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How do you x^2/(1-x^4) in partial fractions?

An expression to simplify:

The solution

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