# How do you integrate int x arctan x  using integration by parts?

Nov 4, 2015

$\int x {\tan}^{- 1} \left(x\right) \mathrm{dx} = \frac{1}{2} \left[\left({x}^{2} + 1\right) {\tan}^{- 1} \left(x\right) - x\right] + c$,
where $c$ is the constant of integration.

#### Explanation:

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

$= \frac{1}{2} \left[{x}^{2} {\tan}^{- 1} \left(x\right) - \int {x}^{2} \frac{d}{\mathrm{dx}} \left({\tan}^{- 1} \left(x\right)\right) \mathrm{dx}\right]$

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

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

$= \frac{1}{2} {x}^{2} {\tan}^{- 1} \left(x\right) - \frac{1}{2} \left(x - {\tan}^{- 1} \left(x\right)\right) + c$,
where $c$ is the constant of integration

$= \frac{1}{2} \left[\left({x}^{2} + 1\right) {\tan}^{- 1} \left(x\right) - x\right] + c$