Answer
$\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{Cx+D}{x^2+1}$
Work Step by Step
Simplify $\dfrac{1}{x^4-1}$ into partial fraction decomposition.
This can be written as:
$\dfrac{1}{x^4-1}=\dfrac{1}{(x^2-1)(x^2+1)}$
Also, further
$\dfrac{1}{(x^2-1)(x^2+1)}=\dfrac{1}{(x-1)(x+1)(x^2+1)}$
Applying the partial fraction decomposition rule.
$\dfrac{1}{(x-1)(x+1)(x^2+1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{Cx+D}{x^2+1}$
