Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 354: 98

$-\frac{1}{2}e^{\frac{1}{x^2}}+C$

$\int \frac{e^{\frac{1}{x^2}}}{x^3}dx$ let $x^{-2}=u$ $-2x^{-3}dx=du$ $-\frac{2}{x^3}dx=du$ $\int \frac{e^{\frac{1}{x^2}}}{x^3}dx$ $=-\frac{1}{2} \int e^u du$ $=-\frac{1}{2}e^u+C$ $=-\frac{1}{2}e^{\frac{1}{x^2}}+C$