# What is the integral of int ( 1 / (25 + x^2) ) dx ?

Apr 4, 2016

$\int \left(\frac{1}{25 + {x}^{2}}\right) \mathrm{dx} = \frac{1}{5} {\tan}^{-} 1 \left(\frac{x}{5}\right) + C$

#### Explanation:

$\int \left(\frac{1}{25 + {x}^{2}}\right) \mathrm{dx}$

$\frac{\mathrm{dx}}{d} \left(\theta\right) = 5 \tan \theta$

$\mathrm{dx} = 5 {\sec}^{2} \theta \cdot \left(d\right) \theta$

$\int \left(\frac{1}{25 + 25 {\tan}^{2} \theta}\right) \cdot 5 {\sec}^{2} \theta \cdot \left(d\right) \theta$

$\int \left(\frac{1}{25 \left(1 + {\tan}^{2} \theta\right)}\right) \cdot 5 {\sec}^{2} \theta \cdot \left(d\right) \theta$

$1 + \tan \theta = {\sec}^{2} \theta$

$\int \left(\frac{1}{25 \left({\sec}^{2} \theta\right)}\right) \cdot 5 {\sec}^{2} \theta \cdot \left(d\right) \theta$

$\int \left(\frac{1}{5}\right) \cdot \left(d\right) \theta + C$

$x = 5 \tan \theta$
$\frac{x}{5} = \tan \theta$
${\tan}^{-} 1 \left(\frac{x}{5}\right) = \theta$

Plug in:

$= \frac{1}{5} {\tan}^{-} 1 \left(\frac{x}{5}\right) + C$