integration - Elementary way to calculate the series $sumlimits_{n=1}^{infty}frac{H_n}{n2^n}$

I want to calculate the series of the Basel problem $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{n^2}}$ by applying the Euler series transformation. With some effort I got that

$$\displaystyle{\frac{\zeta (2)}{2}=\sum_{n=1}^{\infty}\frac{H_n}{n2^n}}.$$

I know that series like the $\displaystyle{\sum_{n=1}^{\infty}\frac{H_n}{n2^n}}$ are evaluated here, but the evaluations end up with some values of the $\zeta$ function, like $\zeta (2),\zeta(3).$

First approach: Using the generating function of the harmonic numbers and integrating term by term, I concluded that

$$\displaystyle{\sum_{n=1}^{\infty}\frac{H_n}{n2^n}=\int_{0}^{\frac{1}{2}}\frac{\ln (1-x)}{x(x-1)}dx},$$

but I can't evaluate this integral with any real-analytic way.

First question: Do you have any hints or ideas to evaluate it with real-analytic methods?

Second approach: I used the fact that $\displaystyle{\frac{H_n}{n}=\sum_{k=1}^{n}\frac{1}{k(n+k)}}$ and then, I changed the order of summation to obtain

$$\displaystyle{\sum_{n=1}^{\infty}\frac{H_n}{n2^n}=\sum_{k=1}^{\infty}\frac{2^k}{k}\left(\sum_{m=2k}^{\infty}\frac{1}{m2^m}\right)}.$$

To proceed I need to evaluate the

$$\int_{0}^{\frac{1}{2}}\frac{x^{2k-1}}{1-x}dx,$$

since $\displaystyle{\sum_{m=2k}^{\infty}\frac{1}{m2^m}=\int_{0}^{\frac{1}{2}}\frac{x^{2k-1}}{1-x}dx}.$

Second question: How can I calculate this integral?

Thanks in advance for your help.