Prove $\int_0^{\pi/2}{\frac{1+2\cos 2x\cdot\ln\tan x}{1+\tan^{2\sqrt{2}} x}}\tan^{1/\sqrt{2}} x~dx=0$ I'm curious, how one can prove the following integral
$$
\int_0^{\pi/2}{\frac{1+2\cos 2x\cdot\ln\tan x}{1+\tan^{2\sqrt{2}} x}}\tan^{1/\sqrt{2}} x~dx=0
$$
Here is the Wolfram Alpha computation which shows that it is correct to at least 45 digits.
My attempt: I knew the integral
$$
\int_0^{\pi/2}\frac{1}{1+\tan^\alpha\phi}d\phi=\int_0^{\pi/4}\frac{1}{1+\tan^\alpha\phi}d\phi+\int_0^{\pi/4}\frac{\tan^\alpha\phi}{1+\tan^\alpha\phi}d\phi=\frac{\pi}{4}
$$
which can be calculated for all values of $\alpha$. I tried to find an analogous symmetry that will allow me to cancel all the terms also in this case, but so far no luck. I also suspect that this integral might be related to derivative of Herglotz integral. Herglotz showed that 
$$
\int_{0}^{1}
\frac{\ln\left(1 + t^{\,{\large\alpha}}\right)}{1 + t}\,{\rm d}t
$$
can be computed for some algebraic values of $\alpha$, e.g. $\alpha=4+\sqrt{5}$.
If we take derivative of this integral with respect to $\alpha$ then we get
$$
\int_{0}^{1}
\frac{t^\alpha\ln t}{(1 + t)(1+t^\alpha)}\,{\rm d}t
$$
Change of variables $t=\tan^2\phi$ gives
$$
4\int_{0}^{\pi/4}
\frac{\tan^{2\alpha+1}\phi\cdot\ln \tan\phi}{1+\tan^{2\alpha}\phi}\,{\rm d}\phi
$$
which looks quite similar to the integral under consideration.
 A: This answer builds on observations made by user90369 and Yuriy S. They define the functions
$$
Y_1(A,P):=\int_0^\infty \frac{\cosh Ax}{\cosh 2Ax} \frac{1}{\cosh P x} dx,\tag{1}
$$
$$
Y_2(A,P):=-2\int_0^\infty \frac{\cosh Ax}{\cosh 2Ax} \frac{x \sinh P x}{\cosh^2 P x} dx=2\frac{\partial}{\partial P} Y_1(A,P),\tag{2}
$$
and then note that OPs conjecture is equivalent to $Y_1(1/\sqrt2,1)+Y_2(1/\sqrt2,1)=0$.
I happened to notice that $(1)$ is an integral of a product of two self-reciprocal functions $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$ (functions 3 and 4 in this list):
$$
f_1(x)=\frac{1}{\cosh\sqrt{\frac{\pi}{2}}x},\quad f_2(x)=\frac{\cosh \frac{\sqrt{\pi}x}{2}}{\cosh \sqrt{\pi}x}.\tag{3}
$$
For any such two functions Ramanujan shows that (simple proof is outlined in this answer)
$$
\int_0^\infty f_1(x)f_2(\alpha x) dx=
\frac{1}{\alpha}\int_0^\infty f_1(x)f_2(x/\alpha) dx.\tag{4}
$$
Substituting $(3)$ in $(4)$ one obtains
$$
\int_0^\infty \frac{1}{\cosh\sqrt{\frac{\pi}{2}}\alpha x}\frac{\cosh \frac{\sqrt{\pi}x}{2}}{\cosh \sqrt{\pi}x} dx=
\frac{1}{\alpha}\int_0^\infty \frac{1}{\cosh\sqrt{\frac{\pi}{2}}x/\alpha}\frac{\cosh \frac{\sqrt{\pi}x}{2}}{\cosh \sqrt{\pi}x} dx,
$$
and after trivial algebra
$$
\int_0^\infty \frac{1}{\cosh\alpha x}\frac{\cosh \frac{x}{\sqrt{2}}}{\cosh \sqrt{2}x} dx=
\frac{1}{\alpha}\int_0^\infty \frac{1}{\cosh x/\alpha}\frac{\cosh \frac{x}{\sqrt{2}}}{\cosh \sqrt{2}x} dx.
$$
This gives the functional equation 
$$Y_1(1/\sqrt{2},P)=\frac{1}{P}Y_1(1/\sqrt{2},1/P).$$
Differentiating this functional equation with respect to $P$ one obtains
$$
\frac12 Y_2(1/\sqrt{2},P)=-\frac{1}{P^2}Y_1(1/\sqrt{2},1/P)-\frac{1}{2P^3}Y_2(1/\sqrt{2},1/P),
$$
which gives at $P=1$ the required relation
$$
Y_1(1/\sqrt2,1)+Y_2(1/\sqrt2,1)=0.
$$
A: This is not a complete answer, but I think this is a possible solution. If you apply the change of variable $t=\tan(x)$, the original integral can be writen as
$$\int_{0}^{\infty}\frac{1 + 2\left(\frac{1-t^{2}}{1+t^{2}}\right)\log(t)}{1+t^{2\sqrt{2}}}t^{1/\sqrt{2}}\frac{dt}{t^{2} + 1} = \int_{0}^{\infty}\frac{t^{1/\sqrt{2}}}{(t^{2} + 1)(1+t^{2\sqrt{2}})}dt + \int_{0}^{\infty}\frac{2(1-t^{2})\log(t)}{(t^{2} + 1)^{2}(1+t^{2\sqrt{2}})}t^{1/\sqrt{2}}dt,$$
and this integral would be solved by the method of residues for Mellin Transformations.
Checks Pags. 60-62 of http://jacobi.fis.ucm.es/artemio/Notas%20de%20curso/VC.pdf (I couldn't find an english version)
