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