Obtain a closed-form for the series: $$\mathcal{S}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4^nn^2}\binom{2n}{n}$$
From here https://en.wikipedia.org/wiki/List_of_m ... cal_series we know that for $\displaystyle{\left| z \right| \le \frac{1}{4}}$ holds $$\sum\limits_{n = 0}^\infty \binom{2n}{n} z^n = \frac{1}{\sqrt {1 - 4z} }$$
Then \begin{align} &\sum\limits_{n = 0}^\infty \binom{2n}{n} \frac{{{z^n}}}{{{4^n}}} = \frac{1}{\sqrt {1 - z} } \\ &\Rightarrow \sum_{n = 1}^\infty \binom{2n}{n} \frac{z^n}{4^n} = \frac{1}{\sqrt {1 - z} } - 1 \\ &\Rightarrow \sum_{n = 1}^\infty \binom{2n}{n} \frac{z^{n - 1}}{4^n} = \frac{1 - \sqrt {1 - z} }{z \sqrt {1 - z} } \\ &\Rightarrow \sum\limits_{n = 1}^\infty \binom{2n}{n} \frac{y^n}{n 4^n} = \int_0^y {\frac{{1 - \sqrt {1 - z} }} {{z \sqrt {1 - z} }}dz} = 2\left( \log 2 - \log \left(1 + \sqrt {1 - y} \right) \right) \\ &\Rightarrow \sum_{n = 1}^\infty \binom{2n}{n} \frac{y^{n - 1}}{n 4^n} = 2\left( {\frac{{\log 2 - \log \left( {1 + \sqrt {1 - y} } \right)}}{y}} \right) \\ &\Rightarrow \sum_{n = 1}^\infty \binom{2n}{n} \frac{x^n}{n^2 4^n} = 2 \int_0^x {\frac{{\log 2 - \log \left( {1 + \sqrt {1 - y} } \right)}}{y}dy} \end{align}