## Calculus, 10th Edition (Anton)

$$x\ln \left( {3x - 2} \right) - x - \frac{2}{3}\ln \left| {3x - 2} \right| + C$$
\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}