Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 18

Given to solve,
int (ln(x))/x^3 dx
let u = ln(x) => u' = (1/x)
and v' = (x^(-3)) =>
v = x^(-3+1)/(-3+1)
= x^(-2)/(-2)
=(-1)/(2x^2)
by applyinght integration by parts we get,
int uv' dx = uv - int u'v dx
so ,
int (ln(x))/x^3 dx
=(ln(x))((-1)/(2x^2)) - int (1/x)((-1)/(2x^2)) dx
= -ln(x)/(2x^2) + int (1/x)((1)/(2x^2)) dx
= -ln(x)/(2x^2) + int ((1)/(2x^3)) dx
=-ln(x)/(2x^2) + (1/2) int ((1)/(x^3)) dx
= -ln(x)/(2x^2) + (1/2) [x^(-3+1)/(-3+1)]
= -ln(x)/(2x^2) + (1/2) [x^(-2)/(-2)]
=-ln(x)/(2x^2) - 1/4 x^(-2) +c
= 1/(2x^2) (-lnx-1/2) + c