Question Number 20291 by tammi last updated on 25/Aug/17
$$\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:{dx}} \\ $$
Answered by $@ty@m last updated on 25/Aug/17
$$=\int\frac{\mathrm{1}−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$={sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{C} \\ $$