# How do you find the integral of int arctan(x) dx?

Oct 20, 2015

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

#### Explanation:

Integration By Parts: $\int u \mathrm{dv} = u v - \int v \mathrm{du}$

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

$\mathrm{dv} = \mathrm{dx} \implies v = x$

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

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

$I = x \arctan x - \frac{1}{2} \int \frac{d \left({x}^{2} + 1\right)}{{x}^{2} + 1}$

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