Definite integral $ \int _{0}^{\infty } x\cdotp \tanh( 2x) \cdotp \ln(\coth x)\mathrm{d} x$ I want to show that
$\displaystyle \int\limits _{0}^{\infty } x\cdotp \tanh( 2x) \cdotp \ln(\coth x)\mathrm{d} x=\frac{\pi ^{2} \cdotp \ln( 2)}{2^{4}}\tag*{}$
I tried integration by parts, Feynman trick with no luck. It just gets complicated.

$$I=\int x \tanh (2 x) \log (\coth (x))\,dx$$ Letting $x=\log(t)$ leads to
$$I=\int \frac{t^4-1}{t(t^4+1)}\log (t) \log \left(\frac{t^2+1}{t^2-1}\right)\,dt$$ Now $t=\sqrt u$ gives
$$I=\frac 14 \int \frac{u^2-1}{u(u^2+1)} \log (u) \log \left(\frac{u+1}{u-1}\right)\,du$$ Using partial fraction decomposition
$$I=\frac 14 \int \Big[\frac{1}{u+i}+\frac{1}{u-i}-\frac{1}{ u}\Big]\big[ \log(u+1)-\log(u-1)\big]\,du$$
Now, we face six integrals
$$J(a,b)=\int \frac{\log (u+a)}{u+b}\,du=\text{Li}_2\left(\frac{a+u}{a-b}\right)+\log (a+u) \log   \left(\frac{b+u}{b-a}\right)$$
Now, back to $x$ with $u=e^{2x}$ and using the bounds leads to the resired result
