Question

第6题求大神帮忙因式分解。

Answer 1

Answer 2

6. (x^2+1)/[(x+1)^2(x-1)] = A/(x-1)+B/(x+1)+C/(x+1)^2
通分比较分子得 A(x+1)^2+B(x^2-1)+C(x-1) = x^2+1
则 A+B = 1， 2A+C = 0， A-B-C = 1
联立解得 A = B = 1/2, C = -1。
则 (x^2+1)/[(x+1)^2(x-1)] = (1/2)[1/(x-1)+1/(x+1)-2/(x+1)^2]
积分 I = (1/2)[ln|x-1| + ln|x+1| + 2/(x+1)] + C
= (1/2)ln|x^2-1| + 1/(x+1) + C

Answer 3

(x^2+1)/[(x+1)^2(x-1)]=A/(x+1)^2+B/(x+1)+C/(x-1)
A(x-1)+B(x+1)(x-1)+C(x+1)^2=x^2+1
令x=-1
A(-1-1)=(-1)^2+1
A=-1
令x=1
C(1+1)^2=1^2+1
C=1/2
令x=0
(-1)(0-1)+B(0+1)(0-1)+1/2(0+1)^2=0^2+1
1-B+1/2=1
B=1/2
∫(x^2+1)/[(x+1)^2(x-1)]
=∫{-1/(x+1)^2+1/[2(x+1)]+1/[2(x-1)}dx
=1/(x+1)+1/2ln(x+1)+1/2ln(x-1)+C
=1/(x+1)+1/2ln[(x+1)(x-1)]+C
=1/(x+1)+1/2ln(x^2-1)+C