Prove $\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$ How to prove without using Euler sums that 

$$I=\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$$

where $\zeta$ is the Riemann zeta function.
We can relate this integral to some Euler sum as follows:
\begin{align}
I&=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^n\ln x}{1-x}\ dx\\
&=-\sum_{n=1}^\infty\frac{(-1)^n}{n}(H_n^{(2)}-\zeta(2))\\
&=-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n}-\ln2\zeta(2)
\end{align}
Also the integral $I$ can be related to $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}$. So I am looking for a different way to evaluate $I$ besides using these two sums.
 A: Start off with the substitution $x\to \frac{1-x}{1+x}$ to get:
$$\require{cancel} I=\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{x}dx-\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{1+x}dx$$
$$X=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{x}dx$$
$$Y=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{1+x}dx$$

$$I_1=\ln 2\int_0^1 \frac{\ln(1-x)}{x}dx=\color{blue}{-\ln 2 \zeta(2)}$$
$$I_2=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx=\color{red}{\frac{5}{8}\zeta(3)}$$
$$I_3=-\ln 2 \int_0^1 \frac{\ln(1+x)}{x}dx=\color{blue}{-\frac{\ln 2}{2}\zeta(2)}$$
$$I_4=\int_0^1 \frac{\ln^2(1+x)}{x}dx=\color{red}{\frac{\zeta(3)}{4}}$$
$$I_5=\ln 2\int_0^1 \frac{\ln(1-x)}{1+x}dx=\cancel{\frac{\ln^3 2}{2}}-\cancel{\ln 2\frac{\zeta(2)}{2}}$$
$$I_6=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{1+x}dx =\cancel{-\frac{\ln^3 2}{3}}+\cancel{\ln 2\frac{\zeta(2)}{2}}-\color{red}{\frac{\zeta(3)}{8}}$$
$$I_7=-\ln 2 \int_0^1 \frac{\ln(1+x)}{1+x}dx=\cancel{-\frac{\ln^3 2}{2}}$$
$$I_8=\int_0^1 \frac{\ln^2(1+x)}{1+x}dx=\cancel{\frac{\ln^3 2}{3}}$$

$$I=X-Y=(I_1+I_2+I_3+I_4)-(I_5+I_6+I_7+I_8)=\boxed{\zeta(3)-\frac32 \ln 2 \zeta(2)}$$
A: \begin{align}J&=\int_0^1 \frac{\ln x\ln(1+x)}{1-x}\,dx\end{align}
Always the same story...
For $x\in [0;1]$ define the function $R$ by,\begin{align}R(x)&=\int_0^x \frac{\ln t}{1-t}\,dt\\
&=\int_0^1 \frac{x\ln(tx)}{1-tx}\,dt\\
J&=\Big[R(x)\ln(1+x)\Big]_0^1-\int_0^1 \frac{R(x)}{(1+x)} dx\\
&=-\zeta(2)\ln 2-\int_0^1 \int_0^1 \frac{x\ln(tx)}{(1-tx)(1+x)}\,dt\,dx\\
&=-\zeta(2)\ln 2-\int_0^1 \left(\int_0^1 \frac{x\ln t}{(1-tx)(1+x)}\,dx\right)\,dt-\int_0^1 \left(\int_0^1 \frac{x\ln x}{(1-tx)(1+x)}\,dt\right)\,dx\\
&=-\zeta(2)\ln 2+\int_0^1\left[\frac{\ln(1-tx)}{t(1+t)}+\frac{\ln(1+x)}{1+t}\right]_{x=0}^{x=1}\ln t\,dt+\int_0^1\left[\frac{\ln(1-tx)}{1+x}\right]_{t=0}^{t=1}\ln x\,dx\\
&=-\zeta(2)\ln 2+\int_0^1 \frac{\ln(1-t)\ln t}{t(1+t)}\,dt+\ln 2\int_0^1 \frac{\ln t}{1+t}\,dt+\int_0^1 \frac{\ln(1-x)\ln x}{1+x}\,dx\\
&=-\zeta(2)\ln 2+\int_0^1 \frac{\ln(1-t)\ln t}{t}\,dt+\ln 2\int_0^1 \frac{\ln t}{1+t}\,dt\\
&=-\zeta(2)\ln 2+\frac{1}{2}\left(\Big[\ln^2 x\ln(1-x)\Big]+\int_0^1 \frac{\ln^2 t }{1-t}\,dt\right)+\ln 2\left(\int_0^1 \frac{\ln t}{1-t}\,dt-\int_0^1 \frac{2t\ln t}{1-t^2}\,dt\right)
\end{align}
In the last integral perform the change of variable $y=t^2$,
\begin{align}J&=-\zeta(2)\ln 2+\frac{1}{2}\int_0^1 \frac{\ln^2 t}{1-t}\,dt+\ln 2\left(\int_0^1 \frac{\ln t}{1-t}\,dt-\frac{1}{2}\int_0^1 \frac{\ln t}{1-t}\,dt\right)\\
&=-\frac{3}{2}\zeta(2)\ln 2+\frac{1}{2}\times 2\zeta(3)\\
&=\boxed{-\frac{3}{2}\zeta(2)\ln 2+\zeta(3)}
\end{align}
NB: i assume,\begin{align}R(1)&=\int_0^1 \frac{\ln t}{1-t}\,dt\\
&=-\zeta(2)\\
\int_0^1 \frac{\ln^2 t}{1-t}\,dt&=2\zeta(3)\end{align}
A: You might need the first generalization from the preprint A simple strategy of calculating two alternating harmonic series generalizations by Cornel Ioan Valean. All the calculations are accomplished by avoiding the evaluation of Euler sums.
$$
\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x}{2}\right)}{1-x}\textrm{d}x
$$
$$=\frac{1}{2}\biggr(m\zeta (m+1)-2\log (2) \left(1-2^{1-m}\right) \zeta (m)$$
$$-\sum_{k=1}^{m-2} \left(1-2^{-k}\right)\left(1-2^{1+k-m}\right)\zeta (k+1)\zeta (m-k)\biggr).$$
A: Consider the integral $$K=\int_0^1\frac{\operatorname{Li}_2(x)}{1+x}\ dx$$
By applying IBP we have
$$K=\ln(2)\zeta(2)+\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx\tag{1}$$
.

On the other hand
\begin{align}
K&=\int_0^1\frac{\operatorname{Li}_2(x)}{1+x}\ dx=\int_0^1\frac{1}{1+x}\left(\int_0^1-\frac{x\ln u}{1-xu}\ du\right)\ dx\\
&=\int_0^1\ln u\left(\int_0^1\frac{-x}{(1+x)(1-ux)}\ dx\right)\ du\\
&=\int_0^1\ln u\left(\frac{\ln2}{1+u}+\frac{\ln(1-u)}{u}-\frac{\ln(1-u)}{1+u}\right)\ du\\
&=\ln2\int_0^1\frac{\ln u}{1+u}\ du+\int_0^1\frac{\ln u\ln(1-u)}{u}\ du-\color{blue}{\int_0^1\frac{\ln u\ln(1-u)}{1+u}\ du}\\
&\overset{\color{blue}{IBP}}{=}\ln2\left(-\frac12\zeta(2)\right)+\zeta(3)\color{blue}{-\int_0^1\frac{\ln u\ln(1+u)}{1-u}du+\int_0^1\frac{\ln(1-u)\ln(1+u)}{u}du}\tag{2}
\end{align}

Combining (1) and (2) we get

$$\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$$

