How to find out the partial fraction decomposition form (or pattern ) of a rational function $\dfrac{P(X)}{Q(x)}$ , for example we know that the partial fraction decomposition form of $$\frac{x^{2}+3x+1}{(x-1)^{2}(x+3)},$$ would be something like this : $$\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+3}, \qquad A,B,C \in \mathbb{R}$$
but how did we "guess" that pattern ? and how to guess it in general case ?
We are just doing polynomial long division on the numberator, and collecting the remainders.
$\frac {p(x)}{(x+3)(x-1)^2} = \frac {q(x) (x+3) + r}{(x+3)(x-1)^2} = \frac {r}{(x+3)} + \frac {q(x)+r}{(x-1)^2}$
$r$ will always have degree less than its denominator. $q(x)$ will have degree less than $p(x).$
Next we do polynomial division to $q(x)+r$ breaking into $\frac {r_2(x-1) + r_3}{(x-1)^2} = \frac {r_2}{x-1} + \frac {r_3}{(x-1)^2}$
First start off with the following:
$\frac{x^2+3x+1}{(x-1)^2(x+3)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+3}$
Clear the denominator
$x^2+3x+1=A(x-1)(x+3)+B(x+3)+C(x-1)^2$
Expand and Collect:
$x^2+3x+1=A(x^2+2x-3)+B(x+3)+C(x^2-2x+1)$
$x^2+3x+1=(A+C)x^2+(2A+B-2C)x+(3B+C-3A)$
Use the following Equations:
$A+C=1$
$2A+B-2C=3$
$3B+C-3A=1$
Add up the equations to get $B=5/4$. This one is going to have fractions.
If anything is unclear, I can go into more depth.
For details on picking what would go in the numerator, like when to use an $Ax+B$, go here.
Resolve the denominator into factors, break them,and in numerators of each write an arbitrary polynomial expression with degree one less than the denominator. Find it by comparison.
