∫[-1,3] dx/(x²-2x-3)
= ∫[-1,3] dx/[(x-3)(x+1)]
= (1/4)∫[-1,3] [1/(x-3) - 1/(x+1)] dx
= (1/4)ln|(x-3)/(x+1)|
= (1/4)ln(0/4) - (1/4)ln(-4/0)
= ∞
这个积分发散