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

Your first and second cases are equivalent. You need to understand complex numbers to be able to see that. The complex numbers, written $\mathbb{C}$ are of the form $x+iy$ where $x$ and $y$ are real numbers that you already know and $i$ is a special number with $i^2 = -1$. Usually, we have $i = \sqrt{-1}$. To see that the second case is the same as the first, notice that:
$$\frac{2x+3}{(x-1)(x^2+4)} \equiv \frac{A}{x-1}+\frac{B}{x-2i}+\frac{C}{x+2i}$$
where $A=1$, $B=-\frac{1}{2}-i\frac{1}{4}$ and $C=-\frac{1}{2}+i\frac{1}{4}$. The only reason you are taught the second form is that you don't know about $x\pm 2i$ and they recombine them:
$$\frac{B}{x-2i} + \frac{C}{x+2i} \equiv \frac{1-x}{x^2+4}$$
The third case isn't quite the same. But notice that the denominator is a cubic. When you multiply it out, you'll have an $x^3$ and lower powers. We couldn't get away with just two fraction because
$$\frac{A}{x-a}+\frac{B}{x-b} \equiv \frac{A(x-b)+B(x-a)}{(x-a)(x-b)}$$
and this only has a quadratic expression as the denominator.
