)}=\frac{27}{32} \end{aligned} $

$ \text { Let } I=\int \frac{3 x^6-2 x^4+x^2-2}{x^2+1} d x $

$ \begin{aligned} I & =\int\left(3 x^4-5 x^2+6-\frac{8}{x^2+1}\right) d x \\ & =\frac{3}{5} x^5-\frac{5}{3} x^3+6 x-8 \tan ^{-1} x+c \end{aligned} $

$ \begin{aligned} &\text {We have, }\\ &\begin{array}{r} \int \frac{\sin x \cdot \sec ^2 x-\tan x \cdot \sin x+\cos x}{(1-\cos 2 x)} d x \\ =\int \frac{\sin x \sec ^2 x-\tan x \cdot \sin x+\cos x}{2 \sin ^2 x} d x \\ =\frac{1}{2}\left[\int \frac{\sin x \sec ^2 x}{\sin ^2 x} d x-\int \frac{\tan x \cdot \sin x}{\sin ^2 x} d x\right. \\ \left.+\int \frac{\cos x}{\sin ^2 x} d x\right] \end{array} \end{aligned} $

$ \begin{array}{r} =\frac{1}{2}\left[\int \frac{1+\tan ^2 x}{\sin x} d x-\int \frac{\tan x}{\sin x} d x\right. \\ \left.+\int \cot x \cdot \operatorname{cosec} x d x\right] \\ =\frac{1}{2}\left[\int \operatorname{cosec} x d x+\int \tan x \cdot \sec x d x\right. \\ \left.-\int \sec x d x+\int \cot x \cdot \operatorname{cosec} x d x\right] \end{array} $

$ \begin{aligned} & =\frac{1}{2}\left[\log \left(\tan \frac{x}{2}\right)+\sec x-\log \right. \\ & \left.\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)-\operatorname{cosec} x\right] \\ & \left.=\frac{1}{2}\left[\sec x-\operatorname{cosec} x-\log \left\lvert\, \frac{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}{\tan \left(\frac{x}{2}\right)}\right.\right]\right]+C \end{aligned} $

We have ,

$ f(x)=\int \frac{16 x^7+5 x^{10}}{\left(x^3+2+3 x^8\right)^2} d x $

and $f(0)=1$

$ \begin{aligned} \Rightarrow \int \frac{16 x^7+5 x^{10}}{x^{16}\left(x^{-5}+2 x^{-8}+3\right)^2} d x & =\int \frac{16 x^{-9}+5 x^{-6}}{\left(x^{-5}+2 x^{-8}+3\right)^2} d x \end{aligned} $

$ \begin{aligned} & \text { Let } x^{-5}+2 x^{-8}+3=d t \\ & \begin{array}{l} \Rightarrow\left(-5 x^{-6}+16 x^{-9}\right) d x=d t \\ \Rightarrow \quad\left(5 x^{-6}+16 x^{-9}\right) d x=-d t \end{array} \\ & \begin{array}{lr} \therefore \quad=\int \frac{-d t}{t^2}=-\int t^{-2} d t=\frac{1}{t}+c \end{array} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad f(x)=\frac{1}{x^{-5}+2 x^{-8}+3}+c \\ & =\frac{x^8}{x^3+2+3 x^8}+c \\ & \text { Since, } f(0)=1 \\ & \Rightarrow \quad c=1 \\ & \therefore \quad f(x)=\frac{x^8}{x^3+2+3 x^8}+1 \\ & \Rightarrow \quad f(-1)=\frac{1}{-1+2+3}+1 \\ & \Rightarrow \quad f(-1)=\frac{1}{4}+1=\frac{5}{4} \end{aligned} $

$ \begin{aligned} & \text { } I=\int \frac{(x+3)}{(x-1)^2(2 x-1)} d x \\ & \text { Let } \frac{x+3}{(2 x-1)(x-1)^2}=\frac{P}{2 x-1}+\frac{Q}{x-1} +\frac{R}{(x-1)^2} \end{aligned} $

$ \begin{aligned} \Rightarrow(x+3)=P(x-1)^2+Q(2 x-1) & (x-1) +R(2 x-1) \end{aligned} $

On putting $x=1, R=4$

On putting $x=1 / 2, P=14$

On putting $x=0, Q=-7$

$ \begin{aligned} & \begin{aligned} & \therefore I=\int\left[\frac{14}{2 x-1}+\frac{-7}{x-1}+\frac{4}{(x-1)^2}\right] d x \\ &=14 \times \frac{1}{2} \log (2 x-1)-7 \log (x-1) +4\left(\frac{-1}{x-1}\right)+k \\ &= \frac{-4}{x-1}+7 \log (2 x-1)-7 \log (x-1)+k \\ & \therefore \quad A=-4, B=7, C=-7 \\ & \Rightarrow \quad A+B+C=-4 \end{aligned} \end{aligned} $

$ \begin{aligned} & \int \frac{1+\cos 8 x}{\tan 2 x-\cot 2 x} d x \\ & =\int \frac{2 \cos ^2 4 x}{\frac{\sin 2 x}{\cos 2 x}-\frac{\cos 2 x}{\sin 2 x}} d x \\ & =\int \frac{2 \cos ^2 4 x}{\frac{\sin ^2 2 x-\cos ^2 2 x}{\cos 2 x \cdot \sin 2 x}} d x \\ & =\int \frac{2 \cos ^2 4 x}{\left(\frac{-\cos 4 x}{\sin 4 x}\right) \times 2} d x \\ & =-\int \cos 4 x \cdot \sin 4 x d x \\ & =\frac{-1}{2} \int \sin 8 x d x=\frac{-1}{2}\left(\frac{-\cos 8 x}{8}\right)+c \\ & =\frac{\cos 8 x}{16}+c \end{aligned} $

$ \begin{aligned} &\text { According to the question, }\\ &\begin{aligned} & f(x)=\frac{1}{16} \text { and } g(x)=8 x \\ \Rightarrow & f\left(\frac{1}{4}\right)=\frac{1}{16} \text { and } g\left(\frac{1}{4}\right)=8 \times \frac{1}{4}=2 \\ \therefore & f\left(\frac{1}{4}\right)+g\left(\frac{1}{4}\right)=\frac{1}{16}+2=\frac{33}{16} \end{aligned} \end{aligned} $

$f\left(\frac{2 x+1}{5 x+3}\right)=x+2$

On putting $\frac{2 x+1}{5 x+3}=t$

$ \begin{gathered} \Rightarrow 2 x+1=5 t x+3 t \\ \Rightarrow(5 t-2) x=1-3 t \\ x=\frac{1-3 t}{5 t-2} \end{gathered} $

$ \begin{aligned} & \therefore \quad f(t)=\frac{1-3 t}{5 t-2}+2 \\ & =\frac{1-3 t+10 t-4}{5 t-2} \\ & =\frac{7 t-3}{5 t-2} \\ & \begin{array}{l} \therefore \quad f(x)=\frac{7 x-3}{5 x-2} \\ \text { Now, } \int f(x) d x=\int\left(\frac{7 x-3}{5 x-2}\right) d x \\ =7 \int \frac{x}{5 x-2} d x-3 \int \frac{1}{5 x-2} d x \\ =\frac{7}{5} \int \frac{(5 x-2)+2}{(5 x-2)} d x-3 \int \frac{1}{5 x-2} d x \\ =\frac{7}{5} \int\left(1+\frac{2}{5 x-2}\right) d x-3 \int \frac{1}{5 x-2} d x \\ =\frac{7}{5}\left[x+2 \times \frac{1}{5} \log |5 x-2|\right] \\ -3 \times \frac{1}{5} \log |5 x-2|+c \\ =\frac{7}{5} x-\frac{1}{25} \log |5 x-2|+c \end{array} \end{aligned} $

$I=\int \frac{\cos \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)}{(3-2 x)^2} d x$

On putting $\frac{2 x+3}{3-2 x}=t$

$ \begin{aligned} & \Rightarrow \frac{d x}{(3-2 x)^2}=\frac{d t}{12} \\ & \Rightarrow \quad I=\frac{1}{12} \int \cos (\log t) d t \end{aligned} $

On putting $t=e^k \Rightarrow d t=e^k d k$

$ \begin{aligned} & \therefore I=\frac{1}{12} \int e^k \cdot \cos (k) d k \\ & =\frac{1}{12} \cdot \frac{e^k}{2}[\cos k+\sin k]+c \\ & \Rightarrow I=\frac{\left(\frac{2 x+3}{3-2 x}\right)}{24} \end{aligned} $

$ \begin{aligned} & {\left[\cos \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)+\sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)\right]+c} \\ & \Rightarrow \quad g(x)=\log \left(\frac{2 x+3}{3-2 x}\right) \\ & \quad f(x)=\frac{2 x+3}{3-2 x} \\ & \therefore \quad g(1)=\log (f(1)) \end{aligned} $

$ \begin{aligned} & \text { Given, } \frac{x^2-3 x+2}{(x-4)(x-3)^2} \\ & =\frac{A}{x-4}+\frac{B}{x-3}+\frac{C}{(x-3)^2} \\ & x^2-3 x+2=A(x-3)^2 \\ & +B(x-4)(x-3)+C(x-4) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

This is an identity so, true for every value of $x$.

On putting $x=3$ in Eq. (i), we have

$ C=-2 $

On putting $x=4$ in Eq. (i), we have

$ A=6 $

On putting $x=0$ in Eq. (i) and $A=6$ and $C=-2$, we have $B=-5$

Now, $A+B+C=6-5-2=-1$