tactically, we're going to set the IBP up so that we get to bust up the #ln x# term by differentiating it. integrating #ln x# is a bad way to go.
So we say that
#int lnx/x^5 \ dx =int\ ln x \ (-1/4)d/dx ( 1/x^4) \ dx#
# =int\ ln x d/dx (- 1/(4x^4)) \ dx#
now we apply IBP: #int u v' = uv - int u' v#
so we can crush that #ln x# term
# int\ ln x d/dx (- 1/(4x^4)) \ dx#
#= - 1/(4x^4) ln x - int\ d/dx( ln x) * (- 1/(4x^4)) \ dx#
#= - 1/(4x^4) ln x + 1/4 int\ 1/x * ( 1/(x^4)) \ dx#
#= - 1/(4x^4) ln x + 1/4 int\ 1/(x^5) \ dx#
#= - 1/(4x^4) ln x + 1/4 ( -1/(4x^4) + C)#
#= - 1/(4x^4) ( ln x + 1/4) + C# , et voila!
We can have a quick go the other way by saying that
#int lnx/x^5 \ dx =int\ d/dx (1/2 ln^2 x) (1/x^4) \ dx#
So IBP tells us that equals
# =1/(2x^4) ln^2 x - int\ 1/2 ln^2 x d/dx (1/x^4) \ dx#
# =1/(2x^4) ln^2 x + 2 int\ (ln^2 x )/x^5 \ dx#
dunno where that's going! but i'm stopping there...
