Answer
$$\frac{{2{{\left( {\sqrt x } \right)}^3}}}{3} - x + 2\sqrt x - 2\ln \left( {1 + \sqrt x } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{1 + \sqrt x }}} dx \cr
& {\text{Let }}x = {u^2},\,\,\,\,dx = 2udu.{\text{ Then}}{\text{,}} \cr
& \cr
& \int {\frac{x}{{1 + \sqrt x }}} dx = \int {\frac{{{u^2}}}{{1 + \sqrt {{u^2}} }}} \left( {2udu} \right) \cr
& {\text{Simplifying}} \cr
& = \int {\frac{{2{u^3}}}{{1 + u}}} du \cr
& \cr
& {\text{Use the long division for }}\frac{{2{u^3}}}{{1 + u}} \cr
& \frac{{2{u^3}}}{{1 + u}} = 2{u^2} - 2u + 2 - \frac{2}{{1 + u}} \cr
& \cr
& \int {\frac{{2{u^3}}}{{1 + u}}} du = \int {\left( {2{u^2} - 2u + 2 - \frac{2}{{1 + u}}} \right)} du \cr
& {\text{Integrating}} \cr
& \int {\frac{{2{u^3}}}{{1 + u}}} du = \frac{{2{u^3}}}{3} - {u^2} + 2u - 2\ln \left| {1 + u} \right| + C \cr
& \cr
& {\text{Write in terms of }}x.\,\,x = {u^2} \to \sqrt x = u \cr
& = \frac{{2{{\left( {\sqrt x } \right)}^3}}}{3} - {\left( {\sqrt x } \right)^2} + 2\left( {\sqrt x } \right) - 2\ln \left| {1 + \sqrt x } \right| + C \cr
& = \frac{{2{{\left( {\sqrt x } \right)}^3}}}{3} - x + 2\sqrt x - 2\ln \left( {1 + \sqrt x } \right) + C \cr} $$
