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How do you (x^3+2)/(x^3-4x) in partial fractions?

An expression to simplify:

The solution

You have entered [src]
  3     
 x  + 2 
--------
 3      
x  - 4*x
$$\frac{x^{3} + 2}{x^{3} - 4 x}$$
(x^3 + 2)/(x^3 - 4*x)
Fraction decomposition [src]
1 - 3/(4*(2 + x)) - 1/(2*x) + 5/(4*(-2 + x))
$$1 - \frac{3}{4 \left(x + 2\right)} + \frac{5}{4 \left(x - 2\right)} - \frac{1}{2 x}$$
        3        1        5     
1 - --------- - --- + ----------
    4*(2 + x)   2*x   4*(-2 + x)
General simplification [src]
        3  
   2 + x   
-----------
  /      2\
x*\-4 + x /
$$\frac{x^{3} + 2}{x \left(x^{2} - 4\right)}$$
(2 + x^3)/(x*(-4 + x^2))
Numerical answer [src]
(2.0 + x^3)/(x^3 - 4.0*x)
(2.0 + x^3)/(x^3 - 4.0*x)
Combinatorics [src]
           3      
      2 + x       
------------------
x*(-2 + x)*(2 + x)
$$\frac{x^{3} + 2}{x \left(x - 2\right) \left(x + 2\right)}$$
(2 + x^3)/(x*(-2 + x)*(2 + x))
Combining rational expressions [src]
        3  
   2 + x   
-----------
  /      2\
x*\-4 + x /
$$\frac{x^{3} + 2}{x \left(x^{2} - 4\right)}$$
(2 + x^3)/(x*(-4 + x^2))
Common denominator [src]
    2 + 4*x 
1 + --------
     3      
    x  - 4*x
$$\frac{4 x + 2}{x^{3} - 4 x} + 1$$
1 + (2 + 4*x)/(x^3 - 4*x)