Originally Posted by
ThePerfectHacker Topsquark is correct.
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It is a famous case that,
$\displaystyle \int \frac{e^x}{x}dx$ is non-elementary.
We shall convert this integral into this form.
For simplicity sake we shall use $\displaystyle a=1$, thus,
$\displaystyle \int e^{-1/x}dx$
Manipulate it as,
$\displaystyle \int \frac{x^2e^{1/x}}{x^2}dx$
Let, (going to apply the substitution rule)
$\displaystyle u=-1/x$ then, $\displaystyle u'=1/x^2$
Thus,
$\displaystyle \int \frac{e^u}{u^2}du$
If you use integration by parts with,
$\displaystyle v=e^u$ and $\displaystyle w'=1/u^2$
Then,
$\displaystyle -\frac{e^u}{u}+\int \frac{e^u}{u}du$
We have converted this integral into an integral which has no elementary anti-derivative. This confirms topsquarks guess.