∫[0→2] 1/(x²-4x+3) dx=∫[0→2] 1/[(x-1)(x-3)] dx=∫[0→1] 1/[(x-1)(x-3)] dx + ∫[1→2] 1/[(x-1)(x-3)] dx积分收敛的充分必要条件是以上两个积分都收敛,下面计算第一个∫[0→1] 1/[(x-1)(x-3)] dx=(1/2)...