Integral taken from instagram user @integralsforyou

$$\int\frac{e^{\frac{1}{x}}}{x^2}dx$$

$$=\int e^{(x^{-1})}\cdot x^{-2}dx$$

let $u=x^{-1}$

$du=-x^{-2}$

$$=-\int e^udu$$

$$=-e^u+C$$

$$=-e^{\frac{1}{x}}+C$$