Answer
$$\frac{{{x^3}}}{3}\ln \left( {ax} \right) - \frac{{{x^3}}}{9} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}\ln \left( {ax} \right)} dx \cr
& {\text{Use the integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = \ln \left( {ax} \right),\,\,\,\,du = \frac{1}{x}dx,\,\,\,\,\,dv = {x^2}dx,\,\,\,\,v = \frac{{{x^3}}}{3} \cr
& \cr
& {\text{Then}}{\text{, by using }}\int {udv} = uv - \int {vdu} \cr
& \int {{x^2}\ln \left( {ax} \right)} dx = \frac{{{x^3}}}{3}\ln \left( {ax} \right) - \int {\left( {\frac{{{x^3}}}{3}} \right)} \left( {\frac{1}{x}} \right)dx \cr
& \int {{x^2}\ln \left( {ax} \right)} dx = \frac{{{x^3}}}{3}\ln \left( {ax} \right) - \frac{1}{3}\int {{x^2}} dx \cr
& \cr
& {\text{Integrating gives}} \cr
& \int {{x^2}\ln \left( {ax} \right)} dx = \frac{{{x^3}}}{3}\ln \left( {ax} \right) - \frac{1}{3}\left( {\frac{{{x^3}}}{3}} \right) + C \cr
& \int {{x^2}\ln \left( {ax} \right)} dx = \frac{{{x^3}}}{3}\ln \left( {ax} \right) - \frac{{{x^3}}}{9} + C \cr} $$
