Integral $\int_0^\infty \frac{\arctan(x) dx}{x(1+x^2)}$?

Rewrite

$$\frac{\arctan{x}}{x} = \int_0^1 \frac{du}{1+x^2 u^2} $$

Then plugging this in and reversing the order of integration, we get the integral; value as

$$\begin{align}\int_0^1 \frac{du}{u^2} \, \int_0^{\infty} dx \, \left (\frac1{\frac1{u^2}+x^2} \frac1{1+x^2} \right ) &= \int_0^1 \frac{du}{1-u^2} \, \int_0^{\infty} dx \left ( \frac1{1+x^2}-\frac1{\frac1{u^2}+x^2} \right )\\ &= \int_0^1 \frac{du}{1-u^2} \, \frac{\pi}{2} (1-u) \\ &= \frac{\pi}{2} \log{2}\end{align}$$

Let $x=\tan u$ then $$\int_0^\infty \frac{\arctan(x) \,dx}{x(1+x^2)}=\int_0^{\pi/2} \frac{u}{\tan u}\ du =\int_0^{\pi/2} u\cot u\ du $$ now use $$\int_0^{\pi/2} u\cot u \ du=\dfrac{\pi}{2}\ln2$$

Let

$$ I=\int_0^\infty \frac{\arctan(x) \,dx}{x(1+x^2)}.$$

The change of variable $\arctan x=t$ give us

$$I=\int_0^{\pi/2} \frac{t\cos t \,dt}{\sin t}=\frac{1}{2}\pi\log{2}$$