Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - Review Exercises - Page 332: 70

Answer

$$x + \ln \left| x \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{x + 1}}{x}} dx \cr & {\text{split the numerator}} \cr & = \int {\left( {\frac{x}{x} + \frac{1}{x}} \right)} dx \cr & = \int {\left( {1 + \frac{1}{x}} \right)} dx \cr & {\text{sum rule}} \cr & = \int {dx} + \int {\frac{1}{x}} dx \cr & {\text{integrating}} \cr & = x + \ln \left| x \right| + C \cr} $$
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