Answer
$${e^{{{\sin }^{ - 1}}x}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{{{\sin }^{ - 1}}x}}dx}}{{\sqrt {1 - {x^2}} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = {\sin ^{ - 1}}x,{\text{ so that }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& {\text{then}} \cr
& \int {{e^{{{\sin }^{ - 1}}x}}\frac{1}{{\sqrt {1 - {x^2}} }}dx} = \int {{e^u}} du \cr
& {\text{Integrating}} \cr
& = {e^u} + C \cr
& {\text{write in terms of }}x;{\text{ replace }}{\sin ^{ - 1}}x{\text{ for }}u \cr
& = {e^{{{\sin }^{ - 1}}x}} + C \cr} $$
