Answer
$$\frac{4}{3}{\left( {3 + \sqrt x } \right)^{3/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {3 + \sqrt x } }}{{\sqrt x }}} dx \cr
& {\text{substitute }}u = 3 + \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx,{\text{ }}2du = \frac{1}{{\sqrt x }}dx \cr
& = \int {\frac{{\sqrt {3 + \sqrt x } }}{{\sqrt x }}} dx = \int {\sqrt u } \left( {2du} \right) \cr
& = 2\int {{u^{1/2}}du} \cr
& {\text{find antiderivative}} \cr
& = 2\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \cr
& = \frac{4}{3}{u^{3/2}} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{4}{3}{\left( {3 + \sqrt x } \right)^{3/2}} + C \cr} $$
