Answer
$$2{e^{\sqrt x }} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} dx \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = \sqrt x,\,\,\,\,du = \frac{1}{{2\sqrt x }}dx,\,\,\,\,dx = 2\sqrt x du \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} dx = \int {\frac{{{e^u}}}{{\sqrt x }}} \left( {2\sqrt x du} \right) \cr
& {\text{Cancel common factor }}\sqrt x \cr
& = \int {{e^u}} \left( {2du} \right) \cr
& = 2\int {{e^u}} du \cr
& {\text{integrating}} \cr
& = 2{e^u} + C \cr
& {\text{replacing }}u = \sqrt x \cr
& = 2{e^{\sqrt x }} + C \cr} $$
