Is this integral impossible to solve?

Question

Is possible to express the antiderivative $$\int\frac{-3e^{-x^3}}{x^2}dx$$ in terms of elementary functions?

Answer 1

$\int\dfrac{-3e^{-x^3}}{x^2}dx$

$=\int\dfrac{-3}{x^2}\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{3n}}{n!}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{3(-1)^{n+1}x^{3n-2}}{n!}dx$

$=\sum\limits_{n=0}^\infty\dfrac{3(-1)^{n+1}x^{3n-1}}{n!(3n-1)}+C$

Answer 2

Unfortunately, nope. See this article on the exponential integral.

It's not quite in the same form as your integral, but I'm pretty sure there's a way to get there. At the very least you can get do some tricky substitution to get (assuming my algebra is correct - always a risk!):

$$\frac{9e^{-x^{3}}}{x^3}-6\int \frac{1}{\ln(x)}dx$$

The second term is known as the logarithmic integral, which funily enough is closely related to the exponential integral.