How does one show that $\int_{-\infty}^{+\infty}{\left(x\over 2+2^{x}+2^{-x}\right)^2}\mathrm dx={\zeta(2)-1\over 3}\cdot{1\over \ln^3(2)}?$ Motivated by this question due to the comment of @Olivier Oloa.
$$\int_{-\infty}^{+\infty}{\left(x\over 2+2^{x}+2^{-x}\right)^2}\mathrm dx={\zeta(\color{red}2)-\color{blue}1\over \color{green}3}\cdot{\color{blue}1\over \ln^\color{green}3(\color{red}2)}\tag1$$
Here is my attempt:
$u=2^x\implies du=2^x\ln 2dx$ then $(1)$ becomes
$${1\over \ln^3(2)}\int_{0}^{+\infty}{\ln^2(u)\over (2+u+u^{-1})^2}\cdot{\mathrm du\over u}\tag2$$
$e^v=u$ then $(2)$ becomes
$${1\over \ln^3(2)}\int_{-\infty}^{+\infty}{v^2\over (2+e^v+e^{-v})^2}\mathrm dv\tag3$$
Not sure what is next step...

We begin with the integral with exponentials in the denominator, which is
$$I=\int_{-\infty}^{\infty} \frac{x^2}{(2+e^{x}+e^{-x})^2} \ dx.$$
Let $x=\ln(u),$ and  $dx =\frac{1}{u} \ du,$ in which we have 
$$I=\int_{0}^{\infty} \frac{u \ (\ln(u))^2}{(u+1)^4} \ du.$$
Now split $I$ into
$$I=\int_{0}^{1} \frac{u \ (\ln(u))^2}{(u+1)^4} \ du + \int_{1}^{\infty} \frac{u \ (\ln(u))^2}{(u+1)^4} \ du.$$ Perform a change of variables $u=\frac{1}{z}$ to see that the two integrals are equal to one another, and thus,
$$I=\int_{0}^{1} \frac{2u \ (\ln(u))^2}{(u+1)^4} \ du.$$
Now here is the fun part. Consider the triple integral
$$J=\int_{0}^{1}\int_{1}^{u}\int_{1}^{y} \frac{2u}{zy(1+u)^4} \ dz \ dy \ du.$$ If we integrate this in the order presented, we get $J=I.$ On the other hand, we reverse the order of integration. 
$$J=\int_{0}^{1}\int_{0}^{y}\int_{1}^{y} \frac{4u}{zy(1+u)^4} \ dz \ du \ dy,$$ and integrating (by parts the second time) gives us
$$I= \int_{0}^{1} \frac{2y(3+y) \ln(y)}{3(1+y)^3} \ dy.$$ Expand the integrand with partial fractions and see
$$I= \int_{0}^{1} - \frac{2 \ln(y)}{3(y+1)} \ dy - \int_{0}^{1} \frac{2 \ln(y)}{3(y+1)^2} \ dy + \int_{0}^{1} \frac{4 \ln(y)}{3(y+1)^3} \ dy$$
Now the first term $$\int_{0}^{1} -\frac{2 \ln(y)}{3(y+1)} \ dy=\frac{\zeta(2)}{3}$$ which we can obtain by converting the integrand into a geometric series and using the fact that $$\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+1)^2} =\eta(2)=\frac{\zeta(2)}{2}=\frac{\pi^2}{12}$$ by the Basel Problem. 
The second term  $$\int_{0}^{1} -\frac{2 \ln(y)}{3(y+1)^2} \ dy=\frac{2 \ln(2)}{3},$$ which can be proved by integration by parts $u=\frac{2}{3} \ln(y)$ and $dv = \frac{-1}{(y+1)^2} \ dy$ and a use of partial fractions (on the $\int_{0}^{1} v \ du$ part). 
Similarly, apply the same reasoning to show that $$\int_{0}^{1} \frac{4 \ln(y)}{3(y+1)^3} \ dy=-\frac{1+ 2\ln(2)}{3}$$
Combining the values together we get that $$I= \frac{-1+\zeta(2)}{3}.$$ Finally multiply this value by $\frac{1}{(\ln(2))^3}$ to get the value of your original integral.
