This approach is pretty identical to Cornel's solution posted on his FB page.
using the fact that $\quad\displaystyle \sum_{n=1}^\infty a_{2n}=\frac12\left(\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty (-1)^na_n\right),\ $ we have
\begin{align}
\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}&=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}=\frac12\left(\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)^2}+\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{(n+1)^2}\right)\\
&=\frac12\left(\sum_{n=1}^\infty\frac{x^n}{n^2}-\sum_{n=1}^\infty(-1)^n\frac{x^n}{n^2}\right)=\frac12\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)\right)
\end{align}
then, the first integral:
\begin{align}
I_1&=4\int_0^1\left(\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}\right)\frac{\operatorname{Li}_2(x)}{x}\ dx\\
&=2\sum_{n=1}^\infty\left(\frac1{n^2}-\frac{(-1)^n}{n^2}\right)\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\
&=2\sum_{n=1}^\infty\left(\frac1{n^2}-\frac{(-1)^n}{n^2}\right)\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\
&=\zeta(2)\zeta(3)-2\zeta(2)\operatorname{Li}_3(-1)-2\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}\\
&\boxed{=\frac72\zeta(2)\zeta(3)-2\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}}
\end{align}
and the second integral:
using the following identity proved by Cornel and can be found in his book, (Almost) Impossible Integrals, Sums and Series. $\quad\displaystyle\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n}$.
multiply both sides by $\displaystyle\frac{\ln^2x}{x}$ then integrate from $0$ to $1$, we get
\begin{align}
I_2&=\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln^2x\ dx\\
&=\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(\frac{2}{(2n)^3}\right)\\
&=-4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}+\frac14\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)\\
&=-2\sum_{n=1}^\infty\frac{H_n}{n^4}-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}+\frac14\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)\\
&\boxed{=-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}-\frac74\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)}
\end{align}
Finally
\begin{align}
I&=I_1+I_2\\
&=\frac72\zeta(2)\zeta(3)-\frac18\zeta(5)-\frac{15}4\sum_{n=1}^\infty\frac{H_n}{n^4}\\
&=\frac72\zeta(2)\zeta(3)-\frac18\zeta(5)-\frac{15}4\left(3\zeta(5)-\zeta(2)\zeta(3)\right)\\
&\boxed{=\frac{29}{4}\zeta(2)\zeta(3)-\frac{91}{8}\zeta(5)}
\end{align}