Computing the exact value of the integral $∫_0^∞ \tanh(2x)\ln(\tanh x)dx.$ LATEST EDIT
Thanks to @FDP’s alternative method and @Claude Leibovici’s generalisation on the integral. Meanwhile, I had found a formula for the integral with power n in general.
$$\boxed{I_n=∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx =\frac{(-1)^n n !}{2^n}\left(1-\frac{1}{2^{n+1}}\right)\zeta(n+1)} $$
and its proof is shown below.

Recently, I investigate the integral
$$∫_0^∞ \tanh(2x)\ln (\tanh x)dx,$$
using the substitution $y=\tanh x$.
$$
\begin{aligned}
I &=\int_0^{\infty} \frac{2 \tanh x}{1+\tanh ^2 x} \ln (\tanh x) d x \\
&=\int_0^{\infty} \frac{2 y \ln y}{1+y^2} \cdot \frac{d y}{2\left(1-y^2\right)} \\
&=\int_0^{\infty} \frac{y \ln y d y}{1-y^4} \\
&\stackrel{y^2\mapsto y}{=} \frac{1}{4} \int_0^{\infty} \frac{\ln y}{1-y^2} d y
\end{aligned}
$$
By my post,
$$
\begin{aligned}
&\int_0^{\infty} \frac{\ln y}{1-y^2} d y=-\frac{\pi^2}{4}
\end{aligned}
$$
We now conclude that
$$\boxed{∫_0^∞ \tanh(2x)\ln (\tanh x)dx= -\frac{\pi^2}{16}}$$
Is there any other method to evaluate the integral? Your comments and alternative methods are highly appreciated.
 A: After submitting the question, I was reminded by @ Claude Leibovici that I had added $n\in N$ originally. I am sorry for that and I want to generalise the integral to
$$I_n=∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx,$$
where $n\in N.$
Using the same substitution $y=\tanh x$, we have
$$
\begin{aligned}
I_n &=\int_0^{\infty} \frac{y \ln ^n y d y}{1-y^4} \stackrel{y^2\mapsto y}{=} \frac{1}{2^{n+1}} \int_0^{\infty} \frac{\ln ^n y}{1-y^2} d y= \frac{1}{2^n} \int_0^1 \frac{\ln ^n y}{1-y^2} d y
\end{aligned}
$$
For the last integral, expanding the denominator yields
$$\int_0^1 \frac{\ln ^n y}{1-y^2} d y =\sum_{k=0}^{\infty} \int_0^1 y^{2 k} \ln ^n y d y =\left.\frac{\partial^n}{\partial a^n} \int_0^1 y^a d y\right|_{a=2 k} \\= \sum_{k=0}^{\infty}\frac{(-1)^n n !}{(2 k+1)^{n+1}} =n!\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1) $$
Hence we can conclude that
$$\boxed{∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx =\frac{(-1)^n n !}{2^n}\left(1-\frac{1}{2^{n+1}}\right)\zeta(n+1)} $$
For examples,
$$I_5= -\frac{5!}{2^5}\left(1-\frac{1}{2^6}\right) \zeta(6)=-\frac{\pi^6}{256}$$
which is checked by WA.
A: \begin{align}J&=\int_0^\infty \tanh(2x)\ln (\tanh x)dx\\
&=\int_0^\infty  \frac{\text{e}^{2x}-\text{e}^{-2x}}{\text{e}^{2x}+\text{e}^{-2x}}\ln\left(\frac{\text{e}^{x}-\text{e}^{-x}}{\text{e}^{x}+\text{e}^{-x}}\right)dx\\
&=\int_0^\infty  \frac{1-\text{e}^{-4x}}{1+\text{e}^{-4x}}\ln\left(\frac{1-\text{e}^{-2x}}{1+\text{e}^{-2x}}\right)dx\\
&\overset{u=-\text{e}^{-2x}}=\frac{1}{2}\int_0^1 \frac{1-u^2}{(1+u^2)u}\ln\left(\frac{1-u}{1+u}\right)du\\
&\overset{z=\frac{1-u}{1+u}}=\int_0^1 \frac{2z\ln z}{1-z^4}dz\\
&\overset{t=z^2}=\frac{1}{2}\int_0^1 \frac{\ln t}{1-t^2}dt\\
&=\frac{1}{2}\int_0^1 \frac{\ln t}{1-t}dt-\frac{1}{2}\underbrace{\int_0^1 \frac{t\ln t}{1-t^2}dt}_{y=t^2}\\
&=\frac{3}{8}\int_0^1 \frac{\ln t}{1-t}dt\\
&=\frac{3}{8}\times -\frac{\pi^2}{6}=\boxed{-\frac{\pi^2}{16}}
\end{align}
A: If we consider
$$I_n=\int \tanh(nx)\,\log(\tanh(x))\,dx$$ there is a (nasty) antiderivative involving logarithms of complex arguments and polylogarithms.
Doing the same as you did and using multiple angles formulae
$$I_1=\int_0^1  \frac y{1-y^2}\log(y)\,dy=-\frac{\pi ^2}{24}$$
$$I_2=\int_0^1  \frac {2y}{1-y^4}\log(y)\,dy=-\frac{\pi ^2}{16}$$
$$I_3=\int_0^1 \frac{y \left(y^2+3\right) }{1-2y^2-3 y^4}\log (y)\,dy=-\frac{5 \pi ^2}{72}-\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{6}-\frac{\log ^2(3)}{12}$$
$$I_4=\int_0^1 \frac{4 y \left(y^2+1\right) }{\left(1-y^2\right) \left(1+6 y^2+y^4\right) }\log (y)\,dy=
-\frac{\pi ^2}{16}-\frac{1}{16} \log ^2\left(3-2 \sqrt{2}\right)$$
$$I_5=\int_0^1 \frac{y \left(y^4+10 y^2+5\right) }{1-9y^2+5y^4+5 y^6}\log (y)\,dy=$$
$$\frac{1}{120} \left(-6 \left(2 \text{Li}_2\left(-1+\frac{2}{\sqrt{5}}\right)-2 \text{Li}_2\left(-5+2 \sqrt{5}\right)+\log ^2\left(5+2 \sqrt{5}\right)\right)-7 \pi ^2\right)$$
