Answer
$$\ln x - \ln \left| {1 + \ln x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln x}}{{x + x\ln x}}} dx \cr
& {\text{Factor the denominator}} \cr
& \int {\frac{{\ln x}}{{x + x\ln x}}} dx = \int {\frac{{\ln x}}{{\left( {1 + \ln x} \right)x}}} dx \cr
& \cr
& {\text{Integrate by using the substitution method}}{\text{,}} \cr
& {\text{Let }}u = 1 + \ln x,\,\,\,\ln x = u - 1,\,\,\,\,\frac{1}{x}dx = du \cr
& {\text{Write the integrand in terms of }}u \cr
& \,\,\int {\frac{{\ln x}}{{\left( {1 + \ln x} \right)x}}} dx = \int {\frac{{u - 1}}{u}} du \cr
& = \int {\left( {1 - \frac{1}{u}} \right)} du \cr
& {\text{Integrate}} \cr
& = u - \ln \left| u \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}1 + \ln x{\text{ for }}u \cr
& = 1 + \ln x - \ln \left| {1 + \ln x} \right| + C \cr
& {\text{Combine the constants}} \cr
& = \ln x - \ln \left| {1 + \ln x} \right| + C \cr} $$
