# How do you integrate (1/(16+x^2))?

Sep 1, 2016

$\int \frac{1}{16 + {x}^{2}} \mathrm{dx} = \frac{1}{4} a r c \tan \left(\frac{x}{4}\right) + C$.

#### Explanation:

We know that $\int \frac{1}{{a}^{2} + {x}^{2}} \mathrm{dx} = \frac{1}{a} a r c \tan \left(\frac{x}{2}\right) + C$.

$\therefore \int \frac{1}{16 + {x}^{2}} \mathrm{dx} = \frac{1}{4} a r c \tan \left(\frac{x}{4}\right) + C$.