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: 5

Answer

$\frac{(3x^2+4x)^5}{10}+c$

Work Step by Step

$u=3x^2+4x$, therefore $du=(6x+4)dx=2(3x+2)dx$. We do the substitution! $\int((3x+2)(3x^2+4x)^4) dx=\int u^4\frac{du}{2}=\frac{1}{2}\int u^4du=\frac{1}{2}\frac{u^5}{5}+c=\frac{u^5}{10}+c$ We re-do the substitution! $\frac{u^5}{10}+c=\frac{(3x^2+4x)^5}{10}+c$
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.