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