Question

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

Answer 1

$I=\int \frac{\left(x^2+{\sin }^{2}x\right){\sec }^{2}x}{\left(1+x^2\right)}dx$
$=\int \left[\frac{\left({\sec }^{2}x{\sin }^{2}x+{\sec }^{2}x\right)}{\left(1+x^2\right)}+{\sec }^{2}x\right]dx$
$=\int \left({\sec }^{2}x-\frac{1}{x^2+1}\right)dx$
$=\tan x-ta{n}^{4}x+C$