Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 508: 72

Answer

$$ - \frac{1}{x}\ln \left( {x + 1} \right) + \ln \left| {\frac{x}{{x + 1}}} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx \cr & {\text{Integrate by parts}} \cr & {\text{Let }}u = \ln \left( {x + 1} \right),{\text{ }}du = \frac{1}{{x + 1}} \cr & dv = \frac{1}{{{x^2}}},{\text{ }}v = - \frac{1}{x} \cr & {\text{Using integration by parts formula}} \cr & \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx = - \frac{1}{x}\ln \left( {x + 1} \right) + \int {\frac{1}{{x\left( {x + 1} \right)}}} dx \cr & {\text{Decomposing }}\frac{1}{{x\left( {x + 1} \right)}}{\text{ into partial fractions }} \cr & \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx = - \frac{1}{x}\ln \left( {x + 1} \right) + \int {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)} dx \cr & {\text{Integrating}} \cr & \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx = - \frac{1}{x}\ln \left( {x + 1} \right) + \ln \left| x \right| - \ln \left| {x + 1} \right| + C \cr & \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx = - \frac{1}{x}\ln \left( {x + 1} \right) + \ln \left| {\frac{x}{{x + 1}}} \right| + C \cr} $$
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