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While playing around with Mathematica, I found that

$$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$

Please help me prove this result.

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  • $\begingroup$ Let $u=\log(1+x)$, $du=1/(1+x)\,dx$ and $1-x=2-e^u$ transforms our integral into $$\int_0^{\log2}u\log(2-e^u)\,\mathrm{d}u$$... not sure what to do from there, however. $\endgroup$ – obataku Jun 1 '13 at 7:36
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    $\begingroup$ you can see:math.stackexchange.com/questions/405356/… $\endgroup$ – math110 Jun 1 '13 at 7:47
  • $\begingroup$ @math110: I think I just understood what you are trying to say. We can relate it to the Euler Sum in your link. $$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(H_n)^2$$ $\endgroup$ – Shobhit Bhatnagar Jun 1 '13 at 8:16
  • $\begingroup$ For a more generalized form of this integral, where the integral is over the range 0 to z rather than 0 to 1, see here -->> [mathhelpboards.com/questions-other-sites-52/… $\endgroup$ – user98087 Oct 1 '13 at 19:32

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Use your favorite program to compute the indefinite integral in terms of polylogarithms $$\int\frac{\ln(1+x)\ln(1-x)\,dx}{1+x}=\frac{\ln2}{2}\ln^2(1+x)-\ln(1+x)\,\mathrm{Li}_2\left(\frac{1+x}{2}\right)+\mathrm{Li}_3\left(\frac{1+x}{2}\right).$$ [This can be verified by straightforward differentiation].

To compute the definite integral, it suffices to know $\mathrm{Li}_{2,3}\left(\frac12\right)$ and $\mathrm{Li}_{2,3}(1)$. However, the definition of polylogarithm immediately implies $\mathrm{Li}_s(1)=\zeta(s)$. Also, the values $\mathrm{Li}_{2,3}\left(\frac12\right)$ can be found here (formulas (16), (17)).

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    $\begingroup$ I can verify that the anti-derivative can be computed through some tedious integration by parts. $\endgroup$ – Potato Jun 1 '13 at 8:46
  • $\begingroup$ @Potato It is maybe not easy to guess the form of the antiderivative. But once you have a correct guess, to check it is a one-line calculation using that $\mathrm{Li}_2'(z)=-\ln(1-z)/z$ and $\mathrm{Li}_3'(z)=\mathrm{Li}_2(z)/z$. $\endgroup$ – Start wearing purple Jun 1 '13 at 8:50
  • $\begingroup$ @Potato: I think I understood what you mean. One integration by parts gives roughly $\mathrm{Li}_2(smth)$ instead of $\frac{\ln(1-z)}{1+z}$ and $1/(1+z)$ instead of $\ln(1+z)$. Subsequent integration gives $\mathrm{Li}_3$. $\endgroup$ – Start wearing purple Jun 1 '13 at 8:56
  • $\begingroup$ Thank O.L.! I did not realize that the indefinite integral would be so easy. $\endgroup$ – Shobhit Bhatnagar Jun 1 '13 at 9:40
  • $\begingroup$ It's clear that the OP likes an analytically derivation. Otherwise, the OP can use a symbolic software to get the solution. $\endgroup$ – Felix Marin Aug 13 '14 at 19:45
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}{\ln\pars{1 + x}\ln\pars{1 - x} \over 1+x}\,\dd x:\ {\large}}$

With $\ds{0 < \mu < 1}$: \begin{align}&\color{#c00000}{\int_{0}^{\mu}% {\ln\pars{1 + x}\ln\pars{1 - x} \over 1+x}\,\dd x} \\[3mm]&=\half\,\ln^{2}\pars{1 + \mu}\ln\pars{1 - \mu} +\color{#00f}{\half\int_{0}^{\mu}{\ln^{2}\pars{1 + x} \over 1 - x}\,\dd x} \tag{1} \end{align}

\begin{align}&\color{#00f}{\half\int_{0}^{\mu}% {\ln^{2}\pars{1 + x} \over 1 - x}\,\dd x} =\half\int_{1}^{1 + \mu}{\ln^{2}\pars{x} \over 2 - x}\,\dd x =\half\int_{1/2}^{\pars{1 + \mu}/2}{\ln^{2}\pars{2x} \over 1 - x}\,\dd x \\[3mm]&=-\,\half\ln\pars{1 - {1 + \mu \over 2}}\ln^{2}\pars{1 + \mu} +\half\int_{1/2}^{\pars{1 + \mu}/2}\ln\pars{1 - x}\,{2\ln\pars{2x} \over x}\,\dd x \\[3mm]&=-\,\half\ln\pars{1 - \mu \over 2}\ln^{2}\pars{1 + \mu} -\int_{1/2}^{\pars{1 + \mu}/2}{\rm Li}_{2}'\pars{x}\ln\pars{2x}\,\dd x \\[3mm]&=-\,\half\,\ln^{2}\pars{1 + \mu}\ln\pars{1 - \mu} +\half\,\ln\pars{2}\ln^{2}\pars{1 + \mu} -{\rm Li}_{2}\pars{1 + \mu \over 2}\ln\pars{1 + \mu} \\[3mm]&+\int_{1/2}^{\pars{1 + \mu}/2}{\rm Li}_{2}\pars{x}\,{1 \over x}\,\dd x \\[3mm]&=-\,\half\,\ln^{2}\pars{1 + \mu}\ln\pars{1 - \mu} +\half\,\ln\pars{2}\ln^{2}\pars{1 + \mu} - {\rm Li}_{2}\pars{1 + \mu \over 2}\ln\pars{1 + \mu} \\[3mm]&+\int_{1/2}^{\pars{1 + \mu}/2}{\rm Li}_{3}'\pars{x}\,\dd x \end{align}

\begin{align} &\color{#00f}{\half\int_{0}^{\mu}% {\ln^{2}\pars{1 + x} \over 1 - x}\,\dd x} \\[3mm]&=-\,\half\,\ln^{2}\pars{1 + \mu}\ln\pars{1 - \mu} +\half\,\ln\pars{2}\ln^{2}\pars{1 + \mu} -{\rm Li}_{2}\pars{1 + \mu \over 2}\ln\pars{1 + \mu} \\[3mm]&+{\rm Li}_{3}\pars{1 + \mu \over 2} - {\rm Li}_{3}\pars{\half} \end{align}

Replacing in $\pars{1}$ and taking the limit $\ds{\mu \to 1^{-}}$: \begin{align}&\color{#c00000}{\int_{0}^{1}% {\ln\pars{1 + x}\ln\pars{1 - x} \over 1+x}\,\dd x} = \half\,\ln^{3}\pars{2} - {\rm Li}_{2}\pars{1}\ln\pars{2} +{\rm Li}_{3}\pars{1} - {\rm Li}_{3}\pars{\half} \end{align}

With the values: $$ {\rm Li}_{2}\pars{1} = {\pi^{2} \over 6}\,,\quad {\rm Li}_{3}\pars{1} = \zeta\pars{3}\,,\quad {\rm Li}_{3}\pars{\half} = {\ln^{3}\pars{2} \over 6} -{\ln\pars{2} \over 12}\,\pi^{2} + {7 \over 8}\,\zeta\pars{3} $$ we find

$$ \color{#66f}{\large\int_{0}^{1}% {\ln\pars{1 + x}\ln\pars{1 - x} \over 1+x}\,\dd x = {\ln^{3}\pars{2} \over 3} - {\ln\pars{2} \over 12}\,\pi^{2} +{1 \over 8}\,\zeta\pars{3}} \approx {\tt -0.3088} $$

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  • $\begingroup$ This is a nice answer! $\endgroup$ – Olivier Oloa Aug 13 '14 at 21:42
  • $\begingroup$ @OlivierOloa Thanks. It's nice to rely on PolyLogarithms... $\endgroup$ – Felix Marin Aug 16 '14 at 21:18
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Ms. Chris's sis asked me exactly same question a few days ago in chatroom & I could answer it.

Here is my answer. Let $I$ be the integral. Using magic substitution $2t=1+x$ we get \begin{align} I&=\int_{\frac{1}{2}}^1 \frac{\log(2t)\log(2-2t)}{t}dt\\ &=\int_{\frac{1}{2}}^1 \frac{\log t\log(1-t)}{t}dt+\ln2\int_{\frac{1}{2}}^1 \frac{\log(1-t)}{t}dt+\ln2\int_{\frac{1}{2}}^1 \frac{\log t}{t}dt+\ln^22\int_{\frac{1}{2}}^1\frac{1}{t}dt\\ &=I_1+I_2+I_3+I_4 \end{align} All of the above integrals are trivial. For instance, $I_1$ & $I_2$ can be solved by using elementary way: using series expansion for $$\frac{\log(1-t)}{t}=\sum_{k=1}^\infty\frac{t^{k-1}}{k}$$ We can also use the reflection formula for dilog function $$\text{Li}_2(x)+\text{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x)$$for solving $I_1$ and the integral representation of dilog function $$\text{Li}_2(x)=\int_0^x \frac{\log(1-t)}{t}dt$$for solving $I_2$. And for $I_3$ & $I_4$, of course an high school student can easily solved it. So $$I=\frac{1}{3}\ln^3(2)-\frac{\pi^2}{12}\ln(2)+\frac{\zeta(3)}{8}$$ Done! (>‿◠)✌

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Following Shobhit's comment, some preliminary lemma.
Lemma 1. $$ \sum_{n\geq 1} H_n x^n = \frac{\log(1-x)}{1-x}. $$ Lemma 2. By Lemma $1$, $$ \sum_{n\geq 1}\frac{H_n}{n+1} x^{n+1} = \frac{1}{2}\log^2(1-x),\qquad \sum_{n\geq 1}\frac{H_n+H_{n+1}}{n+1}x^{n}=\frac{-x+\log^2(1-x)+\text{Li}_2(x)}{x}.$$ Lemma 3. Since $H_{n+1}^2-H_n^2 = \frac{H_n+H_{n+1}}{n+1}$, $$ \sum_{n\geq 1}H_{n}^2 x^n = \frac{\log^2(1-x)+\text{Li}_2(x)}{1-x}.$$ Lemma 4. By Lemma 3, $$ \sum_{n\geq 1}\frac{(-1)^{n+1} H_{n}^2}{n+1} = -\int_{0}^{1}\frac{\log^2(1+x)+\text{Li}_2(-x)}{1+x}\,dx=-\frac{\log^3(2)}{3}-\color{red}{\int_{0}^{1}\frac{\text{Li}_{2}(-x)}{1+x}\,dx}.$$ The problem boils down to the evaluation of the last integral. By integration by parts, it is: $$ \color{red}{\int_{0}^{1}\frac{\text{Li}_{2}(-x)}{1+x}\,dx}=-\frac{\pi^2}{12}\log(2)+\color{blue}{\int_{0}^{1}\frac{\log^2(1+x)}{x}\,dx}\tag{1} $$ but: $$ \color{blue}{\int_{0}^{1}\frac{\log^2(1+x)}{x}\,dx} = -2\int_{0}^{1}\frac{\log(1+x)\log(x)}{1+x}\,dx=\color{blue}{\frac{\zeta(3)}{4}}\tag{2}$$ and the proof is complete. $(2)$ ultimately depends on a reflection formula for $\text{Li}_3$. In our case:

Lemma 5. $$\sum_{n\geq 1}\frac{H_n^2}{n+1}\,x^{n+1} = -\frac{\log(1-x)}{3}\left[\frac{\pi^2}{2}+\log^2(1-x)+3\text{Li}_2(1-x)\right]+2\left[\text{Li}_3(1-x)-\zeta(3)\right]$$

is straightforward to prove through differentiation, leading back to Lemma 3.

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We can take $\ 2ab= a^2+b^2-(a-b)^2$ to get: $$I= \frac12 \int_0^1 \frac{\ln^2 (1-x)}{1+x}dx+\frac12\int_0^1\frac{\ln^2(1+x)}{1+x}dx-\frac12 \int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx$$ By letting $\frac{1-x}{1+x}=t$ and expanding into power series the last one we get: $$I=\frac12 J +\frac{\ln^3(1+x)}{6}\bigg|_0^1 -\frac12 \int_0^1 \frac{\ln^2 t}{1+t}dt=\frac12 J +\frac{\ln^3 2}{6}-\frac34\zeta(3) $$ $$J=\int_0^1 \frac{\ln^2(1-x)}{1+x}dx\overset{1-x\to x}=\frac12\int_0^1 \frac{\ln^2 x}{1-\frac{x}{2}}dx=\frac12 \sum_{n=0}^\infty \frac{1}{2^n} \int_0^1 x^{n}\ln^2 xdx$$ $$=\sum_{n=0}^\infty \frac{1}{2^n}\frac{1}{(n+1)^3}=2\operatorname{Li}_3 \left(\frac12\right)\Rightarrow \boxed{I=\operatorname{Li}_3 \left(\frac12\right)+\frac{\ln^3 2}{6}-\frac34\zeta(3)}$$ Of course one can rewrite the trilogarithm's value as seen from $(17)$ in this link, but this form is also valid.

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My approach, here $t=x/2$

$$\begin{align} & \int_0^1 \frac{\ln(1+x)\ln(1-x)}{1+x} \mathrm{d}x = \int_0^1 \frac{\ln(2-x)\ln x}{2-x} \mathrm{d}x\\ = & \int_0^1 \frac{(\ln(1-\tfrac{x}{2})+\ln2)(\ln\tfrac{x}{2}+\ln2)}{2-x} \mathrm{d}x\\ = &\> \ln^22\int_0^1 \frac{\mathrm{d}x}{2-x} + \ln2\int_0^{1/2} \frac{\ln t}{1-t}\mathrm{d}t + \ln2\int_0^{1/2} \frac{\ln(1-t)}{1-t}\mathrm{d}t + \int_0^{1/2} \frac{\ln(1-t)\ln t}{1-t}\mathrm{d}t\\ = &\> \ln^22(-\ln(2-x))|_{x=0}^{1}+\ln2(\operatorname{Li}_{2}(\tfrac1{2})-\operatorname{Li}_{2}(1))-\frac{\ln2\ln^2(1-t)}{2}\bigg|_{0}^{1/2}\\ & -\frac{\ln^2(1-t)\ln t}{2}\bigg|_{0}^{1/2} + \frac1{2}\int_0^{1/2}\frac{\ln^2(1-t)}{t}\mathrm{d}t \end{align}$$

Last integral is not trivial, but you can finish it with series, which is

$$\int_0^{1/2}\frac{\ln^2(1-t)}{t}\mathrm{d}t = -\frac{\ln^32}{3}+\frac{\zeta(3)}{4}$$

and $\operatorname{Li}_{2}(1) = \frac{\pi^2}{6}$, $\operatorname{Li}_{2}(\tfrac1{2}) = \frac{\pi^2}{12}-\frac{\ln^22}{2}$

thus

$$\int_0^1 \frac{\ln(1+x)\ln(1-x)}{1+x} \mathrm{d}x = \frac{\ln^32}{3} - \frac{\pi^2}{12}\ln2 + \frac{\zeta(3)}{8}$$

Some supplementary for Last integral with only elementary resources

$$\int_0^{1/2}\frac{\ln^2(1-t)}{t}\mathrm{d}t = \int_{1/2}^{1}\frac{\ln^2t}{1-t}\mathrm{d}t = \int_{0}^{1}\frac{\ln^2t}{1-t}\mathrm{d}t - \int_{0}^{1/2}\frac{\ln^2t}{1-t}\mathrm{d}t$$

Let $t=e^{-u}$, using $\frac1{1-t} = \sum_{n=0}^{\infty}t^n$

$$\begin{align} \int_{0}^{1/2}\frac{\ln^2t}{1-t}\mathrm{d}t & = \sum_{n=0}^{\infty}\int_{\ln2}^{\infty}u^2e^{-(n+1)u}\>\mathrm{d}u\\ & = 2\sum_{n=0}^{\infty}\frac{e^{-(n+1)\ln2}}{(n+1)^3} + 2\ln2\sum_{n=0}^{\infty}\frac{e^{-(n+1)\ln2}}{(n+1)^2} + \ln^22\sum_{n=0}^{\infty}\frac{e^{-(n+1)\ln2}}{n+1}\\ & = 2\operatorname{Li}_{3}(\tfrac1{2}) + 2\ln2\operatorname{Li}_{2}(\tfrac1{2}) + \ln^22\operatorname{Li}_{1}(\tfrac1{2}) \end{align}$$

you can calculate $\int_{0}^{1/2}\frac{\ln^2t}{1-t}\mathrm{d}t$ with $\operatorname{Li}_{3}(\tfrac1{2}) = \frac{\ln^32}{6}-\frac{\pi^2}{12}\ln2+\frac7{8}\zeta(3)$, $\operatorname{Li}_{1}(\tfrac1{2}) = \ln2$. And by the same fashion, easily have $\int_{0}^{1}\frac{\ln^2t}{1-t}\mathrm{d}t = 2\zeta(3)$ to finish this solution.

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\begin{align} I&=\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ dx=\int_0^1\frac{\ln x\ln(2-x)}{2-x}\ dx\\ &=\frac12\int_0^1\frac{\ln x(\ln2+\ln(1-x/2)}{1-x/2}\ dx\\ &=\frac12\ln2\int_0^1\frac{\ln x}{1-x/2}\ dx+\frac12\int_0^1\frac{\ln x\ln(1-x/2)}{1-x/2}\ dx\\ &=\ln2\sum_{n=1}^\infty\frac{1}{2^n}\int_0^1x^{n-1}\ln x\ dx-\frac12\sum_{n=1}^\infty \frac{H_n}{2^n}\int_0^1x^n\ln x\ dx\\ &=-\ln2\sum_{n=1}^\infty\frac{1}{2^nn^2}+\frac12\sum_{n=1}^\infty\frac{H_n}{2^n(n+1)^2}\\ &=-\ln2\operatorname{Li}_2\left(\frac12\right)+\sum_{n=1}^\infty\frac{H_n-1/n}{2^nn^2}\\ &=-\ln2\operatorname{Li}_2\left(\frac12\right)+\sum_{n=1}^\infty\frac{H_n}{2^nn^2}-\operatorname{Li}_3\left(\frac12\right)\\ &=-\ln2\left(\frac12\zeta(2)-\frac12\ln^22\right)+\left(\zeta(3)-\frac12\ln2\zeta(2)\right)-\left(\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32\right)\\ &=\frac18\zeta(3)-\frac12\ln2\zeta(2)+\frac13\ln^32 \end{align}

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different approach. \begin{align} I&=\int_0^1\frac{\ln(1+x)\ln(1-x)}{1+x}\ dx=-\sum_{n=1}^\infty(-1)^nH_n\int_0^1x^n\ln(1-x)\ dx\\ &=\sum_{n=1}^\infty(-1)^n\frac{H_nH_{n+1}}{n+1}=-\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}H_n}{n}\\ &=-\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2} \end{align} I was able here to prove $\displaystyle\sum\frac{(-1)^n H_n^2}{n}=\frac12\ln2\zeta(2)-\frac34\zeta(3)-\frac13\ln^32$

then \begin{align} I&=-\left(\frac12\ln2\zeta(2)-\frac34\zeta(3)-\frac13\ln^32\right)+\left(-\frac58\zeta(3)\right)\\ &=\frac18\zeta(3)-\frac12\ln2\zeta(2)+\frac13\ln^32 \end{align}

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\begin{align}J&=\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx\end{align}Since,\begin{align}ab=\frac{1}{2}\left(a^2+b^2-(a-b)^2\right)\end{align}then,\begin{align}J&=\frac{1}{2}\int_0^1 \frac{\ln^2(1+x)}{1+x}\,dx+\frac{1}{2}\int_0^1 \frac{\ln^2(1-x)}{1+x}\,dx-\frac{1}{2}\int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}\,dx\\ &=\frac{1}{6}\ln^3 2+\frac{1}{2}\int_0^1 \frac{\ln^2(1-x)}{1+x}\,dx-\frac{1}{2}\int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}\,dx \end{align}In the first integral perform the change of variable $y=\dfrac{1-x}{2}$, in the second integral perform the change of variable $y=\dfrac{1-x}{1+x}$,\begin{align}J&=\frac{1}{6}\ln^3 2+\frac{1}{2}\int_0^{\frac{1}{2}}\frac{\ln^2 (2x)}{1-x}\,dx-\frac{1}{2}\int_0^1\frac{\ln^2 x}{1+x}\,dx\\ &=\frac{1}{6}\ln^3 2+\ln 2\int_0^{\frac{1}{2}}\frac{\ln x}{1-x}\,dx+\frac{1}{2}\int_0^{\frac{1}{2}}\frac{\ln^2 x}{1-x}\,dx+\frac{\ln^2 2}{2}\int_0^{\frac{1}{2}}\frac{1}{1-x}\,dx-\frac{1}{2}\int_0^1\frac{\ln^2 x}{1+x}\,dx\\ &=\frac{2}{3}\ln^3 2+\ln 2\int_0^{\frac{1}{2}}\frac{\ln x}{1-x}\,dx+\frac{1}{2}\int_0^{\frac{1}{2}}\frac{\ln^2 x}{1-x}\,dx-\frac{1}{2}\int_0^1\frac{\ln^2 x}{1+x}\,dx \end{align}On the other hand,\begin{align}J&=\frac{1}{2}\Big[\left(\ln^2(1+ x)-\ln^2 2\right)\ln(1-x)\Big]_0^1+\frac{1}{2}\int_0^1 \frac{\ln^2(1+x)-\ln^2 2}{1-x}\,dx\end{align}Perform the change of variable $y=\dfrac{1+x}{2}$,\begin{align}J&=\frac{1}{2}\int_{\frac{1}{2}}^1 \frac{\ln^2(2x)-\ln^2 2}{1-x}\,dx\\ &=\frac{1}{2}\int_{\frac{1}{2}}^1 \frac{\ln^2 x}{1-x}\,dx+\ln 2\int_{\frac{1}{2}}^1 \frac{\ln x}{1-x}\,dx \end{align}Therefore,\begin{align}2J&=\frac{2}{3}\ln^3 2+\ln 2\int_0^1 \frac{\ln x}{1-x}\,dx+\frac{1}{2}\int_0^1\frac{\ln^2 x}{1-x}\,dx-\frac{1}{2}\int_0^1\frac{\ln^2 x}{1+x}\,dx\\ &=\frac{2}{3}\ln^3 2+\ln 2\int_0^1 \frac{\ln x}{1-x}\,dx+\frac{1}{2}\int_0^1\frac{2x\ln^2 x}{1-x^2}\,dx\end{align}In the last integral perform the change of variable $y=x^2$,\begin{align}2J&=\frac{2}{3}\ln^3 2+\ln 2\int_0^1 \frac{\ln x}{1-x}\,dx+\frac{1}{8}\int_0^1\frac{\ln^2 x}{1-x}\,dx\\ &=\frac{2}{3}\ln^3 2-\zeta(2)\ln 2+\frac{1}{8}\times 2\zeta(3)\\ J&=\frac{1}{3}\ln^3 2-\frac{1}{2}\zeta(2)\ln 2+\frac{1}{8}\zeta(3)\\ &=\boxed{\frac{1}{3}\ln^3 2-\frac{\pi^2}{12}\ln 2+\frac{1}{8}\zeta(3)} \end{align}NB: i assume:\begin{align}\int_0^1\frac{\ln^2 x}{1-x}\,dx&=2\zeta(3)\\\int_0^1\frac{\ln x }{1-x}\,dx&=-\zeta(2)\\ &=-\frac{\pi^2}{6} \end{align}

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Substitute $x= \frac{1-t}{1+t}$ to derive

\begin{align} & \int_0^1 \frac{\ln(1+x)\ln(1-x)}{1+x}dx =\int_0^1 \frac{\ln\frac{2t}{1+t}\ln\frac2{1+t}}{1+t}dt\\ =& \int_0^1 \frac{[\ln2-\ln(1+t)]^2}{1+t}dt +\ln2 \int_0^1 \frac{\ln t}{1+t}dt -\int_0^1 \frac{\ln t\ln(1+t)}{1+t}dt\\ =&\frac13\ln^32-\frac{\pi^2}{12}\ln2+\frac18\zeta(3) \end{align} where $\int_0^1 \frac{\ln t}{1+t}dt=-\frac{\pi^2}{12}$ and $\int_0^1 \frac{\ln^2(1+t)}{t}dx={\frac{\zeta(3)}{4}}$

$$\int_0^1 \frac{\ln t\ln(1+t)}{1+t}dt\overset{IBP}= -\frac12\int_0^1 \frac{\ln^2(1+t)}tdt = -\frac18\zeta(3) $$

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