Evaluate $\int\limits_0^{\ln{2}}\ln(1+e^x)\ln(1-e^x)dx$

Question

Wolfram says it is approximately $-0.65111415588...$ but I am hoping for an exact solution. Here is my attempt:

$$\int\limits_0^{\ln{2}}\ln\left(e^{x}+1\right)\ln\left(e^{x}-1\right)dx$$ $$\int\limits_0^{\ln{2}}\ln\left(u\right)\ln\left(e^{x}-1\right)\frac{du}{e^x}$$ $$\int\limits_2^3 e^{-x}\ln\left(u\right)\ln\left(u-2\right)du$$ $$\int\limits_2^3 e^{-\ln\left(u-1\right)}\ln\left(u\right)\ln\left(u-2\right)du$$ $$\int\limits_2^3 e^{-\ln\left(u-1\right)}\ln\left(u\right)\ln\left(u-2\right)du$$ $$\int\limits_2^3 \frac{\ln\left(u\right)\ln\left(u-2\right)}{u-1}du$$

This is where I got stuck. I am thinking a series expansion might help but I am not sure.

Answer 1

Substitute $t=e^x-1$ $$\int\limits_0^{\ln{2}}\ln\left(e^{x}+1\right)\ln\left(e^{x}-1\right)dx=\int_0^1 \frac{\ln(t+2)\ln t}{t+1}\ dt= -\frac{13}{24}\zeta(3)$$ where the last step utilizes the result.