Answer
$$x\ln \left( {3x - 2} \right) - x - \frac{2}{3}\ln \left| {3x - 2} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\ln \left( {3x - 2} \right)} dx \cr
& {\text{substitute }}u = \ln \left( {3x - 2} \right),{\text{ }}du = \frac{3}{{3x - 2}}dx \cr
& dv = dx,{\text{ }}v = x \cr
& {\text{ integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {\ln \left( {3x - 2} \right)} dx = x\ln \left( {3x - 2} \right) - \int {\frac{{3x}}{{3x - 2}}dx} \cr
& \int {\ln \left( {3x - 2} \right)} dx = x\ln \left( {3x - 2} \right) - \int {\frac{{3x - 2 + 2}}{{3x - 2}}dx} \cr
& \int {\ln \left( {3x - 2} \right)} dx = x\ln \left( {3x - 2} \right) - \int {\left( {\frac{{3x - 2}}{{3x - 2}} + \frac{2}{{3x - 2}}} \right)dx} \cr
& \int {\ln \left( {3x - 2} \right)} dx = x\ln \left( {3x - 2} \right) - \int {\left( {1 + \frac{2}{{3x - 2}}} \right)dx} \cr
& \int {\ln \left( {3x - 2} \right)} dx = x\ln \left( {3x - 2} \right) - \int {dx} - \int {\frac{2}{{3x - 2}}dx} \cr
& {\text{find antiderivative}} \cr
& \int {\ln \left( {3x - 2} \right)} dx = x\ln \left( {3x - 2} \right) - x - \frac{2}{3}\ln \left| {3x - 2} \right| + C \cr} $$
