两边求导得:xf(x)=1/√(1-x²)则:f(x)=1/[x√(1-x²)]∫1/f(x)dx=∫x√(1-x²)dx=(1/2)∫√(1-x²)d(x²)=-(1/2)∫√(1-x²)d(-x²)=-(1/2)(2/3)(1-x²)^(3/2)+C=-(1/3)(1-x²)^(3/2)+C希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢。