⇒Indefinite Integration

Correct Option is (D)

Let $I=\int \frac{5\tan x}{\tan x-2}dx$

$I=\int \frac{5\sin x}{\sin x-2\cos x}dx$

Here $\frac{d}{dx}(\sin x-2\cos x)=\cos x+2\sin x$

$\begin{array}{rl}\therefore I& =\int \frac{(2\sin x+2\sin x+\sin x)+(2\cos x-2\cos x)}{\sin x-2\cos x}dx\\ & =\int \frac{(2\sin x+\cos x)+(2\sin x+\cos x)+(\sin x-2\cos x)dx}{\sin x-2\cos x}\\ & =\int \frac{2(2\sin x+\cos x)+(\sin x-2\cos x)}{\sin x-2\cos x}dx\\ & =\int dx+2\int \frac{2\sin x+\cos x}{\sin x-2\cos x}dx\\ & =x+2\log |\sin x-2\cos x|+c\end{array}$

From given data, $\mathrm{a}=2$

Correct Option is (C)

Let $I=\int [1+2\tan x(\tan x+\sec x){]}^{1/2}dx=\int {(1+2{\tan }^{2}x+2\tan x\sec x)}^{1/2}dx$

$\begin{array}{rl}& =\int {[(1+{\tan }^{2}x)+{\tan }^{2}x+2\sec x\tan x]}^{1/2}dx=\int {({\sec }^{2}x+{\tan }^{2}x+2\sec x\tan x)}^{1/2}dx\\ & =\int {[(\sec x+\tan x{)}^2]}^{1/2}dx=\int (\sec x+\tan x)dx=\int \sec xdx+\int \tan xdx\\ & =\log \mid \sec x+\tan x)-\log |\cos x|+c=\log \frac{|\sec x+\tan x|}{|\cos x|}+c\\ & =\log |\sec x(\sec x+\tan x)|+c\end{array}$

Correct Option is (A)

$\begin{array}{rl}\text{ Let I }& ={\tan }^{-1}(\sec x+\tan x)dx\\ & {=\int {\tan }^{-1}\left(\frac{1+\sin x}{\cos x}\right)dx=\int {\tan }^{-1}x\left[\frac{{(\cos \frac{x}{2}+\sin \frac{x}{2})}^2}{(\cos \frac{x}{2}+\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})}\right]]}_{dx}\\ & =\int {\tan }^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right)dx=\int {\tan }^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)dx\\ & =\int {\tan }^{-1}[\tan (\frac{\pi }{4}+\frac{x}{2})]dx=\int (\frac{\pi }{4}+\frac{x}{2})dx\\ & =\frac{\pi x}{4}+\frac{x^2}{4}+c\end{array}$

Correct Option is (A)

Let $\begin{array}{rl}& =\int \frac{x+\sin x}{1+\cos x}dx=\int \frac{x+\sin x}{2{\cos }^2\frac{x}{2}}dx\\ & =\int \frac{x}{2{\cos }^2\frac{x}{2}}dx+\int \frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\cos }^2\frac{x}{2}}dx=\frac{1}{2}\int x{\sec }^2\frac{x}{2}dx+\int \tan \frac{x}{2}dx\\ & =\frac{1}{2}[x\tan \frac{x}{2}(2)-\int 2\tan \frac{x}{2}dx]-2\log |\cos \frac{x}{2}|+c\\ & =x\tan \frac{x}{2}+2\log |\cos \frac{x}{2}|-2\log |\cos \frac{x}{2}|+c=x\tan \frac{x}{2}+c\end{array}$

Correct Option is (C)

Given ${\mathrm{f}}^{\prime }(x)=x-\frac{5}{x^5}$

$\therefore$ Integrating both sides, we get

$\begin{array}{ll}& \mathrm{f}(x)=\int (x-\frac{5}{x^5})\mathrm{d}x\\ & \mathrm{f}(x)=\frac{x^2}{2}+\frac{5}{4}\times \frac{1}{x^4}+\mathrm{c}\\ \therefore & \text{ But }\mathrm{f}(1)=4\\ \therefore & \frac{1}{2}+\frac{5}{4}+\mathrm{c}=4\\ \therefore & \mathrm{c}=\frac{9}{4}\\ \therefore & \mathrm{f}(x)=\frac{x^2}{2}+\frac{5}{4}\frac{1}{x^4}+\frac{9}{4}\end{array}$