Difficult integral involving $\arctan x$ 
Question: Show that$$\int\limits_0^{\infty}\mathrm dx\,\frac {\arctan x\log(1+x^2)}{x(1+x^2)}=\frac {\pi}2\log^22.$$

I can't tell if I'm being an idiot, or if this is a lot more difficult than it looks. First, I tried integration by parts using the fact that $$\frac 1{x(1+x^2)}=\frac 1x-\frac x{1+x^2}.$$
But quickly I gave up as I wasn't sure what to do with the result. I then decided to make the substitution $t=\arctan x$ to get rid of the $1+x^2$ term in the denominator. Therefore$$\begin{align*}\mathfrak{I} & =\int\limits_0^{\pi/2}\mathrm dt\,t\cot t\log\sec^2t=-2\int\limits_0^{\pi/2}\mathrm dt\, t\cot t \log\cos t.\end{align*}$$
However, I'm not exactly sure what to do after this. Should I use integration by parts? Differentiation under the integral sign? I'm having trouble getting started with this integral. Any ideas?

Define
$$ f(a,b) = \int \limits_0^\infty \frac{\arctan(a x) \ln (1+ b^2 x^2)}{x (1+x^2)} \, \mathrm{d} x $$
for $0 \leq a , b \leq 1$ . Then $f(1,1) = \mathfrak{I}$ and $f(0,b) = f(a,0) = 0 $ . For $0< a,b<1$ we can differentiate under the integral sign to find
$$ \partial_a \partial_b f(a,b) = 2 b \int \limits_0^\infty \frac{x^2}{(1+a^2 x^2)(1+b^2 x^2)(1+x^2)} \, \mathrm{d} t = \frac{\pi b}{(1+a)(1+b)(a+b)} \, . $$
The integral can be evaluated using the residue theorem, for example. Now integrate again and exploit the symmetry of the derivative to obtain
\begin{align}
\mathfrak{I} &= f(1,1) = \pi \int \limits_0^1 \int \limits_0^1 \frac{b}{(1+a)(1+b)(a+b)} \, \mathrm{d}a \, \mathrm{d}b = \pi \int \limits_0^1 \int \limits_0^1 \frac{a}{(1+a)(1+b)(a+b)} \, \mathrm{d}a \, \mathrm{d}b \\
&= \frac{\pi}{2} \int \limits_0^1 \int \limits_0^1 \frac{a+b}{(1+a)(1+b)(a+b)} \, \mathrm{d}a \, \mathrm{d}b = \frac{\pi}{2} \int \limits_0^1 \int \limits_0^1 \frac{1}{(1+a)(1+b)} \, \mathrm{d}a \, \mathrm{d}b \\
&= \frac{\pi}{2} \ln^2 (2) \, .
\end{align}
