1/（（x^2-1）(x+1)）怎么拆成两项或三项的和?

Question

1/（（x^2-1）(x+1)）怎么拆成两项或三项的和?

Answer 1

1/（（x^2-1）(x+1)）=1/(x-1)(x+1)(x+1)
=1/(x-1)(x+1)方
=A/(x-1)+B/(x+1)+C/(x+1)方
两边同乘以(x-1),得
1/(x+1)方 =A+[B/(x+1)+C/(x+1)方] (x-1)
令x=1,得
1/4=A
同理,两边同乘以(x+1)方,得
1/(x-1)=[A/(x-1)+B/(x+1)](x+1)方+C
=[A/(x-1)*(x+1)方+B*(x+1)]+C
令x=-1,得
C=-1/2
再令x=0代入,得
-1=-A+B+C
-1=-1/4-1/2+B
B=-1/4
所以
拆开后为：
(1/4)/(x-1)+(-1/4)/(x+1)+(-1/2)/(x+1)方