Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 48



Work Step by Step

Let u=$x^3+1=>du=3x^2dx$ and $x^3=u-1$ so $\int 3x^5 \sqrt{x^3+1}dx$ =$\int (u-1)\sqrt{u}$ =$\int (u^{3/2}-u^{1/2})du$ =$\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}+C$ =$\frac{2}{5}(x^3+1)^{5/2}-\frac{2}{3}(x^3+1)^{3/2}+C$
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