Polylogs!

We have $\begin{array}{rcl} \displaystyle \int_0^1 \frac{\mathrm{Li}_2(x^3)}{x^2}\,dx & = & \displaystyle \int_0^1 \sum_{n=1}^\infty \frac{x^{3n-2}}{n^2}\,dx \\ & = & \displaystyle \sum_{n=1}^\infty \frac{1}{n^2(3n-1)} \; = \; \sum_{n=1}^\infty \left( \frac{9}{3n-1} - \frac{3}{n} - \frac{1}{n^2}\right) \\ & = & \displaystyle 3\sum_{n=1}^\infty \left(\frac{1}{n-\frac13} - \frac{1}{n}\right) - \zeta(2) \\ & = & \displaystyle 3\lim_{n\to\infty} \left(\psi(1) - \psi(n) - \psi(\tfrac23) + \psi(n-\tfrac13)\right) - \zeta(2) \\ & = & \displaystyle 3\big(\psi(1) -\psi(\tfrac23)\big) - \zeta(2) \; = \; 3\left(\tfrac32\ln3 - \tfrac{1}{2\sqrt{3}}\pi\right) - \zeta(2) \\ & = & 9\ln\sqrt{3} - \tfrac12\pi\sqrt{3} - \tfrac16\pi^2 \end{array}$ making the answer $\boxed{19}$ .

Using the definition of Polylog function in terms of series we arrive at:-

$\displaystyle \sum_{r=1}^{\infty}\int_{0}^{1} \frac{x^{3r-2}}{r^{2}}dx$

$\displaystyle =\sum_{r=1}^{\infty}\frac{1}{r^{2}(3r-1)}=9\sum_{r=1}^{\infty}\frac{1}{(3r)^{2}(3r-1)} = 9\sum_{r=1}^{\infty}\left(\frac{1}{3r}\left(\frac{1}{3r-1}-\frac{1}{3r}\right)\right)$

$\displaystyle =\sum_{r=1}^{\infty} \left(\frac{9}{(3r)(3r-1)} - \frac{9}{9r^{2}}\right) =\sum_{r=1}^{\infty}\frac{9}{(3r)(3r-1)} - \sum_{r=1}^{\infty}\frac{1}{r^{2}}$ [Both series are convergent by Limit Comparison test with $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ ]

$\displaystyle \sum_{r=0}^{\infty}\left(\frac{9}{(3r+3)(3r+2)}\right) - \frac{\pi^{2}}{6}$

Let us look at the summation $\displaystyle \sum_{r=0}^{\infty}\frac{1}{(3r+3)(3r+2)} = \sum_{r=0}^{\infty}\left(\frac{1}{3r+2}-\frac{1}{3r+3}\right)$

We know $\displaystyle \frac{1}{1-x} = \sum_{r=0}^{\infty} x^{r}$ for $\displaystyle -1<x<1$

So $\displaystyle \frac{1}{1-x^{3}} = \sum_{r=0}^{\infty} x^{3r}$

So $\displaystyle \frac{x}{1-x^{3}} = \sum_{r=0}^{\infty} x^{3r+1}$

So $\displaystyle \int_{0}^{1}\frac{x}{1-x^{3}}dx = \sum_{r=0}^{\infty} \int_{0}^{1}x^{3r+1}dx = \sum_{r=0}^{\infty}\frac{1}{3r+2}$

Similarly $\displaystyle \int_{0}^{1}\frac{x^{2}}{1-x^{3}}dx = \sum_{r=0}^{\infty}\frac{1}{3r+3}$

So we have $\displaystyle \sum_{r=0}^{\infty}\left(\frac{1}{3r+2}-\frac{1}{3r+3}\right) = \int_{0}^{1}\frac{x-x^{2}}{1-x^{3}}dx =\int_{0}^{1}\frac{x}{1+x+x^{2}}dx$

$\displaystyle = \frac{1}{2}\int_{0}^{1}\left(\frac{2x+1}{1+x+x^{2}} - \frac{1}{1+x+x^{2}}\right)\,dx = \frac{\ln(3)}{2} -\frac{\pi}{6\sqrt{3}}$

So our answer is $\displaystyle 9\ln(\sqrt{3}) - \frac{\pi \cdot \sqrt{3}}{2} - \frac{\pi^{2}}{6}$

Note :- The evaluation of the two integrals and the justification of why $\sum_{r=1}^{\infty}\frac{1}{r^{2}} = \frac{\pi^{2}}{6}$ , I leave both of them to the readers of the solution .