JEE 2023 — Mathematics PYQ

$I(x)=\int \frac{x^2(x{\sec }^{2}x+\tan x)}{(x\tan x+1{)}^2}dx$

Let $x\tan x+1=t$

$I(x)=x^2\left(\frac{-1}{x\tan x+1}\right)<br>+\int \frac{2x}{x\tan x+1}dx$

$I(x)=x^2\left(\frac{-1}{x\tan x+1}\right)<br>+2\ln |x\sin x+\cos x|+C$

$I(0)=0\Rightarrow C=0$

$I\left(\frac{\pi }{4}\right)<br>={\log }_e\frac{(\pi +4{)}^2}{16}<br>-\frac{{\pi }^2}{4(\pi +4)}$