I know that $\int{\frac{1}{x}}dx$ is simply $\ln{(x)}+c$ (-which is clearly unrelated to the problem but I just thought I would share anyway) but I am not sure how to approach $e^{x{^2}}$. Perhaps a substitution?
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$\begingroup$Actually, neither the antiderivative of $e^{x^2}$ nor $e^{-x^2}$ can be expressed in terms of 'elementary functions', so we simply define a new function called the error function by
$$\textrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{-\infty}^x e^{-t^2} dt.$$
We can also define a related function, the imaginary error function, by
$$\textrm{erfi}(z)=\frac{\textrm{erf}(iz)}{i}$$
(where $z\in\mathbb{C}$).
Then of course the map $z\mapsto\frac{\sqrt{\pi}}{2}\textrm{erfi}(z)$ is an antiderivative of $z\mapsto e^{z^2}$.
As is alluded to in the comments, the situation is more tractable for (improper) definite integrals of this form, e.g.
$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}.$$
$\endgroup$ 5 $\begingroup$For the sake of accuracy and regarding some of the comments made:
1) $\int e^xdx=e^x+C$ and not $\ln(x)$. 2) The indefinite integral $\int e^{x^2}dx$ exists on any finite interval simply because the integrand is continuous. However, a primitive function can't be expresses as a combination of elementary functions (it is not a trivial proof that that is the case). 3) Using the Taylor expansion of $e^{x^2}$ one can integrate term by term to obtain a power series expansion for a primitive function and to obtain approximations of it. 4) The function $e^{x^2}$ is not integrable on $(-\infty ,\infty)$.
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