Integral ∫10ln(1−x)ln(1+x)ln2xdx{\large\int}_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx

This answer is split into 3 main steps.

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Step 1: Expressing the integral as a sum

$$\begin{align} &\ \ \ \ \ \int^1_0\ln(1+x)\ln(1-x)\ln^2{x} \ {\rm d}x\\ &=\sum^\infty_{j=1}\frac{(-1)^j}{j}\sum^\infty_{k=1}\frac{1}{k}\int^1_0x^{j+k}\ln^2{x} \ {\rm d}x\\ &=2\sum^\infty_{j=1}\frac{(-1)^j}{j}\sum^\infty_{k=1}\frac{1}{k(k+j+1)^3}\\ &=\small{2\sum^\infty_{j=1}\frac{(-1)^j}{j}\sum^\infty_{k=1}\frac{1}{(j+1)^3k}-\frac{1}{(j+1)^3(k+j+1)}-\frac{1}{(j+1)^2(k+j+1)^2}-\frac{1}{(j+1)(k+j+1)^3}}\\ &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}-2\sum^\infty_{j=1}\frac{(-1)^j\left[\zeta(2)-H_{j+1}^{(2)}\right]}{j(j+1)^2}-2\sum^\infty_{j=1}\frac{(-1)^j\left[\zeta(3)-H_{j+1}^{(3)}\right]}{j(j+1)} \end{align}$$

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Step 2a: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n}$
$$\begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n} &=\frac{1}{2}\ln^2{2}-\frac{\pi^2}{12} \end{align}$$
See here for the details.

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Step 2b: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2}$
$$\begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2} &=-\frac{5}{8}\zeta(3) \end{align}$$
See here for the details.

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Step 2c: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}$
$$\begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3} &=\int^{-1}_0\frac{1}{y}\left[\int^y_0\frac{1}{x}\left[\int^x_0\frac{\ln(1-t)}{t(t-1)}{\rm d}t\right]{\rm d}x\right]{\rm d}y\\ &=2{\rm Li}_4\left(\frac{1}{2}\right)-\frac{11\pi^4}{360}+\frac{1}{12}\ln^4{2}+\frac{7}{4}\zeta(3)\ln{2}-\frac{\pi^2}{12}\ln^2{2} \end{align}$$
Tunk-Fey did a calculation of this type here.

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Step 2d: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n}$
$$\begin{align} \sum^\infty_{n=1}\frac{H_n^{(2)}}{n}x^n &=\int^x_0\frac{{\rm Li}_2(t)}{t(1-t)}{\rm d}t\\ &={\rm Li}_3(x)+\int^x_0\frac{{\rm Li}_2(t)}{1-t}{\rm d}t\\ &={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\int^x_0\frac{\ln^2(1-t)}{t}{\rm d}t\\ &={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)+\int^{1-x}_1\frac{\ln^2{t}}{1-t}{\rm d}t\\ &={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\ln^2(1-x)\ln{x}+\int^{1-x}_1\frac{2\ln(1-t)\ln{t}}{t}{\rm d}t\\ &\small{={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\ln^2(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln(1-x)+\int^{1-x}_1\frac{2{\rm Li}_2(t)}{t}{\rm d}t}\\ &\small{={\rm Li}_3(x)-{\rm Li}_2(x)\ln(1-x)-\ln^2(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln(1-x)+2{\rm Li}_3(1-x)-2\zeta(3)} \end{align}$$
Therefore
$$\begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n} &={\rm Li}_3(-1)-{\rm Li}_2(-1)\ln{2}-\ln^2{2}\ln(-1)-2{\rm Li}_2(2)\ln{2}+2{\rm Li}_3(2)-2\zeta(3)\\ &=-\zeta(3)+\frac{\pi^2}{12}\ln{2} \end{align}$$
You can use polylogarithm identities to simplify the last equation. I took the easy way out and used Wolfram Alpha. Note that contour integration is a slightly more efficient method to solve this sum, however this method is required if I want to solve $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}$ as well.

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Step 2e: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(3)}}{n}$

$$\begin{align} \sum^\infty_{n=1}\frac{H_n^{(3)}}{n}x^n &=\int^x_0\frac{{\rm Li}_3(t)}{t(1-t)}{\rm d}t\\ &={\rm Li}_4(x)+\int^x_0\frac{{\rm Li}_3(t)}{1-t}{\rm d}t\\ &={\rm Li}_4(x)-{\rm Li}_3(x)\ln(1-x)-\int^x_0\frac{-\ln(1-t){\rm Li}_2(t)}{t}{\rm d}t\\ &={\rm Li}_4(x)-{\rm Li}_3(x)\ln(1-x)-\frac{1}{2}{\rm Li}^2_2(x) \end{align}$$
Therefore
$$\begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n^{(3)}}{n} &={\rm Li}_4(-1)-{\rm Li}_3(-1)\ln{2}-\frac{1}{2}{\rm Li}^2_2(-1)\\ &=-\frac{19\pi^4}{1440}+\frac{3}{4}\zeta(3)\ln{2} \end{align}$$

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Step 2f: Value of $\displaystyle \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}$

This part is rather similar to Tunk-Fey’s answer, so he certainly deserves credit.
$$\begin{align} &\ \ \ \ \ \sum^\infty_{n=1}\frac{H_n^{(2)}}{n^2}x^n\\ &=\small{{\rm Li}_4(x)-2\zeta(3)\ln{x}+\frac{1}{2}{\rm Li}_2^2(x)+\color{blue}{\int\frac{-\ln^2(1-x)\ln{x}}{x}{\rm d}x}+\color{\orange}{2\int\frac{{\rm Li}_3(1-x)-{\rm Li}_2(1-x)\ln(1-x)}{x}{\rm d}x}} \end{align}$$
The blue integral is
$$\begin{align} &\ \ \ \ \ \color{blue}{\int\frac{-\ln^2(1-x)\ln{x}}{x}{\rm d}x}\\ &=-\frac{1}{2}\ln^2{x}\ln^2(1-x)-\int\frac{\ln^2{x}\ln(1-x)}{1-x}{\rm d}x\\ &=-\frac{1}{2}\ln^2{x}\ln^2(1-x)+\sum^\infty_{n=1}H_n\int x^n\ln^2{x} \ {\rm d}x\\ &=-\frac{1}{2}\ln^2{x}\ln^2(1-x)+\sum^\infty_{n=1}H_n\partial^2_n\frac{x^{n+1}}{n+1}\\ &=\color\grey{-\frac{1}{2}\ln^2{x}\ln^2(1-x)+\ln^2{x}\sum^\infty_{n=1}\frac{H_nx^{n+1}}{n+1}}-2\ln{x}\sum^\infty_{n=1}\frac{H_nx^{n+1}}{(n+1)^2}+2\sum^\infty_{n=1}\frac{H_{n}x^{n+1}}{(n+1)^3}\\ &=\color{blue}{2\ln{x}{\rm Li}_3(x)-2{\rm Li}_4(x)-2\ln{x}\sum^\infty_{n=1}\frac{H_n}{n^2}x^n+2\sum^\infty_{n=1}\frac{H_n}{n^3}x^n} \end{align}$$
The orange integral is
$$\begin{align} &\ \ \ \ \ \ \color{orange}{2\int\frac{{\rm Li}_3(1-x)-{\rm Li}_2(1-x)\ln(1-x)}{x}{\rm d}x}\\ &=2{\rm Li}_3(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln{x}\ln(1-x)+2\int\frac{\ln(1-x)\ln^2{x}}{1-x}{\rm d}x\\ &=\color{orange}{2{\rm Li}_3(1-x)\ln{x}-2{\rm Li}_2(1-x)\ln{x}\ln(1-x)-\ln^2{x}\ln^2(1-x)-4\ln{x}{\rm Li}_3(x)+4{\rm Li}_4(x)+4\ln{x}\sum^\infty_{n=1}\frac{H_n}{n^2}x^n-4\sum^\infty_{n=1}\frac{H_n}{n^3}x^n} \end{align}$$
So
$$\begin{align} & \ \ \ \ \ \sum^\infty_{n=1}\frac{H_n^{(2)}}{n^2}x^n\\ &=3{\rm Li}_4(x)+2{\rm Li}_3(1-x)\ln{x}-2{\rm Li}_3(x)\ln{x}-2\zeta(3)\ln{x}+\frac{1}{2}{\rm Li}_2^2(x)-2{\rm Li}_2(1-x)\ln{x}\ln(1-x)-\ln^2{x}\ln^2(1-x)+2\ln{x}\sum^\infty_{n=1}\frac{H_n}{n^2}x^n-2\sum^\infty_{n=1}\frac{H_n}{n^3}x^n+C \end{align}$$
Therefore
$$\begin{align} & \ \ \ \ \ \sum^\infty_{n=1}\frac{(-1)^nH_n^{(2)}}{n^2}\\ &=3{\rm Li}_4(-1)+\color\grey{2{\rm Li}_3(2)\ln(-1)-2{\rm Li}_3(-1)\ln(-1)-2\zeta(3)\ln(-1)}\\ &+\frac{1}{2}{\rm Li}_2^2(-1)\color\grey{-2{\rm Li}_2(2)\ln(-1)\ln(2)-\ln^2(-1)\ln^2{2}+2\ln(-1)\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^2}}-2\sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}\\ &=\frac{17\pi^4}{480}-4{\rm Li}_4\left(\frac{1}{2}\right)-\frac{1}{6}\ln^4{2}-\frac{7}{2}\zeta(3)\ln{2}+\frac{\pi^2}{6}\ln^2{2} \end{align}$$
The grey terms miraculously cancel.

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Step 3a: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}$
$$\begin{align} & \ \ \ \ \ 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j(j+1)^3}\\ &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}}{j}-\frac{(-1)^jH_{j+1}}{(j+1)^3}-\frac{(-1)^jH_{j+1}}{(j+1)^2}-\frac{(-1)^jH_{j+1}}{j+1}\\ &=\small{2\sum^\infty_{j=1}\frac{(-1)^jH_{j}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)}+2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^3}+2+2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^2}+2+2\sum^\infty_{j=1}\frac{(-1)^jH_j}{j^3}+2}\\ &=4{\rm Li}_4\left(\frac{1}{2}\right)-\frac{11\pi^4}{180}+\frac{1}{6}\ln^4{2}+\frac{7}{2}\zeta(3)\ln{2}-\frac{5}{4}\zeta(3)-\frac{\pi^2}{6}\ln^2{2}-\frac{\pi^2}{3}+2\ln^2{2}-4\ln{2}+8 \end{align}$$

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Step 3b: Evaluating $\displaystyle -\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}$
$$\begin{align} -\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2} &=-\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j}-\frac{(-1)^j}{(j+1)^2}-\frac{(-1)^j}{j+1}\\ &=\frac{\pi^2}{3}\ln{2}+\frac{\pi^4}{36}-\frac{\pi^2}{3}+\frac{\pi^2}{3}\ln{2}-\frac{\pi^2}{3}\\ &=\frac{\pi^4}{36}+\frac{2\pi^2}{3}\ln{2}-\frac{2\pi^2}{3} \end{align}$$

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Step 3c: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j(j+1)^2}$
$$\begin{align} & \ \ \ \ \ 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j(j+1)^2}\\ &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j}-\frac{(-1)^jH_{j+1}^{(2)}}{(j+1)^2}-\frac{(-1)^jH_{j+1}^{(2)}}{j+1}\\ &=4\sum^\infty_{j=1}\frac{(-1)^jH_{j}^{(2)}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}+2\sum^\infty_{j=1}\frac{(-1)^jH_j^{(2)}}{j^2}+2+2\\ &=-8{\rm Li}_4\left(\frac{1}{2}\right)+\frac{17\pi^4}{240}-\frac{1}{3}\ln^4{2}-7\zeta(3)\ln{2}-4\zeta(3)+\frac{\pi^2}{3}\ln^2{2}+\frac{\pi^2}{3}\ln{2}-\frac{\pi^2}{6}-4\ln{2}+8\\ \end{align}$$

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Step 3d: Evaluating $\displaystyle -2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)}$
$$\begin{align} -2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)} &=-2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j}+2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j+1}\\ &=4\zeta(3)\ln{2}-2\zeta(3)\\ \end{align}$$

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Step 3e: Evaluating $\displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j(j+1)}$
$$\begin{align} 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j(j+1)} &=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j}-2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j+1}\\ &=4\sum^\infty_{j=1}\frac{(-1)^jH_{j}^{(3)}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^3}+2\\ &=-\frac{19\pi^4}{360}+3\zeta(3)\ln{2}-\frac{3}{2}\zeta(3)-\frac{\pi^2}{6}-4\ln{2}+8 \end{align}$$

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Step 4: Obtaining the final result

Summing the results from steps 3a, 3b, 3c, 3d and 3e gives
$$\int^1_0\ln(1+x)\ln(1-x)\ln^2{x} \ {\rm d}x=24-\frac{4\pi^2}3-\frac{11\pi^4}{720}-12\ln2\\+2\ln^22-\frac16\ln^42+\pi ^2\ln2+\frac{\pi^2}6\ln^22-4\operatorname{Li}_4\!\left(\tfrac12\right)-\frac{35}4\zeta(3)+\frac72\zeta(3)\ln2.$$
hence completing the proof.