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Find A and B so that: ((1/(x-2)(x+3))=((A/(x-2))+ (B/(x+3))
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basic partial fractions
multiply both sides by the denominator of the the LHS so you get 1 = A(x+3) + B ( x-2)
now sub in brackets which makes the brackets zero sub x=-3 --> 1 = -5B --> B= -1/5 sub x=2 --> 1 = 5A --> A= 1/5 A=1/5, B=-1/5
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