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How do you (2x+3)/(x^2-9) in partial fractions?

An expression to simplify:

The solution

You have entered [src]
2*x + 3
-------
  2    
 x  - 9
2x+3x29\frac{2 x + 3}{x^{2} - 9}
(2*x + 3)/(x^2 - 9)
Fraction decomposition [src]
1/(2*(3 + x)) + 3/(2*(-3 + x))
12(x+3)+32(x3)\frac{1}{2 \left(x + 3\right)} + \frac{3}{2 \left(x - 3\right)}
    1           3     
--------- + ----------
2*(3 + x)   2*(-3 + x)
Combinatorics [src]
    3 + 2*x     
----------------
(-3 + x)*(3 + x)
2x+3(x3)(x+3)\frac{2 x + 3}{\left(x - 3\right) \left(x + 3\right)}
(3 + 2*x)/((-3 + x)*(3 + x))
Numerical answer [src]
(3.0 + 2.0*x)/(-9.0 + x^2)
(3.0 + 2.0*x)/(-9.0 + x^2)