积分∫[x^2/√(1-x^2)]dx=Ax^2/√(1-x^2)+B∫[1/√(1-x^2)]dx,求A、B.

出现不了这种公式,请检查原题,A后面是x而非x^2∫[x^2/√(1-x^2)]dx=∫[(x^2-1+1)/√(1-x^2)]dx=-∫√(1-x^2)]dx+∫1/√(1-x^2)]dx而∫√(1-x^2)]dx=x√(1-x^2)-∫xd√(1-x^2)]=x√(1-x^2)+∫x^2/√(1-x^2)]dx所以2∫x^2/√(1-x^2)]dx=-x√(1-x^2)+∫1/√(1-x^2)]dx∫x^2/√(1-x^2)]dx=-（1/2）x√(1-x^2)+(1/2)∫1/√(1-x^2)]dxA=-1/2 ,B=1/2