University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises: 3

Answer

$\frac{(x^2+5)^{-3}}{-3}+c$

Work Step by Step

$u=x^2+5$, therefore $du=2xdx$. Let's rewrite the integral! $\int (2x(x^2+5)^{-4})dx=\int ((x^2+5)^{-4})2xdx$ We do the substitution! $\int ((x^2+5)^{-4})2xdx=\int u^{-4}du=\frac{u^{-3}}{-3}+c$ We re-do the substitution! $\frac{u^{-3}}{-3}+c=\frac{(x^2+5)^{-3}}{-3}+c$
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