Integration of $\int \frac {x\arctan x}{(\sqrt{1+x^2})^3}$

Question

How can I evaluate $$\int \frac{x \arctan x}{(\sqrt{1+x^2})^3}\,dx$$ I think it must be done by parts, but I can't get any appropriate substitution or such.

Answer 1

Substituting $x=\tan\theta$ gives $$I=\int \frac{\theta\tan\theta\sec^2\theta\,d\theta}{\sec^3\theta} =\int\theta\sin\theta\,d\theta$$ which is now easily done by parts.

(Note that to do this carefully we should specify $-\frac12\pi<\theta<\frac12\pi$, so that $\sec\theta$ is positive.)

Answer 2

There is no magic in this calculation, the key is familiarity, familiar with $\int\frac{dx}{1+x^2}=\arctan(x)+C$.

Let $\theta=\arctan(x)$, then we have $d\theta=\frac{dx}{1+x^2}$, $\frac{x}{\sqrt{1+x^2}}=\sin\theta$ and $\frac{1}{\sqrt{1+x^2}}=\cos\theta$, so

$$ I=\int\frac{x\theta}{\sqrt{1+x^2}}\centerdot\frac{dx}{1+x^2}=\int\theta(\sin\theta) d\theta = -\int\theta d\cos\theta = $$ $$ -\theta\cos\theta + \int(\cos\theta) d\theta = -\theta\cos\theta+\sin\theta = \frac{-\arctan(x)+x}{\sqrt{1+x^2}}+C $$