University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 4


$\ln |4r^2-5|+c$

Work Step by Step

Solve $\int\dfrac{8rdr}{4r^2-5}$ Let $p=4r^2-5$ and $dp=8rdr$ This implies $\int\dfrac{8rdr}{4r^2-5}=\int\dfrac{dp}{p}=\ln|p|+c$ Since, $p=4r^2-5$ Thus, $\int\dfrac{8rdr}{4r^2-5}=\ln |4r^2-5|+c$
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