How to integrate $\int_{0}^{\infty} \frac{x \arctan(x) \ dx}{(1+x^2)^2}$ I am asked to solve the following integral:

$$\int_{0}^{\infty} \frac{x \arctan(x) \ dx}{(1+x^2)^2}$$

I've made a couple of guesses but none of them took me anywhere. One of my ideas was to do, at first, an u-substitution:
$$
u = 1+x^2\\
du = 2x dx\\
\frac{du}{2} = x dx\\
\\
x = 0 \Rightarrow u = 1\\
x = a \Rightarrow u = 1+a^2 
$$
but then there's a $\arctan(x)$ left there untouched. Does anyone have an idea I can pursue?
Source: James Stewart's Calculus book.
Thank you.
 A: Set $u = \arctan(x)$, $du = \frac{dx}{1+x^2}.$ Your integral becomes $$\int_0^{\pi/2}\frac{u\tan(u)}{1+\tan^2(u)}du = \int_0^{\pi/2}u\tan(u)\cos^2(u)du = \int_0^{\pi/2}u\sin(u)\cos(u)du \\= \frac{1}{2}\int_0^{\pi/2}u\sin(2u)du=\frac{1}{2}\left[-\frac{1}{2}u\cos(2u)+\frac{1}{4}\sin(2u)\right]_0^{\pi/2}=\frac{\pi}{8}.$$
A: With the substitution $x=\frac{1}{t}$ we get
$$ I=\int_{0}^{+\infty}\frac{x\arctan(x)}{(1+x^2)^2}\,dx = \int_{0}^{+\infty}\frac{t\left(\frac{\pi}{2}-\arctan(t)\right)}{(1+t^2)^2}\,dt $$
hence by symmetry
$$ I = \frac{\pi}{8}\int_{0}^{+\infty}\frac{2u}{(1+u^2)^2}\,du = \color{red}{\frac{\pi}{8}}.$$
A: Let us integrate by parts and put
$u'=\frac{x}{(1+x^2)^2}$
and
$v=arctan(x).$
thus
the integral becomes
$$I=[-\frac{1}{2(1+x^2)}arctan(x)]_0^\infty$$
$$+\frac{1}{2}\int_0^\infty\frac{dx}{(1+x^2)^2}$$
$$=\frac{1}{2}\int_0^\infty\frac{1+x^2-x^2}{(1+x^2)^2}dx$$
$$=\frac{\pi}{4}+\frac{1}{4}\int_0^\infty x \frac{-2x}{(1+x^2)^2}dx$$
$$=\frac{\pi}{4}+\frac{1}{4}([\frac{x}{1+x^2}]_0^\infty-\int_0^\infty\frac{1}{1+x^2}dx)$$
$$\color{red}{=\frac{\pi}{4}-\frac{\pi}{8}=\frac{\pi}{8}}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
\int_{0}^{\infty}{x\arctan\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x &
\,\,\,\stackrel{x\ \equiv\ \tan\pars{\theta}}{=}\,\,\,
{1 \over 2}\int_{0}^{\pi \over 2}\theta\sin\pars{2\theta}\,\dd\theta
\\[5mm] & =
{1 \over 4}\int_{0}^{\pi \over 2}
\bracks{\theta\sin\pars{2\theta} +
\pars{{\pi \over 2} - \theta}\sin\pars{2\bracks{{\pi \over 2} - \theta}}}
\,\dd\theta
\\[5mm] & =
{1 \over 8}\,\pi\int_{0}^{\pi \over 2}\sin\pars{2\theta}\,\dd\theta =
\bbx{\ds{{1 \over 8}\,\pi}}
\end{align}
