# How do you find the integral of dx/(1+x^2)?

Oct 11, 2015

${\tan}^{-} 1 x + C$

#### Explanation:

Let $x = \tan \theta$, so that $\mathrm{dx} = {\sec}^{2} \theta d \theta$

$\int \frac{1 \mathrm{dx}}{1 + {x}^{2}}$ = $\int \frac{{\sec}^{2} \theta d \theta}{\sec} ^ 2 \theta$

$\int d \theta = \theta + C$

= ${\tan}^{-} 1 x + C$