Question

Solve:
$\int \frac{1}{1+x^2}$ dx

Answer 1

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

put

$x=\tan p$

$dx={\sec }^{2}pdp$

$dx=(1+{\tan }^{2}p)dp$

$dx=(1+x^2)dp$

$\Rightarrow \frac{dx}{(1+x^2)}=dp$

$I=\int dp=p+c={\tan }^{-1}x+c$

$\therefore \int \frac{dx}{1+x^2}={\tan }^{-1}x+c$