## University Calculus: Early Transcendentals (3rd Edition)

$-e^{1/x} +c$
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,$\int \dfrac{e^{1/x} dx}{x^2}=-e^{1/x} +c$