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