1/[（x+1）^2（x-1）]怎么裂项
- 培训职业
- 2025-05-06 20:33:20
可用待定系数法。
1/[(x+1)^2 (x-1)] = A/(x-1) + B/(x+1) + C/(x+1)^2
= [A(x+1)^2+B(x-1)(x+1)+C(x-1)]/[(x-1)(x+1)^2]
分子比较同次幂系数，得
A+B = 0
2A+C = 0
A-B-C = 1
联立解得 A = 1/4， B = -1/4， C = -1/2
1/[(x+1)^2 (x-1)] = (1/4)[1/(x-1)-1/(x+1)-2/(x+1)^2]
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