-∫（1+x-x^2-1)dx/√(1+x-x^2)=-∫√((1+x-x^2-1)+∫dx/√(1+x-x^2)=-∫√[(√5/2）^2-(x-1/2)^2]dx+∫dx/√[(√5/2）^2-(x-1/2)^2]=-(1/2)(x-1/2)√(1+x-x^2)-(5/8)arcsin(x-1/2)/(√5/2)+arcsin(x-1/2)/(√5/2)+C
计算不定积分前面的没打出来求(1-x+x^2)/根号下（1+x-x^2）的不定积分
-∫（1+x-x^2-1)dx/√(1+x-x^2)=-∫√((1+x-x^2-1)+∫dx/√(1+x-x^2)=-∫√[(√5/2）^2-(x-1/2)^2]dx+∫dx/√[(√5/2）^2-(x-1/2)^2]=-(1/2)(x-1/2)√(1+x-x^2)-(5/8)arcsin(x-1/2)/(√5/2)+arcsin(x-1/2)/(√5/2)+C