<s>[INST] Derivation of the general forms of partial fractions I'm learning about partial fractions, and I've been told of 3 types or "forms" that they can take
(1) If the denominator of the fraction has linear factors:
$${5 \over {(x - 2)(x + 3)}} \equiv {A \over {x - 2}} + {B \over {x + 3}}$$
(2) If the denominator of the fraction has quadratic factors that don't factorise:
$${{2x + 3} \over {(x - 1)({x^2} + 4)}} \equiv {A \over {x - 1}} + {{Bx + C} \over {{x^2} + 4}}$$
(3) If the denominator has a factor that repeats:
$${{5x + 3} \over {(x - 2){{(x + 3)}^2}}} \equiv {A \over {x - 2}} + {B \over {x + 3}} + {C \over {{{(x + 3)}^2}}}$$

I'd appreciate it if someone could explain to me how these equivalence relationships are derived, essentially how do they take the forms that they do? I sort of understand the first one, but the other two I don't.
Thank you
 [/INST] To look at your third problem, consider the following example: $${{x^2 + 4} \over {(x)(x + 2){{(3x-2)}}}} = {A \over {x}} + {B \over {x + 2}} + {C \over {{{(3x-2)}}}}$$
Next, set the numerators equal. $$x^2+4 = A(x+2)(3x-2) + Bx(3x-2) + Cx(x+2)$$ Now pick few x values and find the constants. For $x=0,$ $4=A(2)(-2).$ So $A=-1$. Following the same procedure for $x=-2$ and $x=2/3,$ we find that $B = 1/2$ and $C = 5/2.$ Substituting our values give us $${{x^2 + 4} \over {(x)(x + 2){{(3x-2)}}}} = {1 \over {x}} + {1/2 \over {x + 2}} + {5/2 \over {{{(3x-2)}}}}$$
This is not an answer to your question but to show you how its usually done.
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