Calculus: Early Transcendentals 9th Edition

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

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 425: 15

Answer

$\frac{1}{4}\ln \left| {4x + 7} \right| + C$

Work Step by Step

$$\eqalign{ & \text{Let } I=\int {\frac{{dx}}{{4x + 7}}} \cr & {\text{Let }}u = 4x + 7,{\text{ then }}du = 4dx,{\text{ }}dx = \frac{1}{4}du \cr & {\text{Applying the substitution}}{\text{, we obtain}} \cr & I = \int {\frac{1}{u}} \left( {\frac{1}{4}} \right)du \cr & = \frac{1}{4}\int {\frac{1}{u}} du \cr & {\text{Integrating}} \cr & I= \frac{1}{4}\ln \left| u \right| + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}u = 4x + 7 \cr & I = \frac{1}{4}\ln \left| {4x + 7} \right| + C \cr} $$
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