University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Practice Exercises - Page 474: 33


$(\dfrac{-1}{2}) \ln |9-x^2|+c$

Work Step by Step

Consider the integral $\int \dfrac{x dx}{9-x^2}$ The given integral can be re-written as: $\int \dfrac{x dx}{9-x^2}=(\dfrac{-1}{2})\int\dfrac{-2x dx}{(9-x^2)}$ This implies that $(\dfrac{-1}{2})\int\dfrac{-2x dx}{(9-x^2)}=(\dfrac{-1}{2}) \ln |9-x^2|+c$
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