Answer
$\frac 1 2 v^2-\frac 1 {6(3v-1)^2}+C$
Work Step by Step
$\int [v+ \frac 1 {(3v-1)^3}]dv$
$\int vdv+\frac 1 3 \int (3v-1)^{-3}(3)dv$
$\frac {v^2} 2 +C- \frac1 3 (-\frac 1 {2(3v-1)^2}+C)$
$\frac 1 2 v^2-\frac 1 {6(3v-1)^2}+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
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 - Page 512: 19
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