Answer
$$x{\left( {\ln x} \right)^n} - n\int {{{\left( {\ln x} \right)}^{n - 1}}} dx$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {\ln x} \right)}^n}} dx \cr
& {\text{Using the integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}\,\,\,\,\,u = {\left( {\ln x} \right)^n},\,\,\,\,\,\,\,du = n{\left( {\ln x} \right)^{n - 1}}\left( {\frac{1}{x}} \right)dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = dx,\,\,\,\,\,\,\,v = x \cr
& \cr
& {\text{Integration by parts formula }}\int {udv} = uv - \int {vdu.{\text{ T}}} {\text{hen}}{\text{,}} \cr
& \int {udv} = uv - \int {vdu} \,\,\,\, \cr
& \to \int {{{\left( {\ln x} \right)}^n}} dx = \left( x \right){\left( {\ln x} \right)^n} - \int {\left( x \right)\left( {n{{\left( {\ln x} \right)}^{n - 1}}\left( {\frac{1}{x}} \right)dx} \right)} \cr
& \cr
& {\text{simplifying}} \cr
& \int {{{\left( {\ln x} \right)}^n}} dx = x{\left( {\ln x} \right)^n} - \int {n{{\left( {\ln x} \right)}^{n - 1}}} dx \cr
& \int {{{\left( {\ln x} \right)}^n}} dx = x{\left( {\ln x} \right)^n} - n\int {{{\left( {\ln x} \right)}^{n - 1}}} dx \cr} $$
