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

Sep 11, 2016

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

#### Explanation:

We have:

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

And we will use the integration by parts formula:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

So, we set $\int u \mathrm{dv} = \int \arctan \left(x\right) \mathrm{dx}$, so we let:

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

Thus:

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

Solving the second integral:

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

Let $u = 1 + {x}^{2}$ so $\mathrm{du} = 2 x \mathrm{dx}$:

$= x \arctan \left(x\right) - \frac{1}{2} \int \frac{\mathrm{du}}{u}$

$= x \arctan \left(x\right) - \frac{1}{2} \ln \left\mid u \right\mid$

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