# What is the antiderivative of (1)/(1+x^2)?

Jun 26, 2015

$\arctan \left(x\right) + C$

#### Explanation:

This is one to memorize:

$\setminus \int \frac{1}{1 + {x}^{2}} \setminus \mathrm{dx} = \arctan \left(x\right) + C$

It can be derived by differentiation of both sides of the equation $\tan \left(\arctan \left(x\right)\right) = x$ (assuming you know that $\frac{d}{\mathrm{dx}} \left(\tan \left(x\right)\right) = {\sec}^{2} \left(x\right)$) and using the Chain Rule:

${\sec}^{2} \left(\arctan \left(x\right)\right) \cdot \frac{d}{\mathrm{dx}} \left(\arctan \left(x\right)\right) = 1$

$\setminus R i g h t a r r o w \frac{d}{\mathrm{dx}} \left(\arctan \left(x\right)\right) = \frac{1}{\sec} ^ \left[2\right\} \left(\arctan \left(x\right)\right)$

$= \frac{1}{1 + {\tan}^{2} \left(\arctan \left(x\right)\right)} = \frac{1}{1 + {x}^{2}}$ for all $x \setminus \in \mathbb{R}$.