Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 10

Answer

$\int u^{n}du$ $u=x^{2}-4$ $n=-\frac{1}{2}$

Work Step by Step

Try it: $u=x^{2}-4$ $du=2xdx$ $\int u^{n}du=\frac{u^{n+1}}{n+1}+C$ $-\frac{u^{n+1}}{n+1}+C$ $-\frac{(x^{2}-4)^{1/2}}{1/2}+C$ $-2(x^{2}-4)^{1/2}+C$ $-2\sqrt {x^{2}-4}+C$ $\int u^{n}du$ $u=x^{2}-4$ $n=-\frac{1}{2}$
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