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