Answer
$\frac{{{2^{\sqrt x + 1}}}}{{\ln 2}} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{2^{\sqrt x }}}}{{\sqrt x }}} dx \cr
& = \int {{2^{\sqrt x }}\left( {\frac{1}{{\sqrt x }}} \right)} dx \cr
& {\text{Let }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx,{\text{ }}2du = \frac{1}{{\sqrt x }}dx \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& \int {{2^{\sqrt x }}\left( {\frac{1}{{\sqrt x }}} \right)} dx = \int {{2^u}\left( 2 \right)} du \cr
& = 2\int {{2^u}} du \cr
& {\text{Integrating}} \cr
& = 2\left( {\frac{{{2^u}}}{{\ln 2}}} \right) + C \cr
& = \frac{{{2^{u + 1}}}}{{\ln 2}} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\sqrt x {\text{ for }}u \cr
& = \frac{{{2^{\sqrt x + 1}}}}{{\ln 2}} + C \cr} $$