∫√[(1-x)/（1+x)]dx=∫(1-x)/√(1-x^2)dx=∫1/√(1-x^2)-∫x/√(1-x^2)dx=arcsinx+1/2∫(1-x^2)^(-1/2)d(1-x^2)=arcsinx+√(1-x^2)+c原式等于arcsin1+0-arcsin0-1=π/2-1
