Answer
$$\int {{{\left( {\ln u} \right)}^n}} du = u{\left( {\ln u} \right)^n} - n\int {{{\left( {\ln u} \right)}^{n - 1}}du} $$
Work Step by Step
$$\eqalign{
& \int {{{\left( {\ln u} \right)}^n}} du \cr
& {\text{Use integration by parts}} \cr
& t = {\left( {\ln u} \right)^n},{\text{ }}dt = n{\left( {\ln u} \right)^{n - 1}}\left( {\frac{1}{u}} \right)du \cr
& dv = du,{\text{ }}v = u \cr
& \int {tdv} = tv - \int {vdt} \cr
& {\text{Substituting}} \cr
& \int {{{\left( {\ln u} \right)}^n}} du = u{\left( {\ln u} \right)^n} - \int {un{{\left( {\ln u} \right)}^{n - 1}}\left( {\frac{1}{u}} \right)du} \cr
& \int {{{\left( {\ln u} \right)}^n}} du = u{\left( {\ln u} \right)^n} - \int {n{{\left( {\ln u} \right)}^{n - 1}}du} \cr
& \int {{{\left( {\ln u} \right)}^n}} du = u{\left( {\ln u} \right)^n} - n\int {{{\left( {\ln u} \right)}^{n - 1}}du} \cr} $$
