設(shè)x/[(x-1)(x-2)(x-3)]
=A/(x-1)+B/(x-2)+C/(x-3)
=[A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)]/[(x-1)(x-2)(x-3)]
=[A(x^2-5x+6)+B(x^2-4x+3)+C(x^2-3x+2)]/[(x-1)(x-2)(x-3)]
=[(A+B+C)x^2-(5A+4B+3C)x+(6A+3B+2C)]/[(x-1)(x-2)(x-3)]
對(duì)應(yīng)系數(shù)相等,則A+B+C=0,-(5A+4B+3C)=1,6A+3B+2C=0
解得A=1/2,B=-2,C=3/2
故∫x/[(x-1)(x-2)(x-3)]*dx
=∫[(1/2)/(x-1)-2/(x-2)+(3/2)/(x-3)]*dx
=(1/2)ln|x-1|-2ln|x-2|+(3/2)ln|x-3|+C嗯啊……我自己也做出來了謝謝采納！