Topic 10 Partial fractions

If I have several fractions, I might add them together, e.g.

\[\begin{align} \frac{2}{x+1} + \frac{1}{x+2} &= \frac{2(x+2)}{(x+1)(x+2)} + \frac{(x+1)}{(x+1)(x+2)}\\ &= \frac{2(x+2) + (x+1)}{(x+1)(x+2)}\\ &= \frac{2x + 4 + x + 1}{(x+1)(x+2)}\\ &= \frac{3x + 5}{(x+1)(x+2)} \end{align}\]

This combined fraction is good for many things, but would not be easy to integrate, whereas the two original fractions were much simpler, and can be integrated relatively easily.

If you have a polynomial divided by a polynomial, the method of partial fractions allows you to split it into simpler fractions. We will only look at cases where the degree (the highest power of x) in the numerator is smaller than the degree of the denominator, although the method is easily extended to more cases.

As any polynomial can be factorised into the product of linear (\(ax+b\)) and quadratic (\(ax^2+bx+c\)) factors, these form the basis of the method. The next sections look at these cases, and then the case of repeated factors.

10.1 Linear factors

The simplest case is linear factors. The method is most easily explained by an example.

Let’s look at \(\frac{3x + 5}{(x+1)(x+2)}\). We want to split this into two fractions, which when added together give this. Let’s try:

\[\begin{equation} \frac{3x + 5}{(x+1)(x+2)} \equiv \frac{A}{x+1} + \frac{B}{x+2} \tag{10.1} \end{equation}\]

I.e. for each linear factor of the denominator of the fraction we started with, we’ve introduced a term consisting of a constant divided by that factor.

The next step is to find the unknown constants. Let’s treat it as an equation and multiply both sides by the denominator of the original fraction:

\[\begin{align} \frac{3x + 5}{(x+1)(x+2)} &= \frac{A}{x+1} + \frac{B}{x+2}\\ (x+1)(x+2)\frac{3x + 5}{(x+1)(x+2)} &= (x+1)(x+2)\frac{A}{x+1} + (x+1)(x+2)\frac{B}{x+2}\\ 3x + 5 &= (x+2)A + (x+1)B \end{align}\]

This has to be true for every value of x, so, as we have two unknowns, we can substitute any two values of x in to get two equations for those unknowns. In this linear case the substitutions are particularly simple. If we use \(x=-2\), A drops out completely, giving:

\[\begin{align} 3(-2) + 5 &= (-2+1)B\\ -1 &= -B\\ B &= 1 \end{align}\]

Similarly, putting \(x=-1\) in gives us

\[\begin{align} 3(-1) +5 &= (-1+2)A\\ A = 2 \end{align}\]

Putting the values of A and B back into equation (10.1) gives

\[\frac{3x + 5}{(x+1)(x+2)} \equiv \frac{2}{x+1} + \frac{1}{x+2}\]

10.2 Quadratic factors

This is not currently covered at A-level, but is here for completeness.

A quadratic factor has the form \(ax^2+bx+c\), e.g. \(3x^2+2x+2\). Suppose we have

\[ \frac{3x+2}{(3x^2+2x+2)(x+1)}\] As the quadratic term is not factorisable, we can’t write the bottom as three linear factors and use the results from the last section, instead we have to write

\[ \frac{3x+2}{(3x^2+2x+2)(x+1)} \equiv \frac{Ax+B}{3x^2+2x+2} + \frac{C}{x+1}\]

I.e. for a quadratic term in the denominator, we write a fraction with a linear term in the numerator. The method starts as before:

\[\begin{align} \frac{3x+2}{(3x^2+2x+2)(x+1)} &= \frac{Ax+B}{3x^2+2x+2} + \frac{C}{x+1}\\ 3x+2 &= (3x^2+2x+2)(x+1)\frac{Ax+B}{3x^2+2x+2} + (3x^2+2x+2)(x+1)\frac{C}{x+1}\\ 3x+2 &= (x+1)(Ax+B) + (3x^2+2x+2)C \end{align}\]

This must still be true for all values of x, so setting \(x=-1\) still makes sense for getting rid of the first term on the right hand side, which then allows us to get C, however there’s no other obvious choice for x. We could pick some values at random (0 is probably a good choice), however there is an alternative strategy.

If you think about \(x\), \(x^2\), they vary in very different ways, so the only way this expression can be true for every value of x is if the coefficients of each power of x are the same on both sides of the equation. This is very similar to what was done with the harmonic form of the sum of a cosine and a sine in the trigonometry section.

  • Comparing constants

\[2 = B + 2C\]

  • Comparing the coefficients of x

\[3 = A + B + 2C\]

  • Comparing the coefficients of \(x^2\)

\[0 = A + 3C\]

Solving these simultaneous equations gives \(A=1\), \(B=\frac{8}{3}\), \(C=-\frac{1}{3}\).

So the final answer is:

\[ \frac{3x+2}{(3x^2+2x+2)(x+1)} \equiv \frac{x+\frac{8}{3}}{3x^2+2x+2} - \frac{\frac{1}{3}}{x+1}\] Which can be tidied to give:

\[ \frac{3x+2}{(3x^2+2x+2)(x+1)} \equiv \frac{3x+8}{3(3x^2+2x+2)} - \frac{1}{3(x+1)}\]

Or even:

\[ \frac{3x+2}{(3x^2+2x+2)(x+1)} \equiv \frac{1}{3}\left(\frac{3x+8}{3x^2+2x+2} - \frac{1}{x+1}\right)\]

10.3 Repeated linear factors

Suppose we had

\[\frac{2x^2+5x+4}{(x+1)(x+2)^2}\] We have two linear factors here, but the second one is repeated. We could just expand the repeated factor as a quadratic and treat it as a quadratic factor, as in the previous section. There is, however, a better technique in this case.

Expand as

\[\frac{2x^2+5x+4}{(x+1)(x+2)^2} \equiv \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}\] Unlike in the previous section all the numerators are constants.

Proceed as before:

\[\begin{align} \frac{2x^2+5x+4}{(x+1)(x+2)^2} &= \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}\\ 2x^2+5x+4 &= (x+1)(x+2)^2\frac{A}{x+1} + (x+1)(x+2)^2\frac{B}{x+2} + (x+1)(x+2)^2\frac{C}{(x+2)^2}\\ 2x^2+5x+4 &= (x+2)^2A + (x+1)(x+2)B + (x+1)C \end{align}\]

Now either use three different values of x (e.g. 0, \(-1\) and \(-2\) would work well here), or compare coefficients

\[2x^2+5x+4 = (x^2+4x+4)A + (x^2+3x+2)B + (x+1)C\]

  • Constants:

\[4 = 4A + 2B + C\]

  • Coefficients of x:

\[5 = 4A + 3B + C\]

  • Coefficients of \(x^2\)

\[2 = A + B\] Solving these simultaneous equations gives \(A=1\), \(B=1\), \(C=-2\). So the final answer is:

\[\frac{2x^2+5x+4}{(x+1)(x+2)^2} \equiv \frac{1}{x+1} + \frac{1}{x+2} - \frac{2}{(x+2)^2}\]