Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$. I then differentiate to get $$I'(a)=\int_0^1 \frac{ax\ln(1+x)}{1-ax}dx=\int_0^1 ax\ln(1+x)\sum_{n=0}^\infty(ax)^ndx=\sum_{n=1}^\infty a^{n+1}\int_0^1 x^{n+1}\ln(1+x)dx$$
Evaluating this integral by integration by parts and geometric series I get
$$\int_0^1 x^{n+1}\ln(1+x)dx=\frac{x^{n+2}}{n+2}\ln(1+x)|_0^1-\frac{1}{n+2}\int_0^1 \frac{x^{n+2}}{1+x}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\int_0^1 x^{n+2}\sum_{k=0}^\infty(-x)^kdx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty(-1)^k\int_0^1 x^{k+n+2}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty\frac{(-1)^k}{k+n+2}=\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)$$ So I arrive at $$I'(a)=\sum_{n=0}^\infty a^{n+1}\left(\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)\right)$$
Re-indexing I get
$$I'(a)=\frac{\ln(2)}{a}\sum_{n=2}^\infty \frac{a^n}{n}+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n}a^{n-1}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n}a^{n-1}$$Integrating both sides from $0$ to $1$ I recover $I(1)$
$$I(1)=\int_0^1 \frac{\ln(2)}{a}\left(-\ln(1-a)-a\right)da+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ Then using the integral equation for the Dilogarithm I arrive at
$$I(1)=\ln(2)\int_0^1 -\frac{\ln(1-a)}{a}da-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
$$I(1)=\frac{\ln(2)\pi^2}{6}-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ 
At this point I could not continue further as I did not know how to simplify the Digamma terms in the sums. I think that by using the Digamma function's relation to the Harmonic Numbers it could be possible to exploit known values of Harmonic sums to arrive at the answer but I could not get the sums in a form where this would work. If anyone could help me continue further or let me know if I am on the right track I would greatly appreciate it. Thank you in advance.
 A: Another solution, this time using Mike Spivey's alternating Euler sum:
\begin{align*}
I &=-\int_0^1\ln(1-x)\ln(1+x)\,\mathrm dx \\ 
 &= \int_0^1\ln(1-x)\sum_{n=1}^\infty\frac{(-1)^n}{n}x^n\,\mathrm dx,\qquad\text{Mercator series}\\ 
 &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\ln(1-x)x^n\,\mathrm dx\\ 
 &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha}\left[ \int_0^1(1-x)^\alpha x^n\,\mathrm dx\right ]\Bigg\vert_{\alpha=0}\\ 
 &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha} \text{B}\left(\alpha+1,\,n+1 \right )\Bigg\vert_{\alpha=0},\qquad\text{beta function}\\ 
 &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\left[ \frac{\Gamma(\alpha+1)\Gamma(n+1)\left(\psi(\alpha+1)-\psi(\alpha+n+2) \right )}{\Gamma(\alpha+n+2)}\right ]\Bigg\vert_{\alpha=0}\\ 
 &= -\sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{H_{n+1}}{n+1},\qquad\text{used }\psi(1)=-\gamma\text{ and }\psi(m)=H_{m-1}-\gamma\\ 
 &= -\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ 
 &= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ 
 &= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-1+2\ln(2)+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\
 &= -2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-2+2\ln(2)\\
 &= \frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2,\qquad\text{applied Mike's sum.}
\end{align*}
A: \begin{align}J&=\int_0^1 \ln(1+x)\ln(1-x)\,dx\\
&=\Big[\left(\left(1+x\right)\ln(1+x)-x-2\ln 2+1\right)\ln(1-x)\Big]_0^1+\int_0^1\frac{\left(1+x\right)\ln(1+x)-x-2\ln 2+1}{1-x}\,dx\\
&=\int_0^1\frac{\left(1+x\right)\ln(1+x)-x-2\ln 2+1}{1-x}\,dx\\
\end{align}
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
\begin{align}J&=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}-\frac{1-x}{1+x}-2\ln 2+1}{x(1+x)}\\
&=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{x(1+x)}\,dx\\
&=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{x}\,dx-\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{1+x}\,dx\\
&=\int_0^1\frac{\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}+\frac{2x}{1+x}-2\ln 2}{x}\,dx+2+2\ln^2 2-4\ln 2\\
&=\int_0^1 \left(\frac{2\ln\left(\frac{2}{1+x}\right)}{x(1+x)}-\frac{2\ln 2}{x}\right)\,dx+2+2\ln^2 2-2\ln 2\\
&=\int_0^1 \left(\frac{2\ln\left(\frac{2}{1+x}\right)}{x}-\frac{2\ln\left(\frac{2}{1+x}\right)}{1+x}-\frac{2\ln 2}{x}\right)\,dx+2+2\ln^2 2-2\ln 2\\
&=-2\int_0^1 \frac{\ln(1+x)}{x}\,dx-2\int_0^1 \frac{\ln\left(\frac{2}{1+x}\right)}{1+x}\,dx+2+2\ln^2 2-2\ln 2\\
&=-2\int_0^1 \frac{\ln(1+x)}{x}\,dx+2+\ln^2 2-2\ln 2\\
\end{align}
But,
\begin{align}\int_0^1 \frac{\ln(1+x)}{x}\,dx=\frac{1}{2}\int_0^1 \frac{2x\ln(1-x^2)}{x^2}\,dx-\int_0^1 \frac{\ln(1-x)}{x}\,dx\end{align}
In the second integral perform the change of variable $y=x^2$,
\begin{align}\int_0^1 \frac{\ln(1+x)}{x}\,dx&=-\frac{1}{2}\int_0^1 \frac{\ln(1-x)}{x}\,dx\\
&=-\frac{1}{2}\times -\frac{\pi^2}{6}\\
&=\frac{\pi^2}{12}
\end{align}
Thus,
\begin{align}\boxed{J=2+2\ln^2 2-2\ln 2-\dfrac{\pi^2}{6}}\end{align}
A: Proceed as follows
\begin{align}
&\int_0^1 \ln(1-x)\ln(1+x)\,dx \\
=&\int_0^1\ln2 \ln(1-x)dx +\int_0^1 {\ln(1-x)\ln\frac{1+x}2dx}, \>\>\>{IBP} \\
=& -\ln2 +\int_0^1 x\left( \frac{\ln\frac{1+x}2}{1-x} -\frac{\ln(1-x)}{1+x}\right) dx\\=&-\ln2 +1 -\underbrace{\int_0^1 \ln\frac{1+x}2dx}_{=\ln2-1}+\int_0^1\frac{\ln(1-x)}{1+x}dx +\int_0^1  \underbrace{\frac{\ln\frac{1+x}2}{1-x}dx}_{IBP}\\
=& 2-2\ln2 +2 \int_0^1\frac{\ln(1-x)}{1+x}dx, \>\>\>\>\>
{t=\frac{1-x}{1+x}}\\
=&2-2\ln2 +2\int_0^1 \frac{\ln2-\ln(1+t)+\ln t}{1+t}dt\\
=& 2-2\ln2 +\ln^22-\frac{\pi^2}6
\end{align}
A: Some other basic method:
$$\int_0^1 -\log(1+x)\log(1-x) \, {\rm d}x \\
= (1-x)\log(1+x)\log(1-x)\Big|_0^1 - \int_0^1 (1-x) \left[ \frac{\log(1-x)}{1+x} - \frac{\log(1+x)}{1-x}\right] {\rm d}x  \\
\stackrel{t=1-x}{=} (1+x)\left[\log(1+x)-1\right] \Big|_0^1 - \int_0^1 \frac{t/2 \cdot \log t}{1-t/2} \, {\rm d}t \\
=2\log 2 - 1 - \sum_{n=1}^\infty \int_0^1 (t/2)^n \log t \, {\rm d}t \\
=2\log 2-1 + \sum_{n=1}^\infty \frac{2^{-n}}{(n+1)^2} \\
=2\log 2 - 2 + 2\,{\rm Li_2}(1/2) \\
=2\log 2 - 2 + \frac{\pi^2}{6} - \log^22
$$
