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: 6


$\frac{3}{2}(1+\sqrt x)^{4/3}+c$

Work Step by Step

$u=1+\sqrt x$, therefore $du=\frac{1}{2 \sqrt x}dx$. Let's rewrite the integral! $\int \frac{(1+\sqrt x)^{1/3}}{\sqrt x}dx$ We do the substitution! $\int \frac{(1+\sqrt x)^{1/3}}{\sqrt x}dx=\int u^{1/3}(2du)=2\int u^{1/3}du=2\frac{u^{4/3}}{4/3}+c=\frac{3}{2}u^{4/3}+c$ We re-do the substitution! $\frac{3}{2}u^{4/3}+c=\frac{3}{2}(1+\sqrt x)^{4/3}+c$
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