Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Review - Exercises - Page 430: 40

Answer

$x + 3\ln \left| {x - 3} \right| + C$

Work Step by Step

$$\eqalign{ & \int {\frac{x}{{x - 3}}} dx \cr & {\text{Integrate by the substitution method}} \cr & {\text{Let }}u = x - 3,{\text{ }}x = u + 3,{\text{ then }}dx = du \cr & {\text{Applying the substitution}} \cr & \int {\frac{x}{{x - 3}}} dx = \int {\frac{{u + 3}}{u}} du \cr & {\text{Distribute the numerator}} \cr & = \int {\left( {\frac{u}{u} + \frac{3}{u}} \right)} du \cr & = \int {\left( {1 + \frac{3}{u}} \right)} du \cr & = \int {du} + \int {\frac{3}{u}} du \cr & {\text{Integrate }} \cr & = u + 3\ln \left| u \right| + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}x - 3{\text{ for }}u \cr & = \left( {x - 3} \right) + 3\ln \left| {x - 3} \right| + C \cr & {\text{Combining constants}} \cr & = x + 3\ln \left| {x - 3} \right| + C \cr} $$
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