Question

Evaluate $\int \frac{x^2}{1+x^2}dx$.

Answer 1

$\int \frac{x^2}{x^2+1}=\int \frac{x^2+1-1}{x^2+1}$

$=\int 1-\frac{1}{x^2+1}dx$.

$=x-\int \frac{1}{x^2+1}dx$.

$\int \frac{1}{x^2+1}dx$ $x=\tan t$

$dx={\sec }^{2}t.dt$

$\int \frac{1}{{\tan }^{2}t+1}.{\sec }^{2}t.dt$

$\int \frac{{\sec }^{2}t}{{\sec }^{2}t}.dt=\int 1.dt$

$=t=arc\tan x-C$.

So full integral is

$x-arc\tan x+C$.