# How do you integrate x*arctan(x) dx?

Jul 22, 2016

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

#### Explanation:

we can use IBP

$\int x \arctan \left(x\right) \setminus \mathrm{dx} = \int \setminus \frac{d}{\mathrm{dx}} \left({x}^{2} / 2\right) \arctan x \setminus \mathrm{dx}$

$= \left({x}^{2} / 2\right) \arctan x - \int \setminus {x}^{2} / 2 \frac{d}{\mathrm{dx}} \left(\arctan x\right) \setminus \mathrm{dx}$

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

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

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

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

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

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