Answer
$$\int \frac{e^{\sin^{-1}x}dx}{\sqrt{1-x^2}}=e^{\sin^{-1}x}+C$$
Work Step by Step
$$A=\int \frac{e^{\sin^{-1}x}dx}{\sqrt{1-x^2}}$$
We set $u=\sin^{-1}x$
Then $$du=\frac{1}{\sqrt{1-x^2}}dx$$
Therefore, $$A=\int e^udu=e^u+C$$ $$A=e^{\sin^{-1}x}+C$$