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