$\int \frac{1}{x^3}e^{-x^2}dx$
What I did?
put $1/x^2 = t; $
then $\int \frac{1}{x^3}e^{-x^2}dx$ will trasform into $\frac{-1}{2}\int e^{-1/t}dt$
I don't understand how to proceed there after.
EDIT:
This problem was given to me by my student. I inquired her, she corrected the problem as $\int \frac{1}{x^3}e^{-x^{-2}}dx$ that makes the problem very simple. Sorry for troubling you all.