Answer
$$y = x\ln \left| {{x^2} + x} \right| - 2x + \ln \left| {x + 1} \right| + C$$
Work Step by Step
$$\eqalign{
& y' = \ln \left( {{x^2} + x} \right) \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \ln \left( {{x^2} + x} \right) \cr
& {\text{Separate the variables}} \cr
& dy = \ln \left( {{x^2} + x} \right)dx \cr
& {\text{Integrate both sides}} \cr
& y = \int {\ln \left( {{x^2} + x} \right)} dx \cr
& {\text{Use the parts method}} \cr
& u = \ln \left( {{x^2} + x} \right),{\text{ }}du = \frac{{2x + 1}}{{{x^2} + x}} \cr
& dv = dx,{\text{ }}v = x \cr
& y = x\ln \left| {{x^2} + x} \right| - \int {\frac{{2x + 1}}{{{x^2} + x}}\left( x \right)dx} \cr
& y = x\ln \left| {{x^2} + x} \right| - \int {\frac{{2x + 1}}{{x + 1}}dx} \cr
& {\text{By long division }}\frac{{2x + 1}}{{x + 1}} = 2 - \frac{1}{{x + 1}} \cr
& y = x\ln \left| {{x^2} + x} \right| - \int {\left( {2 - \frac{1}{{x + 1}}} \right)dx} \cr
& y = x\ln \left| {{x^2} + x} \right| - 2x + \ln \left| {x + 1} \right| + C \cr} $$
