# What is the integral of int(x tan^-1 x)/(1+x^2)^(3/2) ?

## $\int \frac{x {\tan}^{-} 1 x}{1 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

Jul 13, 2018

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

#### Explanation:

$I = \int \frac{x {\tan}^{-} 1 \left(x\right)}{1 + {x}^{2}} ^ \left(\frac{3}{2}\right) \textcolor{red}{\mathrm{dx}}$

Let $x = \tan \theta$, which implies that $\mathrm{dx} = {\sec}^{2} \theta d \theta$.

Then:

$I = \int \frac{\tan \theta {\tan}^{-} 1 \left(\tan \theta\right)}{1 + {\tan}^{2} \theta} ^ \left(\frac{3}{2}\right) {\sec}^{2} \theta d \theta$

Remember that $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$:

$I = \int \frac{\theta \tan \theta}{\sec} ^ 3 \theta {\sec}^{2} \theta d \theta$

$I = \int \theta \tan \theta \frac{1}{\sec} \theta d \theta$

$I = \int \theta \sin \frac{\theta}{\cos} \theta \cos \theta d \theta$

$I = \int \theta \sin \theta d \theta$

This looks primed for integration by parts. Let:

$\left\{\begin{matrix}u = \theta & \mathrm{du} = d \theta \\ \mathrm{dv} = \sin \theta d \theta & v = - \cos \theta\end{matrix}\right.$

Then:

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

$I = - \theta \cos \theta + \int \cos \theta d \theta$

$I = - \theta \cos \theta + \sin \theta + C$

Remember that $x = \tan \theta$. Imagine the right triangle associated with this: we must have the leg opposite $\theta$ being $x$ and the adjacent leg being $1$. Then the hypotenuse is $\sqrt{{x}^{2} + 1}$.

Thus $\cos \theta = \frac{1}{\sqrt{{x}^{2} + 1}}$ and $\sin \theta = \frac{x}{\sqrt{{x}^{2} + 1}}$. Moreover, note that $\theta = {\tan}^{-} 1 \left(x\right)$.

$I = - {\tan}^{-} 1 \frac{x}{\sqrt{{x}^{2} + 1}} + \frac{1}{\sqrt{{x}^{2} + 1}} + C$

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