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

It comes from two facts for polynomials $f(x)$ and $g(x)$:

  
*
  
*We can find $a(x)$ and $b(x)$ such that $\gcd(f(x),g(x))=a(x)f(x)+b(x)g(x)$.
  
*We can write $f(x)=q(x)g(x)+r(x)$, where $\deg(r)<\deg(g)$.

For your second example, we can use fact 1 to write $1=a(x)(x-1)+b(x)(x^2+4)$. Multiplying by $2x+3$ yields:
$$2x+3=[(2x+3)a(x)](x-1)+[(2x+3)b(x)](x^2+4)$$
If $\deg((2x+3)a(x))\ge \deg(x^2+4)=2$, then we can use fact 2 to write $(2x+3)a(x)=q(x)(x^2+4)+r(x)$, where $\deg(r)<2$. Then:
$$2x+3=r(x)(x-1)+[q(x)(x-1)+(2x+3)b(x)](x^2+4)$$
Comparing degrees, we see that:
$$2x+3=r(x)(x-1)+s(x)(x^2+4)$$
where $\deg(r)<2$, and $\deg(s)<1$. Then:
$$\frac{2x+3}{(x-1)(x^2+4)}=\frac{r(x)(x-1)+s(x)(x^2+4)}{(x-1)(x^2+4)}=\frac{r(x)}{x^2+4}+\frac{s(x)}{x-1}=\frac{Ax+B}{x^2+4}+\frac{C}{x-1}$$
