Answer
$$ - 2{e^{ - \sqrt x }}\left( {\sqrt x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{ - \sqrt x }}} dx \cr
& {\text{Make an appropiate }}u{\text{ - substitution }} \cr
& u = - \sqrt x ,\,\,\,\,\,\,\,du = - \frac{1}{{2\sqrt x }}dx,\,\,\,\,\,\,dx = - 2\sqrt x du,\,\,\,\,\,\,dx = 2udu\,\,\, \cr
& {\text{write in terms of }}u \cr
& \int {{e^{ - \sqrt x }}} dx = \int {{e^u}} \left( {2udu\,\,\,} \right) \cr
& = 2\int {u{e^u}} du \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 51}} \cr
& \left( {51} \right):\,\,\,\,\,\,\int {u{e^u}} du = {e^u}\left( {u - 1} \right) + C \cr
& 2\int {u{e^u}} du = 2{e^u}\left( {u - 1} \right) + C \cr
& {\text{write in terms of }}x{\text{; replace }} - \sqrt x {\text{ for }}u \cr
& = 2{e^{ - \sqrt x }}\left( { - \sqrt x - 1} \right) + C \cr
& = - 2{e^{ - \sqrt x }}\left( {\sqrt x + 1} \right) + C \cr} $$
