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

Start off with the substitution $x\to \frac{1-x}{1+x}$ to get: $$\require{cancel} I=\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{x}dx-\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{1+x}dx$$

$$X=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{x}dx$$ $$Y=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{1+x}dx$$

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

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$$I=X-Y=(I_1+I_2+I_3+I_4)-(I_5+I_6+I_7+I_8)=\boxed{\zeta(3)-\frac32 \ln 2 \zeta(2)}$$

\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}

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.

$$ \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x}{2}\right)}{1-x}\textrm{d}x $$ $$=\frac{1}{2}\biggr(m\zeta (m+1)-2\log (2) \left(1-2^{1-m}\right) \zeta (m)$$ $$-\sum_{k=1}^{m-2} \left(1-2^{-k}\right)\left(1-2^{1+k-m}\right)\zeta (k+1)\zeta (m-k)\biggr).$$