Answer
$-e^{1/x} +c$
Work Step by Step
Solve $\int \dfrac{e^{1/x} dx}{x^2}$
Let us $p=\dfrac{1}{x}$ and $dp=-\dfrac{1}{x^2}dx$
This implies $\int \dfrac{e^{1/x} dx}{x^2}=-\int e^p dp$
$=-e^p +c$
$=-e^{1/x} +c$
Hence,$\int \dfrac{e^{1/x} dx}{x^2}=-e^{1/x} +c$