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: 37

Answer

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

Work Step by Step

$$\eqalign{ & \int {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} dx \cr & {\text{Integrating by parts}} \cr & {\text{Let }}u = \ln \left( {x + \sqrt {{x^2} - 1} } \right) \cr & du = \frac{{1 + \frac{{2x}}{{2\sqrt {{x^2} - 1} }}}}{{x + \sqrt {{x^2} - 1} }}dx \cr & du = \frac{{1 + \frac{x}{{\sqrt {{x^2} - 1} }}}}{{x + \sqrt {{x^2} - 1} }}dx \cr & du = \frac{{\sqrt {{x^2} - 1} + x}}{{\sqrt {{x^2} - 1} \left( {x + \sqrt {{x^2} - 1} } \right)}}dx \cr & du = \frac{1}{{\sqrt {{x^2} - 1} }}dx \cr & and \cr & dv = dx \to v = x \cr & \cr & {\text{Using the integration by parts formula}} \cr & \int {udv} = uv - \int v du \cr & \int {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} dx = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \int {\frac{x}{{\sqrt {{x^2} - 1} }}} dx \cr & \int {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} dx = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \frac{1}{2}\int {\frac{{2x}}{{\sqrt {{x^2} - 1} }}} dx \cr & \int {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} dx = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \frac{1}{2}\int {{{\left( {{x^2} - 1} \right)}^{ - 1/2}}\left( {2x} \right)} dx \cr & {\text{Integrating}} \cr & = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \frac{1}{2}\frac{{{{\left( {{x^2} - 1} \right)}^{1/2}}}}{{1/2}} + C \cr & = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \frac{1}{2}\frac{{{{\left( {{x^2} - 1} \right)}^{1/2}}}}{{1/2}} + C \cr & = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - {\left( {{x^2} - 1} \right)^{1/2}} + C \cr & = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \sqrt {{x^2} - 1} + C \cr} $$
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