A definite integral involving logarithmic functions: $\int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1-x} \, \mathrm{d} x$ I'm trying to show $$
\int_0^1 \frac{\ln x \cdot \ln(1+x)}{1-x}dx=-\frac{1}{4}\pi^2 \ln(2)+\zeta(3).
$$
I am unsure how to approach this integral as I do not know how to use a power series representation for the integrand.  I cannot use the generating function
$$
\frac{\ln(1-x)}{1-x}=-\sum_{n=1}^\infty H_n x^n
$$
since I do not have this in my integrand so it is not so easy to approach.  Thanks for the help!
 A: EDIT:
I modified my answer so that it no longer involves dealing with polylogarithms.
There is also a very nice evaluation HERE.

The generating function of the alternating harmonic numbers (i.e., $\bar{H}_{n} = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k})$ is $$\frac{\ln(1+x)}{1-x} = \sum_{n=1}^{\infty} \bar{H}_{n} x^{n}. $$
This can be derived using the Cauchy product.
(This was mentioned in a partial answer that was deleted shortly after it was posted.)
Using this generating function, we find that
$$ \begin{align} \int_{0}^{1} \frac{\ln (x) \ln(1+x)}{1-x} \, dx &= \int_{0}^{1} \ln(x) \sum_{n=1}^{\infty} \bar{H}_{n} \, x^{n} \, dx \\ &= \sum_{n=1}^{\infty} \bar{H}_{n} \int_{0}^{1} \ln(x) x^{n} \, dx \\ &= -\sum_{n=1}^{\infty} \frac{\bar{H}_{n}}{(n+1)^{2}} \\&= -\sum _{n=1}^{\infty} \frac{\bar{H}_{n+1}}{(n+1)^{2}} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(n+1)^{3}} \\&= \left( -\sum_{n=1}^{\infty}\frac{\bar{H}_{n}}{n^{2}} +1 \right) + \left( \frac{ 3\zeta(3)}{4} -1 \right) \\ &= -\sum_{n=1}^{\infty}\frac{\bar{H}_{n}}{n^{2}} + \frac{3\zeta(3)}{4}.  \end{align}$$
In general, $$2 \sum_{n=1}^{\infty} \frac{\bar{H}_{n}}{n^{q}} = 2 \zeta(q) \ln(2)-q \zeta(q+1) + 2 \eta(q+1) + \sum_{k=1}^{q} \eta(k)\eta(q-k+1) ,  $$
where $\eta(z)$ is the Dirichlet eta function.
(See Theorem 7.1 in this paper for details about how to derive this formula using contour integration.)
Therefore, $$\sum_{n=1}^{\infty} \frac{\bar{H}_{n}}{n^{2}} = \zeta(2) \ln(2) - \zeta(3) + \eta(3) + \eta(1) \eta(2) = \frac{\pi^{2}}{4} \ln(2) - \frac{\zeta(3)}{4},  $$ and 
$$\int_{0}^{1} \frac{\ln (x) \ln(1+x)}{1-x} \, dx = \frac{\zeta(3)}{4} - \frac{\pi^{2}}{4} \log(2) + \frac{3\zeta(3)}{4} = - \frac{\pi^{2}}{4} \ln(2) + \zeta(3).$$
A: Use Feynman’s trick
\begin{align}
J(a) &= \int_0^1\frac{\ln x}{1+x}\left( \frac{ \ln(1-ax)}x+\ln(1-x) \right)dx\\
J’(a) &=\int_0^1\frac{-\ln xdx}{(1+x)(1-ax)}= \frac{-1}{1+a}\int_0^1 \left(\frac{\ln x}{1+x}+ \frac{a\ln x}{1-ax} \right)dx \\
&= \frac{\zeta(2)}{2(1+a)}- \frac1{1+a}\int_0^a\frac{\ln y}{1-y}dy - \frac{\ln a\ln(1-a)}{1+a}\\
\end{align}
Note
$$J(1) = \int_0^1 \frac{\ln x\ln(1-x)}{x}dx \overset{IBP} = 
\frac12\int_0^1 \frac{\ln^2x}{1-x}dx= \zeta(3) \tag1 $$
On the other hand
\begin{align}
J(1)& =J(0)+ \int_0^1 J’(a)da = \frac{\zeta(2)}2\ln2-\int_0^1 d[\ln(1+a)]\int_0^a \frac{\ln y}{1-y}dy \\
&\overset{IBP}= \frac{3}2\ln2\zeta(2)+ \int_0^1 \frac{\ln a \ln(1+a)}{1-a}da \tag2 \\\end{align}
Combine (1) and (2) to obtain
$$\int_0^1 \frac{\ln x \ln(1+x)}{1-x}dx= -\frac{3}2\ln2\zeta(2)+\zeta(3)= -\frac{\pi^2}{4}\ln2+\zeta(3) $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over
1 - x}\,\dd x =
-\,{1 \over 4}\,\pi^{2}\ln\pars{2} + \zeta\pars{3}} \approx -0.5082:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{%
\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x}
\\[5mm] = &\
{1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x +
{1 \over 2}\int_{0}^{1}{\ln^{2}\pars{1 + x} - \ln^{2}\pars{2}\over 1 - x}\,\dd x
\\[2mm] &
\!\!\!\!\! -\,{1 \over 2}\int_{0}^{1}
\bracks{\ln^{2}\pars{x \over 1 + x} - \ln^{2}\pars{2}}\,{\dd x \over 1 - x}
\end{align}
In the second integral I'll make the change $\ds{{1 + x \over 2} \mapsto x}$ while I'll set $\ds{{x \over x + 1} \mapsto x}$ in the last one:
\begin{align}
&\bbox[5px,#ffd]{%
\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x}
\\[5mm] = &\
{1 \over 2}\ \overbrace{\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x}
^{\ds{I_{1}}}\ +\
{1 \over 2}\ \overbrace{\int_{1/2}^{1}
{\ln^{2}\pars{2x} - \ln^{2}\pars{2}\over 1 - x}\,\dd x}^{\ds{I_{2}}}
\\[2mm] &
\!\!\!\!\! -\,{1 \over 2}\bracks{%
\underbrace{\int_{0}^{1}{\ln^{2}\pars{x/2} - \ln^{2}\pars{2} \over 1 - x}\,\dd x}_{\ds{I_{3}}}\ -\
\underbrace{\int_{0}^{1/2}{\ln^{2}\pars{x} - \ln^{2}\pars{2} \over 1 - x}\,\dd x}_{\ds{I_{4}}}}\label{1}\tag{1}
\end{align}
In the last line first integral I did the additional scaling
$\ds{x \mapsto {x \over 2}}$.

In (\ref{1}), all the integrals are reduced to simple ones by integrating twice by parts. Namely,
\begin{equation}
\left\{\begin{array}{rcl}
\ds{I_{1}} & \ds{=} &  \ds{\phantom{-\,\,}2\zeta\pars{3}}
\\[2mm]
\ds{I_{2}} & \ds{=} &  \ds{\phantom{-\,\,}{2\ln^{3}\pars{2} \over 3} -
{\pi^{2}\ln\pars{2} \over 6} + {\zeta\pars{3} \over 4}}
\\[2mm]
\ds{I_{3}} & \ds{=} &  \ds{\phantom{-\,\,}{\pi^{2}\ln\pars{2} \over 3} + 2\zeta\pars{3}}
\\[2mm]
\ds{I_{4}} & \ds{=} & \ds{-\,{2\ln^{3}\pars{2} \over 3} +
{7\zeta\pars{3} \over 4}}
\end{array}\right.
\end{equation}
Then,
\begin{align}
&\bbox[5px,#ffd]{%
\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x} =
{I_{1} + I_{2} - I_{3} + I_{4} \over 2}
\\[5mm] = &\
\bbx{-\,{1 \over 4}\,\pi^{2}\ln\pars{2} + \zeta\pars{3}} \approx -0.5082
\\ &
\end{align}
A: \begin{align}J&=\int_{0}^{1} \frac{\ln(x) \ln (1+x)}{1-x} \, \mathrm{d} x\\&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln t}{1-t}dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x \frac{\ln t}{1-t}dt\right)dx\\&=-\zeta(2)\ln 2-\int_0^1\int_0^1 \frac{x\ln(tx)}{(1+x)(1-tx)}dtdx\\
&=-\zeta(2)\ln 2-\int_0^1\int_0^1 \frac{\ln(tx)}{(1+t)(1-tx)}dtdx+\int_0^1\int_0^1 \frac{\ln(tx)}{(1+x)(1+t)}dtdx\\
&\overset{u(x)=tx}=-\zeta(2)\ln 2-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln u}{1-u}du\right)dt+2\ln 2\int_0^1 \frac{\ln x}{1+x}dx\\
&\overset{\text{IBP}}=-\zeta(2)\ln 2-\left[\ln\left(\frac{t}{1+t}\right)\left(\int_0^t \frac{\ln u}{1-u}du\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln t}{1-t}dt+\\&2\ln 2\int_0^1 \frac{\ln x}{1+x}dx\\
&=-2\zeta(2)\ln 2+\int_0^1 \frac{\ln^2 t}{1-t}dt-J+2\ln 2\int_0^1 \frac{\ln x}{1+x}dx\\
&=-2\zeta(2)\ln 2+2\zeta(3)-J+2\ln 2 \times -\frac{1}{2}\zeta(2)\\
&=-3\zeta(2)\ln 2+2\zeta(3)-J\\
&=-3\times \frac{\pi^2}{6}\ln^ 2+2\zeta(3)-J\\
J&=\boxed{\zeta(3)-\frac{1}{4}\pi^2\ln 2}
\end{align}
NB:
\begin{align}\int_0^1 \frac{\ln x}{1+x}dx&=\int_0^1 \frac{\ln x}{1-x}dx-\int_0^1 \frac{2t\ln t}{1-t^2}dt\\
&\overset{x=t^2}=\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx\\
&=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx\\
&=-\frac{1}{2}\zeta(2)\\
\end{align}
Moreover, i assume that:
\begin{align}\int_0^1 \frac{\ln x}{1-x}dx&=-\zeta(2)\\
\zeta(2)&=\frac{1}{6}\pi^2\\
\int_0^1 \frac{\ln x}{1-x}dx&=2\zeta(3)\\
\end{align}
