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`
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