Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 408: 6

Answer

$$A=\frac{4\sqrt[4]{x^9}}{9}+C$$

Work Step by Step

$$A=\int{\sqrt[4] {x^5}}dx$$ $$A=\int{x^{5/4}}dx$$ According to Table 1, we know that $\int{x^n}dx=\frac{x^{n+1}}{n+1}$. Therefore, $$A=\frac{x^{9/4}}{\frac{9}{4}}+C=\frac{4x^{9/4}}{9}+C$$ $$A=\frac{4\sqrt[4]{x^9}}{9}+C$$
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