Basel problem type sum, proposed by prof. Skordev

Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$

Proof. The function
$$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$
is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that
$$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$
If we make change of variables in the latter integral $t\to 1-t$ we obtain
$$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$
On the other hand, using integration by parts, we obtain
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$
Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$
Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain
$$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$
On the other hand  $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t+\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t=\int_{0}^1\frac{\log(1-t)}{t}\text{d} t=-\frac{\pi^2}{6}$$
Substracting the last two equalities we obtain  $$\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t=-\frac{\pi^2}{12}-\frac{1}{2}\log\left(\frac{1}{2}\right)^2$$
and thus $\displaystyle S\left(\frac{1}{2}\right)=\frac{\pi^2}{12}+\frac{1}{2}\log\left(\frac{1}{2}\right)^2$.