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: 2

Answer

$\frac{(7x-1)^{3/2}}{3/2}+c$

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! $\frac{u^{3/2}}{3/2}+c=\frac{(7x-1)^{3/2}}{3/2}+c$
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