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: 26



Work Step by Step

$$A=\int\frac{dx}{ax+b}(a\ne0)$$ Let $u=ax+b$. We would have $du=(ax+b)'dx=adx$. Therefore, $dx=\frac{1}{a}du$ Substitute into $A$, we have $$A=\frac{1}{a}\int\frac{1}{u}du$$ $$A=\frac{1}{a}\ln|u|+C$$ $$A=\frac{\ln|ax+b|}{a}+C$$
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