Answer
$$\frac{4}{{15}}{\left( {1 + \sqrt x } \right)^{3/2}}\left( {3\sqrt x - 2} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {1 + \sqrt x } } dx \cr
& {\text{Let }}{u^2} = x,{\text{ 2}}udu = dx \cr
& {\text{Substituting}} \cr
& \int {\sqrt {1 + \sqrt x } } dx = \int {\sqrt {1 + \sqrt {{u^2}} } } \left( {{\text{2}}u} \right)du \cr
& = \int {2u\sqrt {1 + u} } du \cr
& {\text{Let }}z = 1 + u,{\text{ }}u = z - 1,{\text{ }}du = dz \cr
& \int {2u\sqrt {1 + u} } du = \int {2\left( {z - 1} \right)\sqrt z } dz \cr
& = 2\int {\left( {{z^{3/2}} - {z^{1/2}}} \right)} dz \cr
& {\text{Integrating}} \cr
& {\text{ = }}2\left( {\frac{{{z^{5/2}}}}{{5/2}} - \frac{{{z^{3/2}}}}{{3/2}}} \right) + C \cr
& {\text{ = }}4\left( {\frac{{{z^{5/2}}}}{5} - \frac{{{z^{3/2}}}}{3}} \right) + C \cr
& {\text{ = }}\frac{4}{{15}}{z^{3/2}}\left( {3z - 5} \right) + C \cr
& {\text{Write in terms of }}u,{\text{ substitute }}1 + u{\text{ for }}z \cr
& {\text{ = }}\frac{4}{{15}}{\left( {1 + u} \right)^{3/2}}\left( {3\left( {1 + u} \right) - 5} \right) + C \cr
& {\text{ = }}\frac{4}{{15}}{\left( {1 + u} \right)^{3/2}}\left( {3u - 2} \right) + C \cr
& {\text{Write in terms of }}x,{\text{ substitute}}\sqrt x {\text{ for }}u \cr
& {\text{ = }}\frac{4}{{15}}{\left( {1 + \sqrt x } \right)^{3/2}}\left( {3\sqrt x - 2} \right) + C \cr} $$
