# How do you integrate int xarctanx by integration by parts method?

Nov 6, 2016

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

#### Explanation:

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

Integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. So, for $\int u \mathrm{dv} = \int x \arctan \left(x\right) \mathrm{dx}$, we should let:

$\left\{\begin{matrix}u = \arctan \left(x\right) & \implies & \mathrm{du} = \frac{\mathrm{dx}}{1 + {x}^{2}} \\ \mathrm{dv} = x \mathrm{dx} & \implies & v = {x}^{2} / 2\end{matrix}\right.$

Thus:

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

Rewrite the numerator or perform polynomial long division of the integrand. Both will result in equivalent simplifications:

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

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

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

Both of these are simply integrated:

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

Or:

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