Answer
$$\frac{{{2^{\sqrt y }}}}{{\ln 2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{2^{\sqrt y }}}}{{2\sqrt y }}dy} \cr
& {\text{Integrate by the substitution method}}{\text{,}} \cr
& \,\,{\text{Let }}u = \sqrt y ,\,\,\,du = \frac{1}{{2\sqrt y }}dy, \cr
& \cr
& \,\,{\text{Write the integrand in terms of }}u \cr
& \int {\frac{{{2^{\sqrt y }}}}{{2\sqrt y }}dy} = \int {{2^u}du} \cr
& {\text{Use the formula }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& \int {{2^u}du} = \frac{{{2^u}}}{{\ln 2}} + C \cr
& \cr
& {\text{Write in terms of }}y;{\text{ substitute }}\sqrt y {\text{ for }}u \cr
& = \frac{{{2^{\sqrt y }}}}{{\ln 2}} + C \cr} $$
