Calculation of $$\sum_{n=1}^{\infty}\frac{\psi_1(n)}{2^nn^2}$$
My attempt
\begin{align*} \sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n^2} &= -\sum_{n=1}^\infty\psi_1(n)\left(\frac{\log(2)}{2^n n}+\int_0^{\frac{1}{2}} x^{n-1}\log(x)dx\right) \\ &= \color{green}{-\log(2)\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n}}\color{blue}{-\int_0^{1\over 2}\frac{\log(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)dx }\tag{1} \end{align*}
I don’t know how to evaluate $$\log(2)\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n}$$ and $$-\int_0^{1\over 2}\frac{\log(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)dx$$
