Answer
$$x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^{ - 1}}x} dx \cr
& {\text{substitute }}u = {\sin ^{ - 1}}x,{\text{ }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& dv = dx,{\text{ }}v = x \cr
& {\text{using integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x - \int {x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)dx} \cr
& \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x - \int {x\left( {\frac{1}{{{{\left( {1 - {x^2}} \right)}^{1/2}}}}} \right)dx} \cr
& \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x - \int {x{{\left( {1 - {x^2}} \right)}^{ - 1/2}}dx} \cr
& {\text{find antiderivative}}{\text{, }}t = 1 - {x^2},{\text{ }}dt = - 2xdx \cr
& \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + \frac{1}{2}\int {{t^{ - 1/2}}dt} \cr
& \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + \frac{1}{2}\frac{{{t^{1/2}}}}{{1/2}} + C \cr
& \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + {t^{1/2}} + C \cr
& \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + {\left( {1 - {x^2}} \right)^{1/2}} + C \cr
& \int {{{\sin }^{ - 1}}x} dx = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C \cr} $$
