Answer
$$2\left( {\sqrt x - {{\tan }^{ - 1}}\sqrt x } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt x }}{{x + 1}}} dx \cr
& {\text{Taking the hint on the book making a substitution of the form }}u = {x^{1/n}} \cr
& {\text{Take }}\,\,\,u = {x^{1/2}} \to \,\,\,\,\,\,\,\,{u^2} = x,\,\,\,\,\,\,\,2udu = dx\,\,\,\, \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{\sqrt x }}{{x + 1}}} dx = \int {\frac{u}{{{u^2} + 1}}} \left( {2udu} \right) \cr
& = \int {\frac{{2{u^2}}}{{{u^2} + 1}}} du = 2\int {\frac{{{u^2} + 1 - 1}}{{{u^2} + 1}}} du \cr
& = 2\int {\left( {1 - \frac{1}{{{u^2} + 1}}} \right)} du \cr
& \cr
& {\text{Integrate}} \cr
& = 2\left( {u - {{\tan }^{ - 1}}u} \right) + C \cr
& \cr
& {\text{write the integrand in terms of }}x; {\text{ replace }}u = {x^{1/2}} \cr
& = 2\left( {{x^{1/2}} - {{\tan }^{ - 1}}{x^{1/2}}} \right) + C \cr
& = 2\left( {\sqrt x - {{\tan }^{ - 1}}\sqrt x } \right) + C \cr} $$
