Question

How do you integrate `(ln x)^2/ x^3 dx`?

Answer 1

let's start by `u = ln(x)`
`du = 1/x`

`int ln^2(x)/x^3dx =int 1/x^2*1/x*ln^2(x) =int 1/x^2*u^2du`

`u = ln(x)`
`e^u=x`
`e^(2u) = x^2`
`e^(-2u)= 1/x^2`

So now we have :

`=>inte^(-2u)*u^2du`

By part :

`dv= e^(-2u)`
`v =-1/2e^(-2u)`

`w = u^2`
`dw = 2u`

`=>-1/2[u^2*e^(-2u)]+intu*e^(-2u)du`

By part again :

`dv = e^(-2u)`
`v = -1/2e^(-2u)`
`w = u`
`dw = 1`

`=>-1/2[u^2*e^(-2u)]-1/2[e^(-2u)*u]+1/2inte^(-2u)du`

`=>-1/2[u^2*e^(-2u)]-1/2[e^(-2u)*u]-1/4[e^(-2u)]`

Substitute back for `u = ln(x)`

`=>-1/2[ln^2(x)*1/x^2]-1/2[ln(x)*1/x^2]-1/4[1/x^2]+C`

`=>-1/(4x^2)(2ln^2(x)+2ln(x)+1)+C`