Answer
$$\frac{2}{3}\left( {\sqrt {3s + 9} {e^{\sqrt {3s + 9} }} - {e^{\sqrt {3s + 9} }}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {{e^{\sqrt {3s + 9} }}} ds \cr
& {\text{use }}t = \sqrt {3s + 9},\,\,\,\,dt = \frac{3}{{2\sqrt {3s + 9} }}ds,\,\,\,\,ds = \frac{{2\sqrt {3s + 9} }}{3}dt \cr
& \int {{e^{\sqrt {3s + 9} }}} ds = \int {{e^{\sqrt t }}} \left( {\frac{{2\sqrt {3s + 9} }}{3}} \right)dt \cr
& = \frac{2}{3}\int {t{e^t}} dt \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}u = t,\,\,\,\,du = dt,\,\,\,\,\,dv = {e^t}dt,\,\,\,\,v = {e^t} \cr
& \cr
& {\text{Integration by parts then gives}} \cr
& \frac{2}{3}\int {t{e^t}} dt = \frac{2}{3}\left( {t{e^t} - \int {{e^t}} dt} \right) \cr
& \frac{2}{3}\int {t{e^t}} dt = \frac{2}{3}t{e^t} - \frac{2}{3}\int {{e^t}} dt \cr
& \frac{2}{3}\int {t{e^t}} dt = \frac{2}{3}t{e^t} - \frac{2}{3}{e^t} + C \cr
& \cr
& {\text{To write in terms of }}s{\text{, substitute }}\sqrt {3s + 9} {\text{ for }}t \cr
& = \frac{2}{3}\sqrt {3s + 9} {e^{\sqrt {3s + 9} }} - \frac{2}{3}{e^{\sqrt {3s + 9} }} + C \cr
& = \frac{2}{3}\left( {\sqrt {3s + 9} {e^{\sqrt {3s + 9} }} - {e^{\sqrt {3s + 9} }}} \right) + C \cr} $$
