Calculus 10th Edition

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: 19

Answer

$\frac 1 2 v^2-\frac 1 {6(3v-1)^2}+C$

Work Step by Step

$\int [v+ \frac 1 {(3v-1)^3}]dv$ $\int vdv+\frac 1 3 \int (3v-1)^{-3}(3)dv$ $\frac {v^2} 2 +C- \frac1 3 (-\frac 1 {2(3v-1)^2}+C)$ $\frac 1 2 v^2-\frac 1 {6(3v-1)^2}+C$
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