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 - 8.2 Exercises: 18


$-\frac{lnx}{2x^2}-\frac 1 {4x^2}+C$

Work Step by Step

$\int \frac{lnx}{x^3}dx$ $u=lnx, du=\frac 1 x dx, v=-\frac 1 {2x^2}, dv=\frac 1 {x^3}dx$ $(lnx)(-\frac 1 {2x^2})-\int (-\frac 1 {2x^2})(\frac 1 x)dx$ $-\frac {lnx} {2x^2}+\frac 1 2(-\frac 1 {2x^2})$ $-\frac{lnx}{2x^2}-\frac 1 {4x^2}+C$
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