Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - Chapter Review - Review Exercises - Page 418: 39

Answer

$$\frac{{{{\left( {3\ln x + 2} \right)}^5}}}{{15}} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{{\left( {3\ln x + 2} \right)}^4}}}{x}} dx \cr & {\text{set }}u = 3\ln x + 2{\text{ then }}\frac{{du}}{{dx}} = \frac{3}{x},\,\,\,\,\,\,\frac{1}{3}du = \frac{1}{x}dx \cr & {\text{write the integrand in terms of }}u \cr & \int {\frac{{{{\left( {3\ln x + 2} \right)}^4}}}{x}} dx = \int {{u^4}} \left( {\frac{1}{3}du} \right) \cr & = \frac{1}{4}\int {{u^4}} du \cr & {\text{integrate by using the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr & = \frac{1}{3}\left( {\frac{{{u^5}}}{5}} \right) + C \cr & = \frac{{{u^5}}}{{15}} + C \cr & {\text{replace }}3\ln x + 2{\text{ for }}u \cr & = \frac{{{{\left( {3\ln x + 2} \right)}^5}}}{{15}} + C \cr} $$
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