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?