Answer
$\frac{1}{x+1}-\frac{1}{(x-1)}+\frac{1}{(x-1)^3}$
Work Step by Step
Step 1. Factor the denominator as:
$(x^4-1)-2x(x^2-1)=(x^2-1)(x^2+1-2x)=(x+1)(x-1)^3$
Step 2. Assume the end results as: $\frac{A}{x+1}+\frac{B}{(x-1)}+\frac{C}{(x-1)^2}+\frac{D}{(x-1)^3}$
Step 3. Combine the above functions as:
$\frac{A(x-1)^3+B(x+1)(x-1)^2+C(x+1)(x-1)+D(x+1)}{(x+1)(x-1)^3}=\frac{(A+B)x^3+(-3A-B+C)x^2+(3A-B+D)x+(-A+B-C+D)}{(x+1)(x-1)^3}$
Step 4. Compare the above result with the original expression to set up the following system of equations:
\begin{cases} A+B=0 \\ -3A-B+C=-2 \\3A-B+D=5 \\-A+B-C+D=-1 \end{cases}
Step 5. Use substitution $B=-A$ to get:
\begin{cases} -2A+C=-2 \\4A+D=5 \\-2A-C+D=-1 \end{cases}
Step 6. Use substitution $C=2A-2$ for the above and solve the remain equations to get: $A=1,B=-1,C=0,D=1$
Step 7. Write the final results as:
$\frac{1}{x+1}-\frac{1}{(x-1)}+\frac{1}{(x-1)^3}$
