Why can't you find all antiderivatives by integrating a power series? if  
$f(x) = \sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}x^n$  
why can't you do the following to find a general solution  
$F(x) \equiv \int f(x)dx$  
$F(x) = \int (\sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}x^n) dx = \sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}(\int x^n dx) = \sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}(\frac{x^{n+1}}{n+1}) = 
\sum\limits^{\infty}_{n=0}\frac{f^{(n)}(0)}{(n+1)!}x^{n+1}$  
I was wondering about this because I tried this approach to finding the antiderivative $\int e^{x^2} dx$
 A: If the integrand $f(x)$ can be represented as a power series (let's say with infinite radius of convergence) like $f(x)=e^{x^2}$ can, then you can use that power series representation, just like you say, to obtain a power series representation of the integral $\int f(x)dx$. But that does not mean you will be able to figure out an elementary expression for that power series. Sometimes, the integral simply does not have such an expression. 
However, certainly for purposes of numerical integration but also for purposes of studying the solution, this method is very useful. Power series are great! 
A: First of all, $\int f(x) dx$ is a collections of functions, not a function.
You should rather define $F(x) = \int_{t = 0}^x f(t) dt$.
Then, you should be careful about interchanging $\int$ and $\sum_{n=0}^\infty$.
This is true for finite sums, but not always for infinite series of integral (look e.g. Fatou-Lebesgue theorem) — it is the same as interchanging a limit for integration.
Finally, you have to be able to calculate the infinite series you end up with.
