Explaining why we can't "find" an antiderivative of $f(t) = e^{t^2}$. We can't find 
$$
\int e^{t^2} \; dt
$$
using basic tools from a calculus class. That is, we can't express an antiderivative of $f(t) = e^{t^2}$ using the basic operations. We can of course just define 
$$
F(t) = \int_{a}^t e^{s^2}\; ds.
$$
I am looking for a way to explain to a student why we can't express the antiderivative using basic operations (addition, subtraction, root, powers, etc.). In particular I am interested in getting at answering the "why we can't do that". I know that one could "just" prove it, but is there a simple argument or something that illustrates this?
 A: First, it is not that we cannot find an antiderivative: we can, in fact: the function $$F(t)=\int_0^t\exp(s^2)\,\mathrm d s$$ is a perfectly good antiderivative. What is true is that you cannot find an expression of an antiderivative as an elementary function (for some specific meaning of «elementary function»)
Now, to prove this one needs to make precise what is meant by elementary and then prove it. The proof is not something you'd be able to present to a student learning integration, sadly. I guess there are several ways to do it, but the standard argument involves the theory of differential field extensions, as explained in the little book by Andy Majid on the subject.
The argument is not unsimilar to that which shows that there is no general formula for the roots of a polynomial —this gives you something to compare it with.
A: You can't prove it easily, but you can give some intuition :


*

*derivative of $e^x$ is $e^x$

*Trying to find $f$ such that $f'(x)=e^{x^2}$, we can guess that $f(x)=g(x)e^{h(x)}$

*So $f'(x)=(g'(x)+g(x)h'(x))e^{h(x)}=e^{x^2}$

*So we want to find a solution such that $h(x)=x^2$ and $g'(x)+2xg(x)=1$


Note the last point seems easier, but if they try, they will not find elementary function $g$ such that $g(x)=\frac{1-g'(x)}{2x}$. You can at least show this is not a polynomial function, nor a quotient of polynomials function, nor a usual trigonometric function.
But let $g(x)=\sum a_ix^i$, then $a_1=1$ and $(i+1).a_{i+1}+2a_{i-1}=0$, so you can at least give some expression to $g$ by solving $a_i$ (you can suppose that $a_0=0$).
