**Partial fraction expansion** or a partial fraction decomposition is a process in which we can separate one complicated fraction into a sum of few smaller ones. This is a process that has many applications – most importantly in integration.

Let’s say you have a rational expression $ f(x) = \frac{1}{x^2 – 1}$. You know that $ x^2 – 1 = (x – 1)(x + 1)$. Now you want to know how you can write this fraction in a form of sum of two different fractions with denominators $ x – 1$ and $ x + 1$.

First you write your fraction in the following way.

$\frac{1}{(x – 1)(x + 1)} = \frac{A}{x – 1} + \frac{B}{x + 1}$

Now you multiply whole equation with the common denominator: $ (x – 1)(x + 1)$

$ 1 = (x + 1)A + (x – 1)B$

Next step is to get rid of the brackets.

$ 1 = xA + A + xB – B$

The easiest way to solve this is by using method of contrary coefficients- we’ll simply add these two equations.

$ 2A = 1 \rightarrow A = \frac{1}{2}$

Since we got one unknown we can easily find the other one.

$ B = – A\rightarrow B = – \frac{1}{2}$

When we put these values we got into the original equation

$\frac{1}{(x – 1)(x + 1)} = \frac{A}{x – 1} + \frac{B}{x + 1}$

We get:

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

$\frac{1}{(x – 1)(x + 1)} = \frac{1}{2(x – 1)} + \frac{1}{2(x + 1)}$

And that is it. We can easily check our solution by doing the inverse procedure.

$\frac{1}{2(x – 1)} + \frac{1}{2(x + 1)} $

$= \frac{x + 1 – x + 1}{2(x – 1)(x + 1)} $

$= \frac{2}{2(x – 1)(x + 1)} $

$= \frac{1}{(x – 1)(x + 1)}$

What are the things you should pay attention to? First of all, the first example we made is a rational expression where we could factorize our denominator into product of linear factors. The procedure is quite different when you can’t factorize them like that.

One example of an expression like that is $\frac{2x + 5}{x^3 + 2x}$.

First, we can extract $x$ from the denominator.

$\frac{2x + 5}{x^3 + 2x} = \frac{2x + 5}{x(x^2 + 2)}$

This is how you should do it. For every linear member you put a fraction with numerator that has only one letter. When the powers of the expression rise, so does your numerator. For quadratic expression you put $ Ax + C$, for cube expression you put $ Ax^2 + Bx + C$ and so on. Generally you always put in for the numerator polynomial with degree that is for $1$ smaller than the expression you are trying to separate from the given fraction.

$\frac{2x + 5}{x(x^2 + 2)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2}$

From this step everything is the same. Now we have to multiply this whole equation with the denominator of the given fraction.

$ 2x + 5 = (x^2 + 2)A + x(Bx + C)$

Next step is to get rid of the brackets and, again, equalize coefficients that multiply the unknowns with same powers.

$ 2x + 5 = x^2A + 2A + x^2B + xC$

$A + B = 0$, $ C = 2$, $2A = 5$

$ A = \frac{5}{2}$, $ B = – \frac{5}{2}$, $ C = 2$

$\frac{2x + 5}{x(x^2 + 2)} = \frac{\frac{5}{2}}{x} + \frac{\frac{-5}{2} + 2}{x^2 + 2} = \frac{5}{2x} + \frac{\frac{-5x + 4}{2}}{x^2 + 2} = \frac{5}{2x} + \frac{-5x + 4}{2(x^2 + 2)}$

One more thing can appear in these kind of tasks that is solved differently than these examples before. This is the case when you can write the given denominator as the product of linear polynomials, but two or more of them are equal.

For example $\frac{x^3 + x^2 + 1}{(x + 1)(x – 1)^3}$.

Since we can express denominator as the product of finitely many linear polynomials, all expressions in the separate fractions’ numerators will be linear. Since we have third degree polynomial we have to go through all the powers to get to the third degree. This means that for the polynomial $ (x – 1)^3$ we’ll have three different fractions – those whose denominators are $ (x – 1)$, $ (x – 1)^2$ and $ (x – 1)^3$.

$ x^3 + x^2 + 1 = \frac{A}{x + 1} + \frac{B}{x – 1} + \frac{C}{(x – 1)^2} + \frac{D}{(x – 1)^3}$

$ x^3 + x^2 + 1 = (x^3 – 3x^2 + 3x – 1)A + (x^3 – x^2 – x + 1)B + (x^2 – 1)C + (x + 1)D$

$ 1 = A + B$

$ 1 = – 3A – B + C$

$ 0 = 3A – B + D$

$ 1 = – A + B – C + D$

$ B = 1 – A$

$ 1 = 3A – 1 + A + C \rightarrow 2 = 4A + C$

$ 0 = 3A – 1 + A + D \rightarrow 1 = 4A + D$

$ 1 = – A + 1 – A – C + D \rightarrow 0 = – 2A – C + D$

$ 2A = D – C$

$ 2 = 2D – 2C + C \rightarrow 2 = 2D – C$

$ 1 = 2D – 2C + D \rightarrow 1 = 3D – 2C$

$ D = 3, C = 4, A = – \frac{1}{2}, B = – \frac{3}{2}$

$\frac{x^3 + x^2 + 1}{(x + 1)(x – 1)^3} = \frac{-1}{2(x + 1)} + \frac{3}{2(x – 1)} + \frac{4}{(x – 1)^2} + \frac{3}{(x – 1)^3}$