Calculus: Early Transcendentals 9th Edition

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

Chapter 5 - Review - Exercises - Page 429: 25

Answer

$\frac{1}{2}\ln \left( {{x^2} + 1} \right) + C$

Work Step by Step

$$\eqalign{ & \int {\frac{x}{{{x^2} + 1}}} dx \cr & {\text{Integrate by the substitution method}} \cr & {\text{Let }}u = {x^2} + 1,{\text{ then }}du = 2xdx,{\text{ }}dx = \frac{1}{{2x}}du \cr & {\text{Applying the substitution}} \cr & \int {\frac{x}{{{x^2} + 1}}} dx = \int {\frac{x}{u}\left( {\frac{1}{{2x}}} \right)} du \cr & = \int {\frac{1}{{2u}}} du \cr & = \frac{1}{2}\int {\frac{1}{u}} du \cr & {\text{Integrate }} \cr & = \frac{1}{2}\ln \left| u \right| + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}{x^2} + 1{\text{ for }}u \cr & = \frac{1}{2}\ln \left| {{x^2} + 1} \right| + C \cr & {\text{or}} \cr & = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C \cr} $$
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