Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 521: 18

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

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