Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 13

Answer

$$\int\frac{dx}{5-3x}=-\frac{1}{3}\ln|5-3x|+C$$

Work Step by Step

$$A=\int\frac{dx}{5-3x}$$ Let $u=5-3x$ Then $du=-3dx$. So $dx=-\frac{1}{3}du$ Substitute into $A$, we have $$A=-\frac{1}{3}\int\frac{1}{u}du$$ $$A=-\frac{1}{3}\ln|u|+C$$ $$A=-\frac{1}{3}\ln|5-3x|+C$$
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