Question

Solve
$\int \frac{{\sec }^2\left(2{\tan }^{-1}x\right)}{(1+x^2)}dx$

Answer 1

$\int \frac{{\sec }^2\left(2{\tan }^{-1}x\right)}{1+x^2}dx$

${\tan }^{-1}x=t$

$\Rightarrow dt=\frac{1}{1+x^2}dx$

$\therefore \int {\sec }^2\left(2t\right)dt$

$\Rightarrow \frac{\tan \left(2t\right)}{2}+c$

$\Rightarrow \frac{1}{2}\tan \left(\frac{2}{1+x^2}\right)+c$.