Answer
$$ - \sqrt {1 - {x^2}} + \frac{1}{2}{\left( {\arcsin x} \right)^2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{x + \arcsin x}}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{Distribute the numerator}} \cr
& = \int {\left( {\frac{x}{{\sqrt {1 - {x^2}} }} + \frac{{\arcsin x}}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr
& {\text{ = }}\int {\frac{x}{{\sqrt {1 - {x^2}} }}} dx + \int {\arcsin x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr
& {\text{ = }} - \frac{1}{2}\int {\frac{{ - 2x}}{{\sqrt {1 - {x^2}} }}} dx + \int {\arcsin x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr
& {\text{Integrating by substitution, we obtain}} \cr
& {\text{ = }} - \sqrt {1 - {x^2}} + \frac{1}{2}{\left( {\arcsin x} \right)^2} + C \cr} $$
