Answer
$$ x\ln \left( {x + {x^2}} \right) - 2x + \ln \left| {x + 1} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\ln \left( {x + {x^2}} \right)} dx \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = \ln \left( {x + {x^2}} \right),\,\,\,\,du = \frac{{1 + 2x}}{{x + {x^2}}}dx,\,\,\,\,dv = dx,\,\,\,\,v = x \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& = x\ln \left( {x + {x^2}} \right) - \int {x\left( {\frac{{1 + 2x}}{{x + {x^2}}}} \right)dx} \cr
& = x\ln \left( {x + {x^2}} \right) - \int {\frac{{1 + 2x}}{{1 + x}}dx} \cr
& {\text{by long division: }}\frac{{1 + 2x}}{{1 + x}} = 2 - \frac{1}{{x + 1}} \cr
& = x\ln \left( {x + {x^2}} \right) - \int {\left( {2 - \frac{1}{{x + 1}}} \right)dx} \cr
& {\text{then}}{\text{,}} \cr
& = x\ln \left( {x + {x^2}} \right) - 2x + \ln \left| {x + 1} \right| + C \cr} $$
