高数积分,划线部分的拆分怎么做的 能给个详细过程吗
2个回答
展开全部
1/(1-t^2)^2 ≡ A/(1-t) + B/(1-t)^2 +C/(1+t) + D/(1+t)^2
=>
1 ≡ A(1-t)(1+t)^2 + B(1+t)^2 +C(1+t)(1-t)^2 + D(1-t)^2
t=1, => B=1/4
t=-1, => D=1/4
coef. of t^3
-A+C =0
A-C =0 (1)
coef. of constant
A+B+C+D =1
A+C = 1/2 (2)
(1)+(2)
A= 1/4
from (1)
C= 1/4
1/(1-t^2)^2 = (1/4) [1/(1-t) + 1/(1-t)^2 +1/(1+t) + 1/(1+t)^2]
=>
1 ≡ A(1-t)(1+t)^2 + B(1+t)^2 +C(1+t)(1-t)^2 + D(1-t)^2
t=1, => B=1/4
t=-1, => D=1/4
coef. of t^3
-A+C =0
A-C =0 (1)
coef. of constant
A+B+C+D =1
A+C = 1/2 (2)
(1)+(2)
A= 1/4
from (1)
C= 1/4
1/(1-t^2)^2 = (1/4) [1/(1-t) + 1/(1-t)^2 +1/(1+t) + 1/(1+t)^2]
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询