TITLE: Prove $\int_0^1 \ln(x+1) \ln(x)dx=2-2\ln(2)-\frac{\pi^2}{12}$
QUESTION [6 upvotes]: Prove that
$$I:=\int_0^1 \ln(x+1) \ln(x)dx=2-2\ln(2)-\frac{\pi^2}{12}$$

I've tried integration by parts and u substitution and cannot make it work but I have one potential path
I use
$$\ln(x+1)=\sum^{\infty}_{n=1} \frac{(-1)^{n-1}x^n}{n}
$$
$$\Rightarrow I=\int_0^1 \ln(x)\sum^{\infty}_{n=1} \frac{(-1)^{n-1}x^n}{n}dx=\sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \int_0^1 x^{n-1} \ln(x)dx= 
$$
How does one proceed from here?
Thank you for your time

REPLY [4 votes]: Integrate by parts
\begin{align}
&\int_0^1 \ln(x+1) \ln x \>dx\\
=& -\int_0^1 \ln(1+x)\>dx -\int_0^1 \ln x \> dx+\int_0^1 \frac{\ln x}{1+x}dx\\
=&-(2\ln2-1)-(-1)+(-\frac{\pi^2}{12}) =2-2\ln2 -\frac{\pi^2}{12}
\end{align}
where $\int_0^1 \frac{\ln x}{1+x}dx= -\frac{\pi^2}{12}$