# How do you integrate int tan^-1x by integration by parts method?

Oct 31, 2016

The integral $= x \arctan x - \ln \frac{1 + {x}^{2}}{2} + C$

#### Explanation:

The integration by parts is $\int u ' v = u v - \int u v '$
Here we have $u ' = 1$$\implies$$u = x$
and $v = \arctan x$$\implies$$v ' = \frac{1}{1 + {x}^{2}}$
If $v = \arctan x$$\implies$$x = \tan v$
So by differentiating, $1 = \left(\frac{1}{\cos} ^ 2 v\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right)$
$v ' = \frac{\mathrm{dv}}{\mathrm{dx}} = {\cos}^{2} v = \frac{1}{1 + {x}^{2}}$

The integration
$\int \arctan x \mathrm{dx} = x \arctan x - \int \frac{x \mathrm{dx}}{1 + {x}^{2}}$
$\int \frac{x \mathrm{dx}}{1 + {x}^{2}} = \left(\frac{1}{2}\right) \int \frac{2 x \mathrm{dx}}{1 + {x}^{2}}$
Let $u = 1 + {x}^{2}$$\implies$$\mathrm{du} = 2 x \mathrm{dx}$
$\int \frac{2 x \mathrm{dx}}{1 + {x}^{2}} = \int \frac{\mathrm{du}}{u} = \ln u$
$\int \frac{x \mathrm{dx}}{1 + {x}^{2}} = \frac{1}{2} \ln \left(1 + {x}^{2}\right)$

And finally $\int \arctan x \mathrm{dx} = x \arctan x - \ln \frac{1 + {x}^{2}}{2} + C$