University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 19


$-e^{1/x} +c$

Work Step by Step

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