Today I met on Facebook the following integral $$\int_{0}^{1}\frac{\text{Li}_3(x)\,\text{Li}_4(x)}{x^2}\,dx$$ which I proved to be equal to $$\frac{10 \pi ^2}{3}-\frac{17 \pi ^4}{180}-\frac{\pi ^6}{540}-10\, \zeta(3)-\frac{\pi^2}{3}\,\zeta(3)-\frac{\pi^4}{90}\,\zeta(3)-\zeta(3)^2-2\,\zeta(5).$$ The trick I used is pretty nice: by integration by parts, the computation of $\int_{0}^{1}\frac{\text{Li}_p(x)\,\text{Li}_q(x)}{x^2}\,dx$ boils down to the computation of $\int_{0}^{1}\frac{\text{Li}_{p-1}(x)\,\text{Li}_q(x)}{x^2}\,dx$ and $\int_{0}^{1}\frac{\text{Li}_p(x)\,\text{Li}_{q-1}(x)}{x^2}\,dx$. It follows that through many steps of integration by parts the evaluation of the generalized $(p,q)$-integral boils down to the evaluation of
$$ \int_{0}^{1}\frac{-\log(1-x)\,\text{Li}_r(x)}{x^2}\,dx =\zeta(2)+\sum_{n\geq 2}\frac{H_{n-1}}{(n-1)n^r},$$ where by partial fraction decomposition the RHS only depends on standard Euler sums $\sum_{n\geq 1}\frac{H_n}{n^s}$, always expressible in terms of values of the $\zeta$ function. See, for instance, Flajolet and Salvy, Theorem 2.2.
I would use this question for collecting alternative/shorter/slicker solutions, both for the starting integral or for the generalized one $\int_{0}^{1}\frac{\text{Li}_p(x)\,\text{Li}_q(x)}{x^2}\,dx$.