Answer
${e^{\arcsin x}} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{\arcsin x}}}}{{\sqrt {1 - {x^2}} }}} dx \cr
& \int {\frac{{{e^{\arcsin x}}}}{{\sqrt {1 - {x^2}} }}} dx = \int {{e^{\arcsin x}}\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr
& {\text{Use the substitution method}} \cr
& {\text{Let }}u = \arcsin x,{\text{ }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& {\text{Substituting}} \cr
& \int {{e^{\arcsin x}}\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)} dx = \int {{e^u}du} \cr
& {\text{Integrating}} \cr
& = {e^u} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\arcsin x{\text{ for }}u \cr
& = {e^{\arcsin x}} + C \cr} $$