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

Answer

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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.