Where does the $Ax+B$ come from when solving partial fractions with a quadratic factor? I'm only just now learning about partial fractions to solve integrals.  I've read that 
$$\frac{f(x)}{(px^2+qx+r)(x-a)}\equiv \frac{Ax+B}{(px^2+qx+r)}+\frac{C}{(x-a)}$$
I haven't figured out how it was decided that $Ax+B$ ought to be used and why it works.  Can someone explain it?
 A: It's because if $g(x)$ is a polynomial of degree greater than $1$, we can always rewrite $$\frac{g(x)}{px^2+qx+r}=h(x)+\frac{Cx+D}{px^2+qx+r}$$ for some polynomial $h$ and constants $C,D$.
Also, it's worth noting that your statement (as written) is only true when $f$ is a polynomial of degree at most $2$.
A: If $f,g,h$ are polynomials, you want to rewrite ${f(x)\over g(x)h(x)}$.
Of course we assume that $g$ and $h$ have no roots in common. Then the extended Euclidean algorithm produces polynomials $u,v$ such that $u(x)g(x)+v(x)h(x)=1$.
After multiplication with $f$ this becomes $f(x)u(x)g(x)+f(x)v(x)h(x)=f(x)$.
Finally after dividing by $g(x)h(x)$, we have
$${f(x)\over g(x)h(x)} = {f(x)u(x)\over h(x)}+{f(x)v(x)\over g(x)}.$$
We can perform polynomial division with remainder, i.e. write $f(x)u(x) = q_1(x)\cdot h(x)+r_1(x)$ and $f(x)u(x) = q_2(x)\cdot g(x)+r_2(x)$. The process of division with remainder can guarantee $\deg r_1<\deg h$ and $\deg r_2<\deg g$, but not more.
Thus we end up with 
$${f(x)\over g(x)h(x)} = q_1(x)+q_2(x) + {r_1(x)\over h(x)}+{r_2(x)\over g(x)}$$
with $\deg r_1<\deg h$ and $\deg r_2<\deg g$. To allow all possible options for $r_1$ and $r_2$ we must use accordingly many powers of $x$ and unknown coefficients.
A: Well, the denominator is a polynomial of degree three. I'll assume that $f(x)$ is a polynomial of degree $\leq 2$. Actually, the case $f(x)=1$ already explains the philosophy.
The numerator of the first fraction will be multiplied by a linear polynomial, so that its degree must be one, if you hope to compete with the second fraction. Assume that
$$
\frac{1}{(px^2+qx+r)(x-a)}=\frac{E(x)}{px^2+qx+r}+\frac{C}{x-a},
$$
then
$$
1=E(x)(x-a)+C(px^2+qx+r),
$$
and you definitely need a term like $x^2$ on the right-hand side to cancel $Cpx^2$. That's why you need at least a polynomial $E(x)$ of the first order.
