Answer
$$ - {e^{\frac{1}{x}}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{1/x}}}}{{{x^2}}}} dx \cr
& {\text{use the property of negative exponents }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr
& \int {{e^{{x^{ - 1}}}}\left( {{x^{ - 2}}} \right)} dx \cr
& {\text{set }}u = {x^{ - 1}}{\text{ then }}\frac{{du}}{{dx}} = - {x^{ - 2}} \to \frac{{du}}{{ - {x^{ - 2}}}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{e^{{x^{ - 1}}}}\left( {{x^{ - 2}}} \right)} dx = \int {{e^u}\left( {{x^{ - 2}}} \right)} \left( {\frac{{du}}{{ - {x^{ - 2}}}}} \right) \cr
& {\text{cancel common terms}} \cr
& = - \int {{e^u}du} \cr
& {\text{integrating}} \cr
& = - {e^u} + C \cr
& {\text{replace }} {x^{ - 1}}{\text{ for }}u \cr
& = - {e^{{x^{ - 1}}}} + C \cr
& or \cr
& = - {e^{\frac{1}{x}}} + C \cr }$$
