calculus - Prove $zeta(3)=2sum_{n=1}^inftyfrac{H_n}{n}left[frac1{4^n}{2nchoose n}left(H_{2n}-H_n-frac1{2n}-ln2right)+frac1{2n}right]$

How to prove

$$\zeta(3)=2\sum_{n=1}^\infty\frac{H_n}{n}\left[\frac1{4^n}{2n\choose n}\left(H_{2n}-H_n-\frac1{2n}-\ln2\right)+\frac1{2n}\right]$$

where $H_n$ is the harmonic number and $\zeta$ is the Riemann zeta function.

This problem is proposed by Cornel which can be found here and no solution has been submitted yet.

I know the following identity $$H_{2n}-H_n-\ln2=-\int_0^1\frac{x^{2n}}{1+x}dx$$
is related but I do not know how to exploit it.

I prefer a solution without calculating each sum separately because if we seperate, all these sums are calculated here but the first one $\sum_{n=1}^\infty\frac{H_nH_{2n}}{n4^n}{2n\choose n}$.