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!

\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}
