Somewhat more generally, Maple 18 says that

$$ \sum_{k=1}^n \dfrac{(-1)^k k}{(n+k)(2k+v)} {n \choose k} {{n+k} \choose k} =
 -{\frac {\Gamma  \left( v/2+2 \right) \Gamma  \left( n-v/2 \right) }{
 \left( v+2 \right) \Gamma  \left( 1-v/2 \right) \Gamma  \left( 1+v/2+
n \right) }}
$$
which for $v=1$ gives you the desired identity.

It appears that Maple is using Zeilberger's algorithm.