Calculus: Early Transcendentals 9th Edition

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

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 522: 50

Answer

$2{\tan ^{ - 1}}\sqrt {x - 1} + C$

Work Step by Step

$$\eqalign{ & \int {\frac{1}{{x\sqrt {x - 1} }}} dx \cr & {\text{Use the substitution method}} \cr & {\text{Let }}{u^2} = x - 1,{\text{ }}x = {u^2} + 1,{\text{ }}dx = 2udu \cr & {\text{Substituting}} \cr & \int {\frac{1}{{x\sqrt {x - 1} }}} dx = \int {\frac{1}{{\left( {{u^2} + 1} \right)\sqrt {{u^2}} }}} \left( {2u} \right)du \cr & {\text{Simplify the integrand}} \cr & = \int {\frac{{2u}}{{\left( {{u^2} + 1} \right)u}}} du \cr & = 2\int {\frac{1}{{{u^2} + 1}}} du \cr & {\text{Integrating}} \cr & = 2{\tan ^{ - 1}}u + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}\sqrt {x - 1} {\text{ for }}u \cr & = 2{\tan ^{ - 1}}\sqrt {x - 1} + C \cr} $$
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