Answer
$$\left( {{x^2} - x} \right)\ln \left( {3x} \right) - \frac{{{x^2}}}{2} + x + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {2x - 1} \right)\ln \left( {3x} \right)} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = \ln \left( {3x} \right){\text{ then }}du = \frac{1}{x}dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \left( {2x - 1} \right)dx{\text{ then }}v = {x^2} - x \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\left( {2x - 1} \right)\ln \left( {3x} \right)} dx = \left( {{x^2} - x} \right)\ln \left( {3x} \right) - \int {\left( {{x^2} - x} \right)} \left( {\frac{1}{x}dx} \right) \cr
& \int {\left( {2x - 1} \right)\ln \left( {3x} \right)} dx = \left( {{x^2} - x} \right)\ln \left( {3x} \right) - \int {\left( {x - 1} \right)} dx \cr
& \int {\left( {2x - 1} \right)\ln \left( {3x} \right)} dx = \left( {{x^2} - x} \right)\ln \left( {3x} \right) - \int x dx + \int {dx} \cr
& {\text{integrate}} \cr
& \int {\left( {2x - 1} \right)\ln \left( {3x} \right)} dx = \left( {{x^2} - x} \right)\ln \left( {3x} \right) - \frac{{{x^2}}}{2} + x + C \cr} $$
