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: 5



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