题目可能有误。 稍改为
(x+1)/(x^2-6x+5) 解之。 此类问题解法如下：f(x) = (x+1)/(x^2-6x+5) = (x+1)/[(x-1)(x-5)] = A/(x-1) + B/(x-5)= [A(x-5)+B(x-1)]/[(x-1)(x-5)],
得 A+B = 1,
-5A-B = 1,
解得 A = -1/2，
B = 3/2，
f(x) =
(-1/2)/(x-1) + (3/2)/(x-5)
= (-1/2)/(x-2+1) + (3/2)/(x-2-3)= -(1/2)/(1+x-2) - (1/2)/[1-(x-2)/3]= -(1/2)∑<n=0, ∞>(-1)^n(x-2)^n -
(1/2)∑<n=0, ∞>[(x-2)/3]^n= -(1/2)∑<n=0, ∞>[(-1)^n+1/3^n](x-2)^n收敛域 -1 < x-2 < 1,
-1 < (x-2)/3 < 1,即 1 < x < 3，
-1 < x < 5，
得 1 < x < 3。}