# What is int tan^-1 x dx ?

Feb 20, 2018

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

#### Explanation:

We want to solve

$I = \int {\tan}^{-} 1 \left(x\right) \mathrm{dx}$

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Let $u = {\tan}^{-} 1 \left(x\right)$ and $\mathrm{dv} = 1 \mathrm{dx}$

Then $\mathrm{du} = \frac{1}{{x}^{2} + 1} \mathrm{dx}$ and $v = x$

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

Make a substitution $u = {x}^{2} + 1 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2 x$

$I = {\tan}^{-} 1 \left(x\right) x - \frac{1}{2} \int \frac{1}{u} \mathrm{du}$

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

Substitute back $u = {x}^{2} + 1$

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