Answer
$$ - 4\sqrt x - x - 4\ln \left| {1 - \sqrt x } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{1 + \sqrt x }}{{1 - \sqrt x }}} dx \cr
& {\text{Taking the hint on the book making a substitution of the form }}u = {\left( {x + a} \right)^{1/n}} \cr
& {\text{Take }}\,\,\,u = {x^{1/2}} \to \,\,\,\,\,\,\,\,{u^2} = x,\,\,\,\,\,\,\,2udu = dx\,\,\,\, \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{1 + \sqrt x }}{{1 - \sqrt x }}} dx = \int {\frac{{1 + u}}{{1 - u}}} \left( {2udu} \right) \cr
& = \int {\frac{{2{u^2} + 2u}}{{1 - u}}} du \cr
& \cr
& {\text{perform the long division}} \cr
& \frac{{2{u^2} + 2u}}{{1 - u}} = - 4 - 2u + \frac{4}{{1 - u}} \cr
& \int {\frac{{2{u^2} + 2u}}{{1 - u}}} du = \int {\left( { - 4 - 2u + \frac{4}{{1 - u}}} \right)} du \cr
& \cr
& {\text{integrating}} \cr
& = - 4u - {u^2} - 4\ln \left| {1 - u} \right| + C \cr
& \cr
& {\text{write the integrand in terms of }}x;{\text{ replace }}u = {x^{1/2}} \cr
& = - 4{x^{1/2}} - x - 4\ln \left| {1 - {x^{1/2}}} \right| + C \cr
& = - 4\sqrt x - x - 4\ln \left| {1 - \sqrt x } \right| + C \cr} $$
