## Thomas' Calculus 13th Edition

$$x\ln \left( {x + 1} \right) - x + \ln \left| {x + 1} \right| + C$$
\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}