real analysis - Integral$=-frac{4}{3}log^3 2-frac{pi^2}{3}log 2+frac{5}{2}zeta(3)$

Hi I have been trying to prove this
$$
I:=\int \limits_{0}^{1} \left[ \frac{1}{x(x-1)} \bigg(2\mathrm{Li}_2\bigg(\frac{1-\sqrt{1-x}}{2}\bigg)-\log\bigg(\frac{1+\sqrt{1-x}}{2}\bigg)^2 \bigg) -\frac{\zeta(2)-2\log^2 2}{x-1} \right]{dx}=\sum_{k=2}^\infty \binom{2k}{k} \frac{1}{k^2 4^k} \sum_{j=1}^{k-1} \frac{1}{j}=\color{#00f}{\large%
-{4 \over 3}\log^3 2-\frac{\pi^2}{3}\log 2+\frac{5}{2}\zeta(3) }
$$
What a beautiful result!!!! I am trying to prove this.
I am not sure of what to do, perhaps we could start with a change of variables

$$
\xi=\frac{1-\sqrt{1-x}}{2},
$$
but I get stuck shortly after. This is strongly related to Mahler measures and integration. Thanks for your help.

I tried the following substitution but failed,

UPDATE: I tried a change of variables given above by $\xi$, we obtain
$$
I=\int\limits_{0}^{1/2}\big(2\mathrm{Li}_2(\xi)-\log^2(1-\xi)\big)\left(\frac{4}{2\xi-1}-\frac{1}{\xi-1}-\frac{1}{\xi}\right)d\xi-4(\zeta(2)-2\log^2 2) \int\limits_0^{1/2}\frac{d\xi}{2\xi-1}

$$
but the integral on the right diverges so I need to use another method now.

Thanks