We know from integral calculus that by using the substitution method: $$ \int 2xe^{x^2} = e^{x^2} +C $$
Show that this antiderivative result is true if you use Maclaurin series representations in place of the functions. $$ \int 2xe^{x^2} = \sum_{n=0}^\infty {2x^{2n+2}\over n!(2n+2)} $$ I'm able to find the Maclaurin series representation but I'm not sure how to modify them to show they are equal.
On
We compute \begin{align} \int 2xe^{x^2}\ \mathsf dx &= \int \sum_{n=0}^\infty \frac{2x^{2n+1}}{n!}\ \mathsf dx\\ &= \sum_{n=0}^\infty \frac2{n!}\int x^{2n+1}\ \mathsf dx\\ &= \sum_{n=0} \frac{2x^{2n+2}}{n!(2n+2)}, \end{align} where the interchange of integral and summation is justified by the uniform convergence of the power series. Dividing by the common factor of $2$ yields $$ \sum_{n=0}^\infty \frac{x^{2(n+1)}}{(n+1)!} = \sum_{n=1}^\infty \frac{x^{2n}}{(2n)!} = e^{x^2} - 1, $$ which is equal to $e^{x^2}$ up to the constant of integration.
$\sum_{n=0}^\infty \frac{2x^{2n+2}}{(2n+2)n!}=\sum_{n=0}^\infty \frac{(x^2)^{n+1}}{(n+1)!}=e^{x^2}-1$. Add a constant of integration to get desired result.