Not sure how to do the following integral, is it integration by parts?
$\displaystyle \int{\frac{e^{{\frac{-1}{x}}}}{x^2}dx}$
Hello, Jason Bourne!
$\displaystyle \int{\frac{e^{\text{-}{\frac{1}{x}}}}{x^2}dx}$
We have: .$\displaystyle \int e^{\text{-}x^{\text{-}1}}\left( x^{\text{-}2}\,dx\right)$
Let: .$\displaystyle u = \text{-}x^{\text{-}1}\quad\Rightarrow\quad du \:=\:x^{\text{-}2}dx $
Substitute: .$\displaystyle \int e^u\,du\quad\hdots\quad Got\; it?$