# How do you find the integral of tanˉ¹(2x) dx?

Jul 14, 2018

The answer is $= x \arctan \left(2 x\right) - \frac{1}{4} \ln \left(4 {x}^{2} + 1\right) + C$

#### Explanation:

Perform a substitution

$u = 2 x$, $\implies$, $\mathrm{du} = 2 \mathrm{dx}$

The integral is

$I = \int \arctan \left(2 x\right) \mathrm{dx}$

$= \frac{1}{2} \int \arctan u \mathrm{du}$

Perform an integration by parts

$\int f g ' \mathrm{dx} = f g - \int f ' g \mathrm{dx}$

Here,

$f = \arctan u$, $\implies$, $f ' = \frac{1}{{u}^{2} + 1}$

$g ' = 1$, $\implies$, $g = u$

Therefore,

$I = \frac{1}{2} u \cdot \arctan u - \frac{1}{2} \int \frac{u \mathrm{du}}{{u}^{2} + 1}$

$= \frac{1}{2} u \cdot \arctan u - \frac{1}{4} \ln \left({u}^{2} + 1\right)$

$= \frac{1}{2} \cdot 2 x \arctan \left(2 x\right) - \frac{1}{4} \ln \left(4 {x}^{2} + 1\right) + C$

$= x \arctan \left(2 x\right) - \frac{1}{4} \ln \left(4 {x}^{2} + 1\right) + C$