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: 20



Work Step by Step

$\int\frac{x^2+1}{\sqrt{x^3+3x}}dx$ Let $u=x^3+3x$. Then $du=(3x^2+3)dx$, so $(x^2+1)dx=\frac{du}{3}$. $=\int\frac{1}{\sqrt{u}}*\frac{du}{3}$ $=\frac{1}{3}\int u^{-\frac{1}{2}} du$ $=\frac{1}{3}(\frac{u^\frac{1}{2}}{\frac{1}{2}}+C)$ $=\frac{2}{3}\sqrt{u}+C$ $=\frac{2}{3}\sqrt{x^3+3x}+C$
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