<s>[INST] 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.
 [/INST] \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}
</s>