Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 23



Work Step by Step

Let u=3x+2 $=>du=3dx=>\frac{1}{3}du=dx$ =$\int sec^2(3x+2)dx$ =$\int (sec^2u)(\frac{1}{3}du)$ =$\frac{1}{3}\int sec^2udu$ =$\frac{1}{3}tan u+C$ =$\frac{1}{3}tan(3x+2)+C$
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.