First we notice that \begin{split} a_n & = n \int_0^1 x^n (1+x)^{n-1}{\rm d}x \\ & = n \int_0^1 (1-x)^n (2-x)^{n-1}{\rm d}x \\ & = n\sum_{k=0}^{n-1} 2^k (-1)^{n-1-k} \int_0^1 (1-x)^n x^{n-1-k}{\rm d}x \\ &= \sum_{k=0}^{n-1} 2^k (-1)^{n-1-k} \binom{2n}{k} / \binom{2n}n. \\ \end{split} Now, the numerator in the last expression is odd, and thus $\nu_2(a_n)=-\nu_2(\tbinom{2n}n)=-s(n)$ by Kummer's theorem. QED

First we notice that \begin{split} a_n & = n \int_0^1 x^n (1+x)^{n-1}{\rm d}x \\ & = n \int_0^1 (1-x)^n (2-x)^{n-1}{\rm d}x \\ & = n\sum_{k=0}^{n-1} \binom{n-1}{k}2^k (-1)^{n-1-k} \int_0^1 (1-x)^n x^{n-1-k}{\rm d}x \\ &= \sum_{k=0}^{n-1} 2^k (-1)^{n-1-k} \binom{2n}{k} / \binom{2n}n. \\ \end{split} Now, the numerator in the last expression is odd, and thus $\nu_2(a_n)=-\nu_2(\tbinom{2n}n)=-s(n)$ by Kummer's theorem. QED