Integral of $x\arctan x/(1+x^2)^{3/2}$ Find
$$\int \frac{x\arctan x}{(1+x^2)^{3/2}}\operatorname{dx} $$
I tried many ways to do it but I can't resolve it.  What are the steps to  solve this exercise?

One way is to just integrate by parts,
$$
\begin{split}
\int \frac{x}{(1+x^2)^{3/2}}\arctan x&=-\frac{1}{\sqrt{1+x^2}}\arctan x+\int\frac{1}{(1+x^2)^{3/2}}\,dx\\
&=-\frac{1}{\sqrt{1+x^2}}\arctan x+\frac{x}{\sqrt{1+x^2}}.
\end{split}
$$
Here, the last step might seem mystery. I know it by heart, but one way to conclude it is:
$$
\begin{split}
\frac{1}{(1+x^2)^{3/2}}
&=\frac{(1+x^2)-x^2}{(1+x^2)^{3/2}}\\
&=\frac{1}{\sqrt{1+x^2}}-\frac{x^2}{(1+x^2)^{3/2}}
\\
&=D(x)\frac{1}{\sqrt{1+x^2}}+xD\frac{1}{\sqrt{1+x^2}}\\
&=D\frac{x}{\sqrt{1+x^2}}.
\end{split}
$$
