Answer
${e^{\arcsin x}} + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{\arcsin x}}}}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{Let }}u = \arcsin x,{\text{ then }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx,{\text{ then substituting}} \cr
& \int {\frac{{{e^{\arcsin x}}}}{{\sqrt {1 - {x^2}} }}} dx = \int {{e^{\arcsin x}}\frac{1}{{\sqrt {1 - {x^2}} }}} dx = \int {{e^u}} du \cr
& {\text{Integrate }} \cr
& = {e^u} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}u = \arcsin x \cr
& = {e^{\arcsin x}} + C \cr} $$
