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

The antiderivative of $\frac{1}{1 + {x}^{2}}$ is the integral

$\int \frac{1}{1 + {x}^{2}} \mathrm{dx}$ which is equivalent to

$\int \frac{1}{1 + {x}^{2}} \mathrm{dx} = \arctan x + C$

where $\arctan x$ is the inverse of the trigonometric function

$\tan x$ and C is the integration constant.

Feb 3, 2018

$= \arctan \left(x\right) + c$

#### Explanation:

Let

color(blue)(x = tantheta

$\implies \textcolor{red}{\mathrm{dx} = {\sec}^{2} \theta d \theta} \text{ By the use of the quotient rule...}$

int 1/(1+color(blue)(x)^2 ) color(red)(dx) = int 1/(1+color(blue)((tantheta))^2 ) * color(red)(sec^2 theta d theta

We know ${\sin}^{2} x + {\cos}^{2} x \equiv 1$

$\implies {\sin}^{2} \frac{x}{\cos} ^ 2 x + {\cos}^{2} \frac{x}{\cos} ^ 2 x \equiv \frac{1}{\cos} ^ 2 x$

$\implies {\tan}^{2} x + 1 \equiv {\sec}^{2} x$

$\implies \int {\sec}^{2} \frac{\theta}{\sec} ^ 2 \theta d \theta$

$\implies \int 1 d \theta$

$\implies \theta + c$

If $x = \tan \theta \implies \arctan x = \theta$

Substitute back in...

color(blue)( arctan(x ) + c