Running the addition backwards
Proper fraction — One in which the degree of the numerator is less than the degree of the denominator. Partial fractions apply only to proper fractions — anything else must be divided out first.
You already know how to combine 2/(x−1) + 3/(x+2) into the single fraction (5x + 1)/[(x−1)(x+2)]. Partial fractions is that process reversed: given the combined fraction, recover the pieces.
It looks like an odd thing to want until you meet integration, where ∫ 2/(x−1) dx is a standard log and ∫ (5x+1)/(x²+x−2) dx is not. Splitting first turns one hard integral into two easy ones, and the same is true for binomial expansions and for inverse Laplace transforms later on.
Check the degrees before you start
For (x² + 1)/(x² − 4) the numerator is not of lower degree, so the fraction is improper. Divide first: it becomes 1 + 5/(x² − 4), and only the remainder fraction gets decomposed. Skipping this step produces equations with no solution and wastes half the question.
The four cases, decided by the denominator
The shape of the decomposition is fixed entirely by how the denominator factorises. Write the correct shape down first, before finding a single constant — the marks for the shape are separate from the marks for the values.
| Denominator contains | Contribute |
|---|---|
| a distinct linear factor (x − a) | A / (x − a) |
| a repeated linear factor (x − a)² | A/(x − a) + B/(x − a)² |
| a repeated factor (x − a)³ | A/(x−a) + B/(x−a)² + C/(x−a)³ |
| an irreducible quadratic (x² + bx + c) | (Ax + B) / (x² + bx + c) |
Why a quadratic factor needs Ax + B on top
The numerator must be allowed to be one degree lower than its denominator. Over a linear factor that means a constant; over a quadratic it means a linear expression. Putting a bare constant over a quadratic gives too few unknowns, and the equations then have no solution — which is how you find out you got the shape wrong.
Finding the constants: the cover-up shortcut
The standard method is to multiply through by the denominator and compare coefficients. For distinct linear factors there is a much faster route: substitute the value that kills one bracket, and every unknown but one disappears.
Express (5x + 1) / [(x − 1)(x + 2)] in partial fractions.
- Two distinct linear factors, so the shape is
A/(x−1) + B/(x+2).Write the shape before anything else — it is worth a mark on its own. - Multiply through by
(x−1)(x+2):5x + 1 = A(x + 2) + B(x − 1).This identity must hold for every x, which is what lets you choose convenient values. - Put
x = 1:6 = 3A, soA = 2.x = 1 makes the B bracket zero, so B vanishes from the equation entirely. - Put
x = −2:−9 = −3B, soB = 3.Now the A bracket is the one that dies. - Answer:
2/(x−1) + 3/(x+2).Check by recombining: 2(x+2) + 3(x−1) = 5x + 1 ✓
2/(x − 1) + 3/(x + 2)
When substitution runs out
Convenient substitutions only reach the distinct linear factors. A repeated factor or a quadratic factor leaves constants that no substitution can isolate, and for those you compare coefficients.
The routine: after the substitutions you can do, expand both sides fully, then match the coefficient of each power of x. Matching the highest power and the constant term usually settles the rest in two lines.
Express (3x² + 2) / [(x − 1)(x² + 1)] in partial fractions.
x² + 1has no real factors, so the shape isA/(x−1) + (Bx + C)/(x² + 1).An irreducible quadratic takes a linear numerator.- Multiply up:
3x² + 2 = A(x² + 1) + (Bx + C)(x − 1). - Put
x = 1:5 = 2A, soA = 2.5.The one substitution available, because only one factor is linear. - Compare
x²coefficients:3 = A + B, soB = 0.5.The x² terms come from A(x²) and from Bx(x). - Compare constants:
2 = A − C, soC = 0.5.Constants come from A(1) and from C(−1). - Answer:
2.5/(x−1) + (0.5x + 0.5)/(x² + 1).Equivalently 5/[2(x−1)] + (x + 1)/[2(x² + 1)].
5/[2(x − 1)] + (x + 1)/[2(x² + 1)]
Each distinct factor of the denominator is a vertical asymptote of the original fraction, and each one becomes a separate term in the decomposition. Choose (x+1)/(x²−4) to see two of them at once.
Before you leave this chapter
- Only proper fractions decompose. If the numerator degree is not lower, divide first.
- Distinct linear factor → A/(x−a). Repeated → one term per power. Quadratic → (Ax+B) over it.
- Write the shape down before finding any constants; the shape carries its own marks.
- Substitute the value that kills a bracket to get each constant over a distinct linear factor.
- Compare coefficients for anything substitution cannot reach, then recombine to check.
Repeated quadratic factors, and a check that always works
The fourth case on the syllabus is a quadratic factor that appears more than once. It follows the same principle as a repeated linear factor: one term for each power, and each numerator one degree below its denominator.
So (x² + 1)² in the denominator contributes (Ax + B)/(x² + 1) + (Cx + D)/(x² + 1)² — four unknowns from one factor. Count your unknowns before you start: the total must equal the degree of the denominator, and if it does not, the shape is wrong.
Verify by substituting one convenient number
Once you have the constants, do not recombine the whole thing algebraically — it is slow. Instead substitute a single easy value such as x = 0 into both the original fraction and your decomposition. If the two numbers agree, the answer is almost certainly right; if they do not, you have found the error in ten seconds.