<s>[INST] Integration of $\int \frac {x\arctan x}{(\sqrt{1+x^2})^3}$ 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.
 [/INST] 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
$$
I=\int\frac{x\theta}{\sqrt{1+x^2}}\centerdot\frac{dx}{1+x^2}=\int\theta(\sin\theta) d\theta = -\int\theta d\cos\theta = 
$$
$$
-\theta\cos\theta + \int(\cos\theta) d\theta = -\theta\cos\theta+\sin\theta = \frac{-\arctan(x)+x}{\sqrt{1+x^2}}+C
$$
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