Answer
$$\int arcsin\,x\,dx=x\,arcsin\,x+\sqrt{1-x^{2}}+C$$
Work Step by Step
${(x\,arcsin\,x)}'=arcsin\,x+\frac{x}{\sqrt{1-x^{2}}}$
$$\int arcsin\,x\,dx=x\,arcsin\,x- \int\frac{x}{\sqrt{1-x^{2}}}dx$$
$$=x\,arcsin\,x+\frac{1}{2}\int (1-x^{2})^{-\frac{1}{2}}d(1-x^{2})$$
$$=x\,arcsin\,x+(1-x^{2})^{\frac{1}{2}}+C$$
$$=x\,arcsin\,x+\sqrt{1-x^{2}}+C$$
