In preparation for an exam, I must find the primitive of $e^{-x^2}$. But when I look up the equation in an integral list I get a primitive that seems way overcomplicated for the level of math they ask us. Here is the link to the image of the resulting function.
How to get the primitive of $e^{-x^2}$?
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If you are talking about an explicit primitive, like to say that, for example, a primitive of $x$ might be $\frac{1}{2}x^2 + C$, then there is naught.
Risch's algorithm will help you in proving this.
For what concerns instead the possibility of writing a "primitive" in other terms (that is, not through elementary function but with something else, like series or products) then here you are:
$$\int e^{-x^2}\ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} \int x^{2k} \ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} \frac{x^{2k+1}}{2k+1}$$
The latter sum is known, and it's called the Error Function, a special function (not an elementary function). Whence:
$$\sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} \frac{x^{2k+1}}{2k+1} = \frac{1}{2} \sqrt{\pi } \text{erf}(x)$$
Hence if you like:
$$\int e^{-x^2}\ \text{d}x = \frac{1}{2} \sqrt{\pi } \text{erf}(x)$$
As I said: this is not a primitive in the usual sense.
You can find documentation on the Error Function online.
P.s. For the integration, I used nothing but Taylor Series for the exponential.
You can derive the result written in your book in the same way.
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Are you allowed to use that list? If yes, all you have to do is plug in $a=1$ and you get the primitive $\frac{\sqrt{\pi}}2\operatorname{erf}(x)$.
You can be of the opinion that this is useless because $\operatorname{erf}(x)$ is usually defined to be $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}dt. $$ On the other hand, $\operatorname{erf}$ is a perfectly nice function that is easily accessible via tables or calculators.
Usually they should only ask you to integrate $\int_0^{\infty}e^{-x^2}$, which is $\frac{\sqrt \pi}{2}$, as can be calculated by multivariable calculus. The function $\mathrm{erf}$ arises exactly because of this integral.