Here is an elementary proof. We rewrite the series as
$$\frac{1}{4}\int_0^1\frac{1-x^4}{1-x^6}\,dx=\frac{1}{8}\int_0^1\frac{dx}{1-x+x^2}+\frac{1}{8}\int_0^1\frac{dx}{1+x+x^2}.$$
It is straightforward to show that
\begin{align*}
\int_0^1\frac{dx}{1-x+x^2}&=\frac{2\pi}{3\sqrt{3}},\\
\int_0^1\frac{dx}{1+x+x^2}&=\frac{\pi}{3\sqrt{3}},
\end{align*}
so we are done.