The following integral can be characterised as an exotic one!
Evaluate the integral:
$$\int_{0}^{\pi/2} \left ( \frac{\log \left ( \tan x +1 \right )}{\log \tan x} - \frac{{\rm Li}_2 \left ( -\cot x \right )}{\log^2 \cot x} - \frac{\zeta(2)}{2 \log^2 \tan x} \right ) \, {\rm d}x$$
It was proposed by Cornel Ioan Valean, Romania.
I don't have any clue on how to attack this. Also, I don't think I can come up with any strategy that could work. So, I leave it entirely up to the community to come up with a clever approach.
Edit: The name of the proposer is Cornel not Corel that I had written.
$$ I = \int_0^{\frac{\pi}{2}} \left(\frac{\ln(\tan x+1)}{\ln(\tan x)} -\frac{\text{Li}_2(-\cot x)}{\ln^2(\tan x)} - \frac{\zeta(2)}{2\ln^2(\tan x)}\right) dx \\ = \left(\int_0^\frac{\pi}{4} + \int_\frac{\pi}{4}^\frac{\pi}{2}\right) (\ldots)dx = I_1 + I_2 $$ On $I_2$, perform the substitution $x\to\frac{\pi}{2}-x$ to obtain: $$ I_2 = \int_0^{\frac{\pi}{4}} \left(\frac{-\ln(\cot x+1)}{\ln(\tan x)} -\frac{\text{Li}_2(-\tan x)}{\ln^2(\tan x)} - \frac{\zeta(2)}{2\ln^2(\tan x)}\right) dx $$ Summing $I_1$ and $I_2$, we get: $$ I = \int_0^{\frac{\pi}{4}} \left(\frac{\ln(\tan x+1) - \ln(\cot x + 1)}{\ln(\tan x)} -\frac{\text{Li}_2(-\tan x)+\text{Li}_2(-\cot x)+\zeta(2)}{\ln^2(\tan x)}\right) dx $$ By the inversion formula, we have: $$ \text{Li}_2(-\tan x)+\text{Li}_2\left(\frac{1}{-\tan x}\right)+\zeta(2) = - \frac{1}{2}\ln^2(\tan x) $$ Also as : $$ \ln(\tan x+1) - \ln(\cot x + 1) = \ln\left(\frac{\tan x+1}{\cot x + 1}\right) = \ln\left(\tan x \ \frac{\tan x+1}{1 + \tan x}\right) = \ln(\tan x) $$ We can substitute in: $$ I = \int_0^{\frac{\pi}{4}} \left(\frac{\ln(\tan x)}{\ln(\tan x)} -\frac{-\frac{1}{2}\ln^2(\tan x)}{\ln^2(\tan x)}\right) dx = \int_0^{\frac{\pi}{4}}\frac{3}{2} \ dx = \frac{3\pi}{8} $$