Find Taylor series of f around x=0 where

$\displaystyle f(x)=\int_0^{x^2}\arctan(t)dt$.

Well first of I derived the series expansion of $\displaystyle arctan(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}$.

Using that:

$\displaystyle f(x)=\int_0^{x^2}\arctan(t)dt=\int_0^{x^2}\left(\s um_{n=0}^{\infty}\frac{(-1)^n}{2n+1}t^{2n+1}\right)dt=\sum_{n=0}^{\infty}\f rac{(-1)^n}{2n+1}\int_0^{x^2}t^{2n+1}dt$

$\displaystyle =\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\frac{(x^2)^{2n+2}}{2n+2}=\sum_{n=0}^{\ infty}\frac{(-1)^n}{(2n+1)(2n+2)}x^{4n+4}$

Is this correct? I could not make wolfram alpha show me the series expansion of f(x).