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

Jul 27, 2016

I got:

$\frac{{x}^{2}}{2} \arctan x + \arctan \frac{x}{2} - \frac{x}{2} + C$

Just pick the term that is easier to integrate, and use that as your $\mathrm{dv}$.

$\setminus m a t h b f \left(\int u \mathrm{dv} = u v - \int v \mathrm{du}\right)$

So, we can pick our $u$ as $\arctan x$, because who knows the antiderivative of $\arctan x$? Not me. However, I do know that its derivative is $\frac{1}{1 + {x}^{2}}$, so we can do that.

Then, of course, now that we've done that, $\mathrm{dv} = x \mathrm{dx}$.

$u = \arctan x , \mathrm{du} = \frac{1}{1 + {x}^{2}} \mathrm{dx}$
$\mathrm{dv} = x \mathrm{dx} , v = {x}^{2} / 2$

$\textcolor{b l u e}{\int x \arctan x \mathrm{dx}}$

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

Now the challenge is to integrate this one. Fortunately, it's not that bad.

$\implies \frac{{x}^{2}}{2} \arctan x - \frac{1}{2} \left[\int \frac{{x}^{2} + 1 - 1}{{x}^{2} + 1} \mathrm{dx}\right]$

$= \frac{{x}^{2}}{2} \arctan x - \frac{1}{2} \left[\int \mathrm{dx} - \int \frac{1}{{x}^{2} + 1} \mathrm{dx}\right]$

$= \frac{{x}^{2}}{2} \arctan x - \frac{1}{2} \left[x - \arctan x\right]$

And finishing things up:

$= \textcolor{b l u e}{\frac{{x}^{2}}{2} \arctan x + \arctan \frac{x}{2} - \frac{x}{2} + C}$