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: 17

Answer

$(\dfrac{1}{16})\ln |\dfrac{(v-2)^5(v+2)}{v^6}|+c$

Work Step by Step

Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$ Solve $\int \dfrac{v+3}{2v^3-8v}dv$ Write the integral into partial fractions as follows: $\int \dfrac{v+3}{2v^3-8v}dv=(\dfrac{1}{2}) \int [\dfrac{-3}{4v}+\dfrac{5}{8(v-2)}+\dfrac{1}{8(v+2)}] dv$ This implies that $(\dfrac{-3}{8}) \ln |v|+(\dfrac{5}{16}) \ln |v-2|+\dfrac{1}{16} \ln |v+2|+c=(\dfrac{1}{16})\ln |\dfrac{(v-2)^5(v+2)}{v^6}|+c$
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