A primitive function of $ e^{x^{2}} $ I made some efforts to set a closed form of primitive function of $ e^{x^{2}} $ i find 
this  function  : $ f(x)=\frac{x}{2x^{2}-1}e^{x^{2}} $ where :
$f'(x)=(\frac{x}{2x^{2}-1}e^{x^{2}})'$= $\frac{4x^{4}-4x^{2}-1}{4x^{4}-4x^{2}+1}e^{x^{2}}$
my question is : can I take $ f(x)=\frac{x}{2x^{2}-1}e^{x^{2}} $ as a primitive function 
of $ e^{x^{2}} $ if $x$ hold a big values  ?
is there a function better then $ f(x)=\frac{x}{2x^{2}-1}e^{x^{2}} $ ?
thank you for any replies or any comments
 A: As already mentioned, there is "closed form" of $\int e^{x^2}$, at least in terms of a collection of "basic functions". Now, when you ask if it is a good approximation for an antiderivative for large $x$, the answer is that it depends! Note:
$$
\frac{4x^4 - 4x^2 - 1}{4x^4 - 4x^2 + 1} = 1 - \frac{2}{4x^4 - 4x^2 + 1} = 1 + O(x^{-4}).
$$
Now, as a ratio:
$$
\frac{\big(1+O(x^{-4})\big)\, e^{x^2}}{e^{x^2}} = 1+O(x^{-4}) \to 1 ~ \text{ as $x \to \infty$}. 
$$
However, as a difference
$$
\big(1+O(x^{-4})\big)\,e^{x^2} - e^{x^2} = O(x^{-4})\, e^{x^2} \to - \infty ~\text{ as $x \to \infty$}
$$
where we note that term $O(x^{-4})$ is negative. So, as a ratio, reasonably good; as  a difference, it seems reasonably bad.
A: $f$ is not a primitive of $e^{x^2}$ because $$f'(x) = \frac{e^{x^2}(4x^4-4x^2-1)}{(2x^2-1)^2}$$
which is not $e^{x^2}$.
In fact, $e^{x^2}$ does not have an antiderivative in terms of elementary functions. For some notes on this topic, see this talk by Martin Leslie.
A: 
thank you for any replies or any comments

It can be proven, using either Liouville's theorem or the Risch algorithm, that the anti–
derivative of $e^{x^n}$, with $n\in\mathbb N$, cannot be expressed in terms of elementary functions for 
$n\ge2$. Letting $n\in\mathbb Q$ , the case $n=\dfrac1m$ , with $m\in\mathbb N^*$, can also be added to the previous 
list of exceptions. However, it can be expressed in terms of special functions. Thus, for 
$n=2$, it can be expressed in terms of the error function, $\displaystyle\int e^{x^2}dx=\frac{\sqrt\pi}2\cdot\text{erfi}(x)$, and 
for other values of n in terms of the $($incomplete$)$ $\Gamma$ function, $\displaystyle\int e^{x^n}dx=\frac{(-1)^{^{-\tfrac1n}}}n~\cdot$ 
$\cdot~\Gamma\bigg(\frac1n,-x^n\bigg)$. Also, as far as definite integrals are concerned, we have the beautiful 
identity $\displaystyle\int_0^\infty e^{-x^n}dx=\Big(\tfrac1n\Big)!=\Gamma\bigg(1+\frac1n\bigg)$.
