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How do you x^3/(x^4-1) in partial fractions?

An expression to simplify:

The solution

You have entered [src]
   3  
  x   
------
 4    
x  - 1
$$\frac{x^{3}}{x^{4} - 1}$$
x^3/(x^4 - 1)
Fraction decomposition [src]
1/(4*(1 + x)) + 1/(4*(-1 + x)) + x/(2*(1 + x^2))
$$\frac{x}{2 \left(x^{2} + 1\right)} + \frac{1}{4 \left(x + 1\right)} + \frac{1}{4 \left(x - 1\right)}$$
    1           1            x     
--------- + ---------- + ----------
4*(1 + x)   4*(-1 + x)     /     2\
                         2*\1 + x /
Combinatorics [src]
             3           
            x            
-------------------------
        /     2\         
(1 + x)*\1 + x /*(-1 + x)
$$\frac{x^{3}}{\left(x - 1\right) \left(x + 1\right) \left(x^{2} + 1\right)}$$
x^3/((1 + x)*(1 + x^2)*(-1 + x))
Numerical answer [src]
x^3/(-1.0 + x^4)
x^3/(-1.0 + x^4)