How to find $\int_0^\infty\frac{\log\left(\frac{1+x^{4+\sqrt{15}}}{1+x^{2+\sqrt{3}}}\right)}{\left(1+x^2\right)\log x}\mathrm dx$

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?

Answer

This integral can be evaluated in a closed form for arbitrary real exponents, and does not seem to be related to Herglotz-like integrals.

Assume $a,b\in\mathbb{R}$. Note that
$$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\int_0^\infty\frac{\ln\left(\frac{1+x^b}2\right)}{\ln x}\frac{dx}{1+x^2}.\tag1$$
Both integrals on the right-hand side have the same shape, so we only need to evaluate one of them:
$$\begin{align}&\phantom=\underbrace{\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}}_\text{split the region}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}+\underbrace{\int_1^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}}_{\text{change variable}\ y=1/x}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}+\underbrace{\int_1^0\frac{\ln\left(\frac{1+y^{-a}}2\right)}{\ln\left(y^{-1}\right)}\frac1{1+y^{-2}}\left(-\frac1{y^2}\right)dy}_\text{flip the bounds and simplify}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\underbrace{\int_0^1\frac{\ln\left(\frac{1+y^{-a}}2\right)}{\ln y}\frac{dy}{1+y^2}}_{\text{rename}\ y\ \text{to}\ x}\\&=\underbrace{\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\int_0^1\frac{\ln\left(\frac{1+x^{-a}}2\right)}{\ln x}\frac{dx}{1+x^2}}_\text{combine logarithms}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}{1+x^{-a}}\right)}{\ln x}\frac{dx}{1+x^2}=\underbrace{\int_0^1\frac{\ln\left(\frac{x^a\left(x^{-a}+1\right)}{1+x^{-a}}\right)}{\ln x}\frac{dx}{1+x^2}}_{\text{cancel}\ \ 1+x^{-a}}\\&=\int_0^1\frac{\ln\left(x^a\right)}{\ln x}\frac{dx}{1+x^2}=a\int_0^1\frac{dx}{1+x^2}=a\,\Big(\arctan1-\arctan0\Big)\\&=\vphantom{\Bigg|^0}\frac{\pi\,a}4.\end{align}\tag2$$
So, finally,
$$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\frac\pi4(a-b).\tag3$$

Attribution
Source : Link , Question Author : Zakharia Stanley , Answer Author : Vladimir Reshetnikov

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