Partial fractions: undoing a single fraction

Adding two fractions creates one fat fraction:

$\frac{2}{x+1} + \frac{3}{2x-1} = \frac{7x+1}{(x+1)(2x-1)}$

Partial fractions run that backwards. Given the fat fraction, you recover the simple pieces.

The reason to care: simple fractions are far easier to differentiate, integrate and expand than one big fraction.

One condition comes first. The numerator must have lower degree than the denominator. Such a fraction is called a proper fraction.

If it does not, divide first and keep the polynomial part separate.

The symbol $\equiv$ matters. An equation such as $x = 3$ is true only sometimes; an identity is true always.

Match the form to the denominator. That choice is worth a mark before any algebra.

Three rules sit behind the table:

1. One constant per linear factor. Two linear factors need $A$ and $B$.
2. A repeated factor gets a term for every power up to the repeat. So $(cx+d)^2$ needs both $\dfrac{B}{cx+d}$ and $\dfrac{C}{(cx+d)^2}$.
3. An irreducible quadratic needs a linear numerator $Bx + C$, because a constant cannot cover both an $x$ part and a constant part.

1. Write the identity with unknown constants on top.
2. Multiply through by the whole denominator. The fractions disappear and you are left with a polynomial identity.
3. Substitute the root of each factor. Each substitution kills one unknown, so the constants come out one at a time.
4. State the answer and check it with one convenient value.

The forgotten step is step 2. If you clear only part of the denominator, the identity is unbalanced and the constants come out wrong.