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

Answer

$$\int\frac{dx}{ax+b}(a\ne0)=\frac{\ln|ax+b|}{a}+C$$

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 $$A=\frac{1}{a}\int\frac{1}{u}du$$ $$A=\frac{1}{a}\ln|u|+C$$ $$A=\frac{\ln|ax+b|}{a}+C$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.