University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 329: 2



Work Step by Step

$u=7x-1$, therefore $du=7dx$. Let's rewrite the integral! $\int (7\sqrt {7x-1})dx=\int (\sqrt {7x-1})7dx$ We do the substitution! $\int (\sqrt {7x-1})7dx=\int u^{1/2}du=\frac{u^{3/2}}{3/2}+c$ We re-do the substitution! $\frac{u^{3/2}}{3/2}+c=\frac{(7x-1)^{3/2}}{3/2}+c$
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.