Answer
$\frac{4}{3}\left( {8 - 3\sqrt 3 } \right)$
Work Step by Step
$$\eqalign{
& \int_1^4 {\frac{{\sqrt {2 + \sqrt x } }}{{\sqrt x }}} dx \cr
& {\text{Let }}u = 2 + \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx,{\text{ }}2du = \frac{1}{{\sqrt x }}dx \cr
& {\text{The new limits of integration are:}} \cr
& x = 4 \to u = 2 + \sqrt 4 ,{\text{ }}u = 4 \cr
& x = 1 \to u = 2 + \sqrt 1 ,{\text{ }}u = 3 \cr
& {\text{Applying the substitution}} \cr
& \int_1^4 {\frac{{\sqrt {2 + \sqrt x } }}{{\sqrt x }}} dx = \int_3^4 {\sqrt u \left( 2 \right)} du \cr
& = 2\int_3^4 {{u^{1/2}}} du \cr
& {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr
& = 2\left[ {\frac{{{u^{3/2}}}}{{3/2}}} \right]_3^4 \cr
& = \frac{4}{3}\left[ {{u^{3/2}}} \right]_3^4 \cr
& {\text{Evaluating the limits of integration}} \cr
& = \frac{4}{3}\left[ {{4^{3/2}} - {3^{3/2}}} \right] \cr
& = \frac{4}{3}\left( {8 - 3\sqrt 3 } \right) \cr} $$
