∫√[(1-x)/(1+x)]dx = ∫√[(1-x)^2 / (1-x^2)]dx = ∫(1-x) / √(1-x^2)dx= ∫1 / √(1-x^2)dx - ∫x / √(1-x^2)dx= arcsinx - 1/2∫1 / √(1-x^2)d(1-x^2)= arcsinx - √(1-x^2) + C