问题描述：
已知∫xf(x)dx=x/(根号1-x^2)+C,求∫1/f(x)dx
问题解答：
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答：
∫xf(x)dx=x/√(1-x²)+C
两边求导得：
xf(x)=1/√(1-x²)+(-x/2)*(-2x)/[(1-x²)√(1-x²)]
=(1-x²+x²)/[(1-x²)√(1-x²)]
=1/[(1-x²)√(1-x²)]
1/f(x)=x(1-x²)^(3/2)
∫[1/f(x)]dx
=∫[x(1-x²)^(3/2)]dx
=(1/2)∫[(1-x²)^(3/2)]d(x²)
=(-1/2)∫[(1-x²)^(3/2)]d(1-x²)
=(-1/2)*(2/5)*(1-x²)^(5/2)+C
=(-1/5)*(1-x²)^(5/2)+C
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