Answer
$$2\ln \sqrt x - 2\ln \left( {1 + \sqrt x } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x + x\sqrt x }}} dx \cr
& = \int {\frac{1}{{x + {x^{3/2}}}}} dx \cr
& {\text{Let }}{u^2} = x,{\text{ }}2udu = dx \cr
& {\text{Substituting}} \cr
& = \int {\frac{1}{{{u^2} + {{\left( {{u^2}} \right)}^{3/2}}}}} \left( {2u} \right)du \cr
& = \int {\frac{{2u}}{{{u^2} + {u^3}}}} du \cr
& = \int {\frac{2}{{u + {u^2}}}} du \cr
& {\text{Factoring the denominator}} \cr
& = \int {\frac{2}{{u\left( {1 + u} \right)}}} du \cr
& {\text{Decomposing into partial fractions }} \cr
& \frac{2}{{u\left( {1 + u} \right)}} = \frac{A}{u} + \frac{B}{{1 + u}} \cr
& 2 = A\left( {1 + u} \right) + Bu \cr
& u = 0 \to A = 2 \cr
& u = - 1 \to B = - 2 \cr
& \frac{2}{{u\left( {1 + u} \right)}} = \frac{2}{u} - \frac{2}{{1 + u}} \cr
& {\text{Therefore,}} \cr
& \int {\frac{2}{{u\left( {1 + u} \right)}}} du = \int {\left( {\frac{2}{u} - \frac{2}{{1 + u}}} \right)} du \cr
& {\text{Integrating}} \cr
& = 2\ln \left| u \right| - 2\ln \left| {1 + u} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = 2\ln \left| {\sqrt x } \right| - 2\ln \left| {1 + \sqrt x } \right| + C \cr
& = 2\ln \sqrt x - 2\ln \left( {1 + \sqrt x } \right) + C \cr} $$
