Match List I with List II

Choose the correct answer from the options given below:

A-III, B-IV, C-I, D-II

A. $\int\frac{\sin x}{1+\cos x}dx$

let $y=1+\cos x⇒dy=-\sin x\,dx$

$⇒\int-\frac{dy}{y}=-\log|1+\cos x|+C$

B. $\int\frac{1}{1-\tan x}dx⇒\int\frac{1}{1-\frac{\sin x}{\cos x}}dx$

$⇒\int\frac{\cos x}{\cos x-\sin x}dx⇒\frac{1}{2}\frac{2\cos x}{\cos x-\sin x}dx$

$⇒\frac{1}{2}\int\frac{\cos x-\sin x}{\cos x-\sin x}+\int\frac{\cos x+\sin x}{\cos x-\sin x}dx$

$=\frac{x}{2}+\frac{\log|\cos x-\sin x|+C}{2}$

C. $\int\frac{e^{\tan^{-1}x}}{1+x^2}dx$

Let $y=\tan^{-1}x⇒dy=\frac{dx}{1+x^2}$

$⇒\int e^ydy=e^y+c=e^{\tan^{-1}x}+c$

D. $\int\frac{1}{x+x\log x}dx$

Let $y=1+\log x⇒dy=\frac{1}{x}dx$

$⇒\frac{dy}{y}=\log y +c=\log(\log x+1)+c$