∫[1,0] dx/(x²-2x-3)
= ∫[1,0] dx/[(x-3)(x+1)]
= (1/4)∫[1,0] [1/(x-3) - 1/(x+1)] dx
= (1/4)ln|(x-3)/(x+1)|
= (1/4)ln(2/2) - (1/4)ln(3/1)
= -1/4ln3你这个好像是不定积分的求法啊！