A twisted hypergeometric series $\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2$
Solution 1:
One have FL expansion $$\small K(x)=\sum _{n=0}^{\infty } \frac{2 P_n(2 x-1)}{2 n+1},\ \ \frac{\log (1-x)}{x}=\sum _{n=0}^{\infty } 2 (-1)^{n-1} (2 n+1) P_n(2 x-1) \left(\sum _{k=n+1}^{\infty } \frac{(-1)^{k-1}}{k^2}\right)$$ Thus by orthogonal relation of Legendre polynomial and series-integral conversion one have: $$\int_0^1 \frac{K(x) \log (1-x)}{x} \, dx=16 \int_0^1 \frac{\log (x) \tanh ^{-1}(x)}{x^2+1} \, dx$$ RHS has a polylogarithmic primitive (up to trilog) hence trivial.