antiderivative of $\sum _{ n=0 }^{ \infty }{ (n+1){ x }^{ 2n+2 } } $ I've proven that the radius of convergence of  $\sum _{ n=0 }^{ \infty  }{ (n+1){ x }^{ 2n+2 } } $ is $R=1$, and that it doesn't converge at the edges.
Now, I was told that this is the derivative of a function $f(x)$, which holds $f(0)=0$. 
My next step is to find this function in simple terms, and here I get lost. My attempt:
$f(x)=\sum _{ n=0 }^{ \infty  }{ \frac { n+1 }{ 2n+3 } { x }^{ 2n+3 } }  $
and this doesn't seem to help.
I'd like to use the fact that $|x|<1$ so I'll get a nice sum based on the sum of a geometric series but I have those irritating coefficients.
Any tips?

Hint: $$\frac{n+1}{2n+3} = \frac{1}{2}\left(1-\frac{1}{2n+3}\right)$$
