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`