Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 390: 41

Answer

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

Work Step by Step

$$\eqalign{ & \int {\frac{{{e^{1/x}}}}{{{x^2}}}} dx \cr & {\text{use the property of negative exponents }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr & \int {{e^{{x^{ - 1}}}}\left( {{x^{ - 2}}} \right)} dx \cr & {\text{set }}u = {x^{ - 1}}{\text{ then }}\frac{{du}}{{dx}} = - {x^{ - 2}} \to \frac{{du}}{{ - {x^{ - 2}}}} = dx \cr & {\text{write the integrand in terms of }}u \cr & \int {{e^{{x^{ - 1}}}}\left( {{x^{ - 2}}} \right)} dx = \int {{e^u}\left( {{x^{ - 2}}} \right)} \left( {\frac{{du}}{{ - {x^{ - 2}}}}} \right) \cr & {\text{cancel common terms}} \cr & = - \int {{e^u}du} \cr & {\text{integrating}} \cr & = - {e^u} + C \cr & {\text{replace }} {x^{ - 1}}{\text{ for }}u \cr & = - {e^{{x^{ - 1}}}} + C \cr & or \cr & = - {e^{\frac{1}{x}}} + C \cr }$$
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