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.
 A: $$I=\int x \tanh (2 x) \log (\coth (x))\,dx$$ 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
A: With the variable change $t=\tanh x$
\begin{align}
I= &\int_{0}^{\infty } x\tanh( 2x) \ln(\coth x)\ dx\\
=&\int_0^1 \frac {t \ln t \ln \frac{1-t}{1+t}}{1-t^4}\ dt
\overset{t\to\frac{1-t}{1+t}}
=\frac14 \int_0^1 \ln t \ln \frac{1-t}{1+t}\left(\frac1t-\frac{2t}{1+t^2}\right)dt\tag1
\end{align}
Note that
\begin{align}
J=\int_0^1 \frac{t\ln t \ln \frac{1-t}{1+t}}{1+t^2}dt
\overset{t\to\frac{1-t}{1+t}}
= \int_0^1 \frac{\ln t \ln \frac{1-t}{1+t}}{1+t}dt-J=\frac12 \int_0^1 \frac{\ln t \ln \frac{1-t}{1+t}}{1+t}dt
\end{align}
and substitute $J$ into (1) to get
\begin{align}
I=& \ \frac14 \int_0^1 \ln t \ln \frac{1-t}{1+t}\left(\frac1t-\frac{1}{1+t}\right)
\overset{t\to\frac{1-t}{1+t}}{dt}\\
=& \ \frac14 \bigg( \int_0^1 \frac{\ln t \ln (1-t)}{1-t}dt
- \int_0^1 \frac{\ln t \ln (1+t)}{1-t}dt\bigg)\tag2
\end{align}
Next, let
\begin{align}
K(a) &= \int_0^1\frac{\ln t}{1+t}\left(\ln(1-t) + \frac{ 1}t\ln(1-at) \right)dt\\
K’(a) &=-\int_0^1\frac{\ln t}{(1+t)(1-at)}dt= -\frac{1}{1+a}\int_0^1 \bigg(\frac{\ln t}{1+t}+ \overset{y=at}{\frac{a\ln t}{1-at}} \bigg){dt} \\
&= \frac{\zeta(2)}{2(1+a)} - \frac{\ln a\ln(1-a)}{1+a}- \frac1{1+a}\int_0^a\frac{\ln y}{1-y}dy\\
\end{align}
Then
\begin{align}
&\int_0^1 \frac{\ln t\ln(1-t)}{1-t}dt = \int_0^1 \frac{\ln t\ln(1-t)}{t}dt \\
=& \ K(1)=K(0)+ \int_0^1 K’(a)da = \frac{\zeta(2)}2\ln2-\int_0^1 d[\ln(1+a)]\int_0^a \frac{\ln y}{1-y}dy \\ \overset{ibp}= &\ \frac{3}2\ln2\ \zeta(2)+ \int_0^1 \frac{\ln a \ln(1+a)}{1-a}da \tag3\\\end{align}
Substitute (3) into (2) to obtain
$$I=\frac14\cdot \frac{3}2\ln2\ \zeta(2)= \frac{\pi^2}{16}\ln2$$
