Alternative ways to evaluate $\int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$ In the following link here I found the integral & the evaluation of 
$$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$
I'll also include a simpler version together  with the question: is it possible to find some easy
ways of computing both integrals without using complicated sums that require multiple zeta
formulae and "never-ending long" generating functions?
$$i). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^2}{x}\,dx$$
$$ii). \displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$

Here is a slightly different approach to evaluating
$$\int^1_0 \frac{\text{Li}^2_2 (x)}{x} \, dx,$$
which, like the answer given by @Zaid Alyafeai, makes use of the result
$$\sum^\infty_{n = 1} \frac{H_n}{n^4} = 3 \zeta (5) - \zeta (2) \zeta (3).$$
We start by writing
$$I = \int^1_0 \frac{\text{Li}^2_2 (x)}{x} \, dx = \int^1_0 \frac{\text{Li}_2 (x)}{x} \cdot \text{Li}_2 (x) \, dx.$$
Integrating by parts we have
$$I = \text{Li}_2 (1) \text{Li}_3 (1) + \int^1_0 \frac{\text{Li}_3 (x) \ln (1 - x)}{x} \, dx = \zeta (2) \zeta (3)  + \int^1_0 \frac{\text{Li}_3 (x) \ln (1 - x)}{x} \, dx,$$
where we have made use of the well-known results of $\text{Li}_s (0) = 0$ and $\text{Li}_s (1) = \zeta (s)$ respectively.
Integrating by parts again
$$I = \zeta (2) \zeta (3) - \int^1_0 \frac{\zeta (4) - \text{Li}_4 (x)}{1 - x} \, dx.$$
Recalling 
$$\text{Li}_s (x) = \sum^\infty_{n = 1} \frac{x^n}{n^s} \quad \text{and} \quad \zeta (4) = \sum^\infty_{n = 1} \frac{1}{n^4},$$
we can write
\begin{align*}
I &= \zeta (2) \zeta (3) - \int^1_0 \frac{1}{1 - x} \left [\sum^\infty_{n = 1} \frac{1}{n^4} - \sum^\infty_{n = 1} \frac{x^n}{n^4} \right ] \, dx\\
&= \zeta (2) \zeta (3) - \sum^{\infty}_{n = 1} \frac{1}{n^4} \int^1_0 \frac{1 - x^n}{1 - x} \, dx.
\end{align*}
From the integral representation for the harmonic number $H_n$, namely
$$H_n = \int^1_0 \frac{1 - x^n}{1 - x} \, dx,$$
we can rewrite our integral as
$$I = \zeta (2) \zeta (3) - \sum^\infty_{n = 1} \frac{H_n}{n^4} = \zeta (2) \zeta (3) - \left \{3 \zeta (5) - \zeta (2) \zeta (3) \right \},$$
or
$$\int^1_0 \frac{\text{Li}_2^2 (x)}{x} \, dx = 2 \zeta (2) \zeta (3) - 3 \zeta (5).$$
