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?
 A: Let the considered integral be $I$. Just to make it easier to write, let $4+\sqrt{15}=a$ and $2+\sqrt{3}=b$. Use the substitution $x=\tan\theta$ to get:
$$I=\int_0^{\pi/2} \frac{\ln\left(\dfrac{1+(\tan\theta)^a}{1+(\tan\theta)^b}\right)}{\ln\tan\theta}\,d\theta$$
Next, use the substitution $\theta=\pi/2-t$ to obtain:
$$I=\int_0^{\pi/2} \frac{\ln\left(\dfrac{1+(\tan t)^a}{(1+(\tan t)^b)(\tan t)^{a-b}}\right)}{\ln\cot t}\,dt=\int_0^{\pi/2} \frac{-\ln\left(\dfrac{1+(\tan\theta)^a}{1+(\tan\theta)^b}\right)+(a-b)\ln\tan\theta}{\ln\tan\theta}\,d\theta$$
where I used $\ln(\tan(\pi/2-\theta))=\ln(\cot\theta)=-\ln(\tan\theta)$.
Add the two expressions for I and notice that you are left with:
$$2I=\int_0^{\pi/2} \frac{(a-b)\ln \tan\theta}{\ln \tan\theta}\,d\theta=\frac{\pi}{2}(a-b)$$
$$I=\frac{\pi}{4}(a-b)$$
Therefore,
$$\boxed{I=\dfrac{\pi}{4}(2+\sqrt{15}-\sqrt{3})}$$
A: 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
$$ I = (a-b)\frac{\pi}{4}. $$
A: I do not know how to answer your question. However, in order you to challenge your challenger, I give you a few amazing results (obtained using a CAS) for $$f(n)=\int_0^\infty\frac{\log\left(\frac{1+x^a}{1+x^b}\right)}{\left(1+x^2\right)\log x}dx$$ in which $a=2n+\sqrt{4 n^2-1}$ and $b=n+\sqrt{n^2-1}$. 
$$f(1)=\frac{1}{4} \left(1+\sqrt{3}\right) \pi$$
$$f(2)=\frac{1}{4} \left(2-\sqrt{3}+\sqrt{15}\right) \pi$$
$$f(3)=\frac{1}{4} \left(3-2 \sqrt{2}+\sqrt{35}\right) \pi$$
$$f(4)=\frac{1}{4} \left(4+3 \sqrt{7}-\sqrt{15}\right) \pi$$
$$f(5)=\frac{1}{4} \left(5-2 \sqrt{6}+3 \sqrt{11}\right) \pi$$
$$f(6)=\frac{1}{4} \left(6-\sqrt{35}+\sqrt{143}\right) \pi$$
$$f(7)=\frac{1}{4} \left(7-4 \sqrt{3}+\sqrt{195}\right) \pi$$
$$f(8)=\frac{1}{4} \left(8-3 \sqrt{7}+\sqrt{255}\right) \pi$$
$$f(9)=\frac{1}{4} \left(9-4 \sqrt{5}+\sqrt{323}\right) \pi$$
$$f(10)=\frac{1}{4} \left(10-3 \sqrt{11}+\sqrt{399}\right) \pi$$
which are exactly what Pranav Arora answered (what I missed) $$f(n)=\frac \pi 4(a-b)$$
A: Though this isn't an answer, this is interesting enough to me but too large for a comment. Based on Vladimir's solution, if we know
$$f(a) = \int_0^\infty \frac{\ln[(1+x^a)/2]}{\ln x} \frac{1}{1+x^2}dx = a\frac{\pi}{4}$$
then we should have
$$f'(a) = \int_0^\infty \frac{\ln x e^{a\ln x}}{\ln x (1+e^{a \ln x})} \frac{1}{1+x^2} dx = \pi/4$$
or
$$f'(a) = \int_0^\infty \frac{x^a}{1+x^a} \frac{1}{1+x^2}dx = \int_0^\infty \left(1-\frac{1}{1+x^a} \right) \frac{1}{1+x^2}dx = \pi/4.$$
This is quite interesting because I would not expect the integral to be constant as a function of $a$. Furthermore, we should expect
$$f''(a) = \int_0^\infty \frac{\ln x \cdot x^a}{(1+x^a)^2}\frac{1}{1+x^2} dx = 0.$$
I didn't bother to check carefully for convergence issues (passing the derivative through), but I think everything cis okay. Does anyone know how to compute the above integrals without referring to Vladimir's answer?
A: 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,
$$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\frac\pi4(a-b).\tag3$$
