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