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