How do I determine the values of $r$ so $e^{t^r}$ has an antiderivative? It's fairly easy to show for individual values of $r$, just plug in the value and try it, but is there any way to find, in general, which values of $r$ work so that $\int{e^{t^r}dt}$ can be expressed in a way without integrals (using elementary functions)?
 A: If you're willing to use the incomplete Gamma function, it can be done for all $r$.
Note btw that the substitution $\exp(t^r) = u$ gives you 
$$ \dfrac{1}{r} \int (\ln u)^{1/r - 1}\ du$$
and you can do this in elementary functions if $1/r$ is a positive integer. 
A: As Robert Israel showed, if $1/r$ is a positive integer the antiderivative can be coomputed on the basis of elementary functions.  
Suppose we write $1/r=n+1$, I give you below the expressions obtained for some values of $n$ (from $n=1$ to $n=7$).
$$2 e^{\sqrt{t}} \left(\sqrt{t}-1\right)$$
$$e^{\sqrt[3]{t}} \left(3 t^{2/3}-6 \sqrt[3]{t}+6\right)$$
$$e^{\sqrt[4]{t}} \left(4 t^{3/4}-12 \sqrt{t}+24 \sqrt[4]{t}-24\right)$$
$$e^{\sqrt[5]{t}} \left(5 t^{4/5}-20 t^{3/5}+60 t^{2/5}-120 \sqrt[5]{t}+120\right)$$
$$e^{\sqrt[6]{t}} \left(6 t^{5/6}-30 t^{2/3}+120 \sqrt{t}-360 \sqrt[3]{t}+720
   \sqrt[6]{t}-720\right)$$
$$e^{\sqrt[7]{t}} \left(7 t^{6/7}-42 t^{5/7}+210 t^{4/7}-840 t^{3/7}+2520 t^{2/7}-5040
   \sqrt[7]{t}+5040\right)$$  
$$e^{\sqrt[8]{t}} \left(8 t^{7/8}-56 t^{3/4}+336 t^{5/8}+6720 t^{3/8}-1680
   \sqrt{t}-20160 \sqrt[4]{t}+40320 \sqrt[8]{t}-40320\right)$$  
For the most general case, the antiderivative is given by
$$-\frac{t E_{\frac{r-1}{r}}\left(-t^r\right)}{r}$$ where $E_a(x)$ stands for the exponential integral function.
A: Look up "integration in finite terms Liouville".
This will give you a differential equation
from $\int f(t)\, dt$
which has to have a certain type of solution
for the integral to be integrable
in finite terms.
You can then find out which values of $r$
enable the integration to be done.
Also,
I think Chebychev worked on
(and, maybe, solved)
this particular problem.
