Answer
$$ - 2{e^{ - \sqrt r }} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{ - \sqrt r }}}}{{\sqrt r }}} dr \cr
& {\text{Set }}u = - \sqrt r {\text{ then }}\frac{{du}}{{dr}} = - \frac{1}{{2\sqrt r }} \to dr = - 2\sqrt r du \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\frac{{{e^{\sqrt r }}}}{{\sqrt r }}} dr = \int {\frac{{{e^u}}}{{\sqrt r }}} \left( { - 2\sqrt r du} \right) \cr
& = - 2\int {{e^u}} du \cr
& {\text{Integrating}} \cr
& = - 2{e^u} + C \cr
& {\text{Replace -}}\sqrt r {\text{ for }}u \cr
& = - 2{e^{ - \sqrt r }} + C \cr} $$
