TITLE: How to find $\int_0^\infty\frac{\log\left(\frac{1+x^{4+\sqrt{15}}}{1+x^{2+\sqrt{3}}}\right)}{\left(1+x^2\right)\log x}\mathrm dx$
QUESTION [47 upvotes]: I was challenged to prove this identity
$$\int_0^\infty\frac{\log\left(\frac{1+x^{4+\sqrt{15\vphantom{\large A}}}}{1+x^{2+\sqrt{3\vphantom{\large A}}}}\right)}{\left(1+x^2\right)\log x}\mathrm dx=\frac{\pi}{4}\left(2+\sqrt{6}\sqrt{3-\sqrt{5}}\right).$$
I was not successful, so I want to ask for your help. Can it be somehow related to integrals listed in that question?

REPLY [7 votes]: I have a short way to solve this problem. Let $x=\frac{1}{u}, a=4+\sqrt{15},b=2+\sqrt{3}$. Then
\begin{eqnarray*}
I&=&-\int_\infty^0\frac{\ln\frac{1+u^{-a}}{1+u^{-b}}}{(1+u^{-2})\ln(u^{-1})}u^{-2}du\\
&=&-\int_0^\infty\frac{\ln\frac{1+u^{-a}}{1+u^{-b}}}{(1+u^{2})\ln u}du\\
&=&-\int_0^\infty\frac{\ln \left(u^{b-a}\frac{1+u^{a}}{1+u^{b}}\right)}{(1+u^{2})\ln u}du\\
&=&-\int_0^\infty\frac{\ln \left(\frac{1+u^{a}}{1+u^{b}}\right)}{(1+u^{2})\ln u}du-\int_0^\infty\frac{(b-a)\ln u}{(1+u^{2})\ln u}du
\end{eqnarray*}
and hence
$$ 2I = -\int_0^\infty\frac{(b-a)}{1+u^{2}}du=(a-b)\frac{\pi}{2}. $$
So
$$ I = (a-b)\frac{\pi}{4}. $$