Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3 - Page 286: 12



Work Step by Step

Use the substitution $u=5+x^4$. Then $du=4x^3dx$, so $x^3dx=\frac{1}{4}du$. $\int x^3\sqrt{5+x^4} dx$ $=\int \sqrt{u}*\frac{1}{4}du$ $=\frac{u^\frac{3}{2}}{\frac{3}{2}}*\frac{1}{4}+C$ $=\frac{u^\frac{3}{2}}{6}+C$ $=\frac{(5+x^4)^\frac{3}{2}}{6}+C$
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