Evaluation of a dilogarithmic integral 
Problem. Prove that the following dilogarithmic integral has the indicated value:
  $$\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}\stackrel{?}{=}-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$


My attempt:
I began by using the polylogarithmic expansion in terms of generalized harmonic numbers,
$$\frac{\operatorname{Li}_r{(x)}}{1-x}=\sum_{n=1}^{\infty}H_{n,r}\,x^n;~~r=2.$$
Then I switched the order of summation and integration and used the substitution $u=-\ln{x}$ to evaluate the integral:
$$\begin{align}
\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}
&=\int_{0}^{1}\mathrm{d}x\ln^2{(x)}\sum_{n=1}^{\infty}H_{n,2}x^n\\
&=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{1}\mathrm{d}x\,x^n\ln^2{(x)}\\
&=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{\infty}\mathrm{d}u\,u^2e^{-(n+1)u}\\
&=\sum_{n=1}^{\infty}H_{n,2}\frac{2}{(n+1)^3}\\
&=2\sum_{n=1}^{\infty}\frac{H_{n,2}}{(n+1)^3}.
\end{align}$$
So I've reduced the integral to an Euler sum, but unfortunately I've never quite got the knack for evaluating Euler sums. How to proceed from here?

This integral belongs to a wider class of integrals that can always be reduced to single zeta values and to bi-variate zeta values which in turn -- provided the weight of the zeta function in question is not too big-- can also be reduced to single zeta values. My standard way of doing this integrals is as follows. We need to calculate:
\begin{equation}
{\mathcal I}_{0,2}^{(2)} := \int\limits_0^1 \frac{[\log(1/x)]^2}{2!} \cdot \frac{Li_0(x) Li_2(x)}{x} dx
\end{equation}
which is just one half of your integral. Now we use the following identity:
\begin{equation}
\frac{[\log(1/x)]^2}{2!} = \int\limits_{x < \xi_1  < \xi_2 < 1} \prod\limits_{j=1}^2 \frac{d \xi_j}{\xi_j}
\end{equation}
Inserting this into the above and changing order of integration gives:
\begin{eqnarray}
{\mathcal I}_{0,2}^{(2)} = \int\limits_{0 < \xi_1 < \xi_2 < 1} \frac{1}{\xi_1} \frac{1}{\xi_2} \underbrace{\int\limits_0^{\xi_1} \frac{Li_0(x) Li_2(x)}{x} d x}_{\left[Li_1(\xi_1) Li_2(\xi_2) - \int\limits_0^{\xi_1} \frac{[Li_1(x)]^2}{x} dx\right]} \cdot d\xi_1 d\xi_2
\end{eqnarray}
where we integrated by $x$ using integration by parts. Since two minus zero is even we are left with an irreducible integral that we leave unevaluated for the time being. Now we have:
\begin{eqnarray}
{\mathcal I}_{0,2}^{(2)} &=& \int\limits_0^1 \frac{[\log(1/\xi_1)]^1}{1!} \left( \frac{1}{2} [Li_2(\xi_1)]^2 \right)^{'} d\xi_1 - \int\limits_0^1 \frac{[\log(1/x)]^2}{2!} \cdot \frac{[Li_1(x)]^2}{x} dx \\
&=& \frac{1}{2} \left(
\underbrace{\int\limits_0^1 \frac{[Li_2(\xi)]^2}{\xi} d\xi}_{J_1} - 
\underbrace{\int\limits_0^1 [\log(1/\xi)]^2 \cdot \frac{[Li_1(\xi)]^2}{\xi} d\xi}_{J_2}
\right)
\end{eqnarray}
Now from Compute an integral containing a product of powers of logarithms. we have that :
\begin{eqnarray}
J_2&=& -\frac{1}{3} \Psi^{(4)}(1) + 2 \Psi^{(2)}(1) \Psi^{(1)}(1)\\
&=& 8 \zeta(5) - 4 \zeta(3) \zeta(2)
\end{eqnarray}
where $\Psi^{(j)}(1)$ is the polygamma function at unity and $\Psi^{(j)}(1)=(-1)^{j+1} j! \zeta(j+1)$. On the other hand we have:
\begin{eqnarray}
J_1 &=& \sum\limits_{m\ge1,n\ge 1} \frac{1}{m^2} \frac{1}{n^2} \frac{1}{(m+n)}\\
& =& \sum\limits_{m\ge 1} \left(\frac{\zeta(2)}{m^3} - \frac{H_m}{m^4}\right) \\
&=& \zeta(2) \zeta(3) - {\bf H}^{(1)}_4(+1)\\
&=& 2 \zeta(2) \zeta(3) - 3 \zeta(5)
\end{eqnarray}
where in the last line above we used my answer to Calculating alternating Euler sums of odd powers . Therefore:
\begin{equation}
{\mathcal I}^{(2)}_{0,2}= \frac{1}{2} \left(J_1-J_2\right)= \frac{1}{2} \left(-11 \zeta(5)+6 \zeta(3) \zeta(2) \right)
\end{equation}
as expected. Note that exactly the same steps can be performed is we replace the power of the logarithm and the orders of the poly-logarithms by any nonnegative integers. The generic result is then given in An integral involving product of poly-logarithms and a power of a logarithm. .
