Answer
$\Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(x)}^{(4n+2)}}{2n+1}$
Work Step by Step
$f(x)=arctan(x^{2})=tan^{-1}(x^{2})$
Since, $\int \frac{1}{1+t^{2}}dt=tan^{-1}t$
$\frac{1}{1+t^{2}}=\frac{1}{1-(-t^{2})}$
$=\Sigma_{n=0}^{\infty}(-t^{2})^{n}$
$=\Sigma_{n=0}^{\infty}(-1)^{n}(t^{2n})$
$tan^{-1}t=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(t)}^{(2n+1)}}{2n+1}$
Thus, $arctan(x^{2})=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(x)}^{(4n+2)}}{2n+1}$
