Chapter 13 - Section 13.4 - The Definite Integral: Algebraic Viewpoint and the Fundamental Theorem of Calculus - Exercises - Page 998: 35

$e-\sqrt e$

Given: $I=\int_{1}^{2} \dfrac{e^{1/x}}{x^2} \ dx$ Let us consider that $u=1/x \implies dx=-x^2 \ du$ Now, we have $I=\int_{1}^{2} \dfrac{e^{u}}{x^2} \times -x^2 \ du$ or, $=-[e^u]_1^2+C$ or, $=-[e^{1/x}]_1^2+C$ or, $=-[e^{1/2}]+[e^{1/1}]$ or, $=e-\sqrt e$