把3x^2+1/x^4-1化成分子中不含x的若干个分式的和

LZ的题目是不是括号没括出来?
(3x^2+1)/(x^4-1)
如果是这样,解题如下：
原式=(3x^2+1)/(x^2+1)(x^2-1)
=[3(x^2+1)-2]/(x^2+1)(x^2-1)
=3/(x^2-1)-2/(x^2+1)(x^2-1)
=(3/2)[1/{x-1)-1/(x+1)]-(1/2)[1/(x-1)-1/(x+1)]+1/(x^2+1)
=1/(x-1)-1/(x+1)+1/(x^2+1)
这个是不是你要的结果?

3/x^(-2)+1/x^4-1

(3x^2+1)/(x^4-1)=?
x^4-1=(x-1)(x+1)(xx+1)
设
(3x^2+1)/(x^4-1)=A/(x-1)+B/(x+1)+C/(xx+1)
(3x^2+1)=A(x+1)(xx+1)+B(x-1)(xx+1)+C(xx-1)
=A(xxx+xx+x+1)+B(xxx-xx+x-1)+C(xx-1)
=(A+B)xxx+...

（1/(x^2+1)）+（2/(x^2-1)）

=3X^2+x^-4