A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$ 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.
 A: 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.
A: 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.
A: 
It is easy to prove that $$\int\frac{\log(1-x)}{x+1}dx=\log(2)\log(x+1)-\operatorname{Li}_2\left(\frac{x+1}{2}\right)$$

The proof is straightforward  if we set $1+x=y$ and rest it follows by definition of dilogarithm and logarithmic integrals.
Next we do integration by parts giving us $$\log(1+x)\left(\log(2)\log(1+x)-\operatorname{Li}_2\left(\frac{x+1}{2}\right)\right)\Bigg|_0^1-\int_0^1\frac{dx}{1+x}\left(\log(2)\log(1+x)-\operatorname{Li}_2\left(\frac{x+1}{2}\right)\right)=\log^3(2)-\frac{\pi^2}{6}\log(2)-\frac{\log^3(2)}{3}+\int_0^1\frac{\operatorname{Li}_2\left(\frac{x+1}{2}\right)}{x+1}dx=\frac{\log^3(2)}{2}-\frac{\pi^2}{6}\log(2)+\zeta(3)-\hbox{Li}_3\left(\frac{1}{2}\right)$$ Now using the  known value of $\hbox{Li}_3\left(\frac{1}{2}\right)=\frac{\log^3(2)}{6}-\frac{\pi^2}{12}\log(2)+\frac{7}{8}\zeta(3)$ and simplifying gives us  the desires closed form
for it.
$$\frac{\log^3(2)}{3} -\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$.
A: 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! (>‿◠)✌
A: 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)).
A: 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.
A: \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}
A: 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}
A: 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
$\int_0^1 \frac{\ln^2(1+t)}{t}dx={\frac{\zeta(3)}{4}}$
$$\int_0^1 \frac{\ln t\ln(1+t)}{1+t}dt\overset{IBP}=
-\frac12\int_0^1 \frac{\ln^2(1+t)}tdt = -\frac18\zeta(3)
$$
