# How do you integrate 1/(x^2+4)?

Jun 24, 2016

$\frac{1}{2} \arctan \left(\frac{x}{2}\right) + C$

#### Explanation:

Our goal should be to make this mirror the arctangent integral:

$\int \frac{1}{{u}^{2} + 1} \mathrm{du} = \arctan \left(u\right) + C$

To get the $1$ in the denominator, start by factoring:

$\int \frac{1}{{x}^{2} + 4} \mathrm{dx} = \int \frac{1}{4 \left({x}^{2} / 4 + 1\right)} \mathrm{dx} = \frac{1}{4} \int \frac{1}{{x}^{2} / 4 + 1} \mathrm{dx}$

Note that we want ${u}^{2} = {x}^{2} / 4$, so we let $u = \frac{x}{2}$, which implies that $\mathrm{du} = \frac{1}{2} \mathrm{dx}$.

$\frac{1}{4} \int \frac{1}{{x}^{2} / 4 + 1} \mathrm{dx} = \frac{1}{2} \int \frac{\frac{1}{2}}{{\left(\frac{x}{2}\right)}^{2} + 1} \mathrm{dx} = \frac{1}{2} \int \frac{1}{{u}^{2} + 1} \mathrm{du}$

This is the arctangent integral:

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