# What is the antiderivative of 1/(x^2+4)?

Mar 27, 2018

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

#### Explanation:

Calculate the indefinite integral:

$\int \frac{\mathrm{dx}}{{x}^{2} + 4}$

by substituting:

$x = 2 u$

$\mathrm{dx} = 2 \mathrm{du}$

so:

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

$\int \frac{\mathrm{dx}}{{x}^{2} + 4} = 2 \int \frac{\mathrm{du}}{4 {u}^{2} + 4}$

$\int \frac{\mathrm{dx}}{{x}^{2} + 4} = \frac{1}{2} \int \frac{\mathrm{du}}{{u}^{2} + 1}$

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

and undoing the substitution:

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

Mar 27, 2018

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

#### Explanation:

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

Now let $2 u = x$ and $2 \mathrm{du} = \mathrm{dx}$

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