let 6/((x+3)(x-3)
= A/(x+3) + B/(x-3) = ( A(x-3) + B(x+3) )/((x+3)(x-3) )
then A(x-3) + B(x+3) = 6
let x = 3 ---> A(0) + 6B = 6 or B = 1
let x = -3 ----> -6A + 0 = 6 or A = -1
thus:
6/((x+3)(x-3)) = -1/(x+3) + 1/(x-3)
6/(x+3)(x-3)
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