definiteinteggral The integral is given by 
$$\int_0^1 \frac{\ln (1-x)\ln x}{1+x} dx = \frac{1}{8}\big(-\pi^2\ln(4) +13\zeta(3)\big).$$
Any ideas how to prove?  We cannot solve the integral so easily because we cannot use a power series approach or harmonic series generating function since the denominator and numerator of the integrand are different.  You will see convergence problems if we try using the power series for $\ln(1-x)$ and $(1+x)^{-1}$ to integrate term by term, thus this method does not work here as in other cases.
Substitution of the form $x=e^t$ yields
$$
\int_{-\infty}^0 \frac{t\cdot e^t\ln(1-e^t)}{1+e^t}dt.
$$
This integral does not seem to be trivial either however.  The Riemann zeta function is
$$
\zeta(3)=\sum_{n=1}^\infty n^{-3}.
$$
 A: Using the geometric series for $\frac{1}{1+x}$, we can write
\begin{align*}
I := \int_{0}^{1} \frac{\log(1-x)\log x}{1+x} \, dx
&= \sum_{n=1}^{\infty} (-1)^{n-1} \int_{0}^{1} x^{n-1} \log (1-x) \log x \, dx \\
&= \sum_{n=1}^{\infty} (-1)^{n-1} \cdot \frac{1}{n} \left( \frac{H_{n}}{n} - \zeta(2) + H_{n}^{(2)} \right),
\end{align*}
where $H_{n}^{(s)} = \sum_{k=1}^{n} k^{-s}$ is the generalized harmonic number. Introducing the alternating Euler sum as
$$ A_{p}^{(s)} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}H_{n}^{(s)}}{n^{p}}, $$
we can write
$$ I = A_{2}^{(1)} + A_{1}^{(2)} - \zeta(2)\log 2. $$
Now it is relatively well-known that 
$$ A_{1}^{(2)} = \zeta(3) - \frac{1}{2}\zeta(2)\log 2
\quad \text{and} \quad
A_{2}^{(1)} = \frac{5}{8}\zeta(3). $$
(There are several ways of proving these identities. I proved them using dilogarithm and trilogarithm, but I also saw solutions using complex analysis. See here for my old approach.)
Therefore we have
$$ I = \frac{13}{8}\zeta(3) - \frac{3}{2}\zeta(2)\log 2. $$

Addendum. Using partial fraction expansion, you can check that
$$ \frac{1}{k(n+k)^{2}} = \frac{1}{n^{2}} \left( \frac{1}{k} - \frac{1}{k+n} \right) - \frac{1}{n(k+n)^{2}}. $$
So it follows that
\begin{align*}
\int_{0}^{1} x^{n-1} \log (1-x) \log x \, dx
&= \sum_{k=1}^{\infty} \frac{1}{k(n+k)^{2}} \\
&= \sum_{k=1}^{\infty} \frac{1}{n^{2}} \left( \frac{1}{k} - \frac{1}{k+n} \right) - \frac{1}{n(k+n)^{2}} \\
&= \frac{H_{n}}{n^{2}} - \frac{1}{n} \sum_{k=1}^{\infty} \frac{1}{(n+k)^{2}} \\
&= \frac{H_{n}}{n^{2}} - \frac{1}{n} \zeta(2) + \frac{H_{n}^{(2)}}{n}.
\end{align*}
