∫[(x^2- 1+1)/(x-1)^100]dx=∫[(x+1)/(x-1)^99]dx+∫[1/(x-1)^100]dx
=∫[(x - 1+2)/(x-1)^99]dx+∫[1/(x-1)^100]dx
=∫[1/(x-1)^98]dx+∫[2/(x-1)^99]dx+ ∫[1/(x-1)^100]dx
=(- 1/97)[1/(x-1)^97]+(- 2/98)[1/(x-1)^98]+(- 1/99)[1/(x-1)^99]+C
求积分：∫[x^2/(x-1)^100]dx.得的结果和答案总是不一样,就是用的分步积分.