How to prove, in a simple way that
$$I=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{x(1-x^2)}dx=\frac{17}{16}\zeta(4)\ ?$$
Where $\operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$ is the dilogarithmic function and $\zeta(a)$ is the Riemann zeta function.
Here is my approach and would like to see different ones.
\begin{align} I&=\int_0^1\frac{\ln x}{1-x^2}.\frac{\operatorname{Li}_2(-x)}{x}\ dx\\ &=\int_0^1\frac{\ln x}{1-x^2}\left(\int_0^1\frac{\ln u}{1+ux}\ du\right)\ dx\\ &=\int_0^1\ln u\left(\int_0^1\frac{\ln x}{(1-x^2)(1+ux)}\ dx\right)\ du\\ &=\int_0^1\ln u\left(-\frac{\zeta(2)}{4(1-u)}-\frac{\zeta(2)}{2(1+u)}-\frac{u\operatorname{Li}_2(-u)}{1-u^2}\right)\ du\\ &=-\frac{\zeta(2)}{4}\int_0^1\frac{\ln u}{1-u}\ du-\frac{\zeta(2)}{2}\int_0^1\frac{\ln u}{1+u}\ du-\int_0^1\frac{\color{red}{u}\ln u\operatorname{Li}_2(-u)}{\color{red}{1-u^2}}\ du\\ &=-\frac{\zeta(2)}{4}\left(-\zeta(2)\right)-\frac{\zeta(2)}{2}\left(-\frac12\zeta(2)\right)-\underbrace{\int_0^1\frac{\ln u\operatorname{Li}_2(-u)}{\color{red}{u(1-u^2)}}\ du}_{\large I}+\underbrace{\int_0^1\frac{\ln u \operatorname{Li}_2(-u)}{\color{red}{u}}\ du}_{\large\frac{7}{8}\zeta(4)}\\ 2I&=\frac{17}{8}\zeta(4)\quad\Longrightarrow I=\frac{17}{16}\zeta(4) \end{align}
This problem is proposed by a friend on a facebook group.
Different approach:
\begin{align} I&=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{x(1-x^2)}\ dx\\ &=\frac12\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{x(1-x)}\ dx+\frac12\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{x(1+x)}\ dx\\ &=\frac12\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1\frac{x^{n-1}\ln x}{1-x}\ dx+\frac12\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1 x^{n-1}\ln x\ dx\\ &=\frac12\sum_{n=1}^\infty\frac{(-1)^n}{n^2}(H_{n-1}^{(2)}-\zeta(2))-\frac12\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{n^2}, \quad H_{n-1}^{(2)}=H_{n}^{(2)}-\frac1{n^2}\\ &=-\frac12\sum_{n=1}^\infty\frac{(-1)^n}{n^4}-\frac12\zeta(2)\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\\ &=-\frac12\left(-\frac78\zeta(4)\right)-\frac12\zeta(2)\left(-\frac12\zeta(2)\right)\\ &=\frac{17}{16}\zeta(4) \end{align}