Partial Fractions

A2 · P3 · 13 min

Adding fractions is easy: $\dfrac{2}{x + 2} + \dfrac{3}{2x - 1}$ combines to $\dfrac{7x + 4}{(x + 2)(2x - 1)}$ . Partial fractions run that process backwards, splitting one complicated fraction into a sum of simple ones. The reason to bother is that the simple pieces can be integrated (each gives a logarithm) and expanded as binomial series, while the combined fraction cannot. On P3 partial fractions are almost always part (a) of a longer question, followed by an integral or a series expansion, so the constants must be right.

The idea

Every simple fraction you get from a linear factor looks like $\dfrac{A}{ax + b}$ , where $A$ is a constant. When you add a few of these, the denominator is the product of the factors and the numerator is some polynomial of lower degree. So going backwards, you guess the shape of the answer from the factors of the denominator, then find the constants.

Definition

A rational function $\dfrac{p(x)}{q(x)}$ is proper if the degree of the numerator is less than the degree of the denominator, and improper otherwise. Only proper fractions can be written directly as a sum of partial fractions; an improper one must first have a polynomial part taken out.

The forms you must recall

The syllabus restricts denominators to three types. For each, the number of unknown constants equals the degree of the denominator. That is the check that you have the right form.

Key result

Denominator	Partial fractions
$(ax + b)(cx + d)(ex + f)$	$\dfrac{A}{ax + b} + \dfrac{B}{cx + d} + \dfrac{C}{ex + f}$
$(ax + b)(cx + d)^2$	$\dfrac{A}{ax + b} + \dfrac{B}{cx + d} + \dfrac{C}{(cx + d)^2}$
$(ax + b)(cx^2 + d)$	$\dfrac{A}{ax + b} + \dfrac{Bx + C}{cx^2 + d}$

(The first also covers two distinct linear factors, with just $A$ and $B$ .) If the fraction is improper with numerator and denominator of the same degree, add a constant term: for example $\dfrac{p(x)}{(ax + b)(cx + d)} = K + \dfrac{A}{ax + b} + \dfrac{B}{cx + d}$ .

Why these shapes?

Repeated factor. $\dfrac{C}{(cx + d)^2}$ alone is not enough, because combining it with the other terms cannot produce every possible numerator. Including both $\dfrac{B}{cx + d}$ and $\dfrac{C}{(cx + d)^2}$ gives three unknowns for a cubic denominator.
Quadratic factor. Over an irreducible quadratic $cx^2 + d$ , the numerator can be any polynomial of degree less than $2$ , so it is $Bx + C$ , not just a constant. Again three unknowns.

Finding partial fractions

Check the fraction is proper. If the degrees are equal, include a constant term $K$ (or divide first).
Factorise the denominator fully, and write the correct form with unknown constants.
Multiply both sides by the full denominator to get an identity between polynomials.
Substitute the values of $x$ that make each linear factor zero. Each gives one constant immediately.
For any constants left, compare coefficients, usually of the highest power and the constant term, or substitute another easy value such as $x = 0$ .
Write the final answer in full, and check by substituting a spare value of $x$ into both sides.

Distinct linear factors

Two linear factors

Express $\dfrac{7x + 4}{(x + 2)(2x - 1)}$ in partial fractions.

Solution

\frac{7x + 4}{(x + 2)(2x - 1)} \equiv \frac{A}{x + 2} + \frac{B}{2x - 1}

Multiply through by $(x + 2)(2x - 1)$ :

7x + 4 \equiv A(2x - 1) + B(x + 2).

$x = -2$ : $-14 + 4 = A(-5)$ , so $A = 2$ .

$x = \tfrac{1}{2}$ : $\tfrac{7}{2} + 4 = B\left(\tfrac{5}{2}\right)$ , so $\tfrac{15}{2} = \tfrac{5}{2}B$ and $B = 3$ .

\frac{7x + 4}{(x + 2)(2x - 1)} = \frac{2}{x + 2} + \frac{3}{2x - 1}.

Check at $x = 0$ : left $\dfrac{4}{-2} = -2$ ; right $1 - 3 = -2$ .

Three linear factors

Express $\dfrac{5x^2 + 2x + 2}{(x - 1)(x + 2)(2x + 1)}$ in partial fractions.

Solution

5x^2 + 2x + 2 \equiv A(x + 2)(2x + 1) + B(x - 1)(2x + 1) + C(x - 1)(x + 2)

$x = 1$ : $5 + 2 + 2 = A(3)(3)$ , so $9 = 9A$ , $A = 1$ .

$x = -2$ : $20 - 4 + 2 = B(-3)(-3)$ , so $18 = 9B$ , $B = 2$ .

$x = -\tfrac{1}{2}$ : $\tfrac{5}{4} - 1 + 2 = C\left(-\tfrac{3}{2}\right)\left(\tfrac{3}{2}\right)$ , so $\tfrac{9}{4} = -\tfrac{9}{4}C$ , $C = -1$ .

\frac{5x^2 + 2x + 2}{(x - 1)(x + 2)(2x + 1)} = \frac{1}{x - 1} + \frac{2}{x + 2} - \frac{1}{2x + 1}.

Check the $x^2$ coefficients: $2A + 2B + C = 2 + 4 - 1 = 5$ .

Tip

The cover-up rule is a fast way to get the constant over a non-repeated linear factor. To find $A$ over $(x - 1)$ in the example above, cover up $(x - 1)$ in the original fraction and substitute $x = 1$ into what is left: $\dfrac{5 + 2 + 2}{(3)(3)} = 1$ . It is exactly the substitution method done in your head. It does not work for the $\dfrac{B}{cx + d}$ term of a repeated factor, or for $Bx + C$ .

A repeated linear factor

Repeated factor (non-unit coefficient)

Express $\dfrac{4x^2 - x + 4}{(2x - 1)(x + 1)^2}$ in partial fractions.

Solution

\frac{4x^2 - x + 4}{(2x - 1)(x + 1)^2} \equiv \frac{A}{2x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}

Multiply by $(2x - 1)(x + 1)^2$ :

4x^2 - x + 4 \equiv A(x + 1)^2 + B(2x - 1)(x + 1) + C(2x - 1).

$x = -1$ : $4 + 1 + 4 = C(-3)$ , so $C = -3$ .

$x = \tfrac{1}{2}$ : $1 - \tfrac{1}{2} + 4 = A\left(\tfrac{3}{2}\right)^2$ , so $\tfrac{9}{2} = \tfrac{9}{4}A$ , $A = 2$ .

No value of $x$ isolates $B$ , so compare coefficients of $x^2$ : $4 = A + 2B$ , so $B = 1$ .

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

Check with the constant terms: $A - B - C = 2 - 1 + 3 = 4$ .

Watch out

In the identity, the term with $B$ is $B(2x - 1)(x + 1)$ , not $B(2x - 1)(x + 1)^2$ . Each term is multiplied by whatever is missing from its own denominator. For $\dfrac{B}{x + 1}$ the missing factors are $(2x - 1)$ and one $(x + 1)$ .

A quadratic factor

Linear and quadratic factor

Express $\dfrac{5x^2 + x + 6}{(x - 1)(x^2 + 2)}$ in partial fractions.

Solution

\frac{5x^2 + x + 6}{(x - 1)(x^2 + 2)} \equiv \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 2}

5x^2 + x + 6 \equiv A(x^2 + 2) + (Bx + C)(x - 1).

$x = 1$ : $12 = 3A$ , so $A = 4$ .

$x^2$ coefficients: $5 = A + B$ , so $B = 1$ .

Constant terms: $6 = 2A - C$ , so $C = 8 - 6 = 2$ .

\frac{5x^2 + x + 6}{(x - 1)(x^2 + 2)} = \frac{4}{x - 1} + \frac{x + 2}{x^2 + 2}.

Check the $x$ coefficients: $-B + C = -1 + 2 = 1$ .

Improper fractions: equal degrees

The syllabus includes fractions where the numerator has the same degree as the denominator (it excludes cases where the numerator's degree is higher). There are two equally good methods.

Include a constant. Write the answer as $K + (\text{partial fractions})$ . The constant $K$ is the ratio of the leading coefficients, which you can see by comparing the highest power.
Divide first. Divide the numerator by the expanded denominator, then split the proper remainder fraction.

Equal degrees, two linear factors

Express $\dfrac{2x^2 + 5x + 1}{(x + 1)(x + 2)}$ in partial fractions.

Solution

Numerator and denominator both have degree $2$ , so include a constant:

\frac{2x^2 + 5x + 1}{(x + 1)(x + 2)} \equiv K + \frac{A}{x + 1} + \frac{B}{x + 2}

2x^2 + 5x + 1 \equiv K(x + 1)(x + 2) + A(x + 2) + B(x + 1).

$x^2$ coefficients: $K = 2$ .

$x = -1$ : $2 - 5 + 1 = A(1)$ , so $A = -2$ .

$x = -2$ : $8 - 10 + 1 = B(-1)$ , so $B = 1$ .

\frac{2x^2 + 5x + 1}{(x + 1)(x + 2)} = 2 - \frac{2}{x + 1} + \frac{1}{x + 2}.

By division instead: $2x^2 + 5x + 1 = 2(x^2 + 3x + 2) + (-x - 3)$ , and $\dfrac{-x - 3}{(x + 1)(x + 2)} = -\dfrac{2}{x + 1} + \dfrac{1}{x + 2}$ , the same answer.

Check at $x = 0$ : left $\tfrac{1}{2}$ ; right $2 - 2 + \tfrac{1}{2} = \tfrac{1}{2}$ .

Equal degrees with a repeated factor

Express $\dfrac{2x^3 + 6x^2 - 2x - 2}{(x - 1)(x + 1)^2}$ in partial fractions.

Solution

Both have degree $3$ , and the ratio of leading coefficients is $2$ :

2x^3 + 6x^2 - 2x - 2 \equiv 2(x - 1)(x + 1)^2 + A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1).

$x = 1$ : $2 + 6 - 2 - 2 = 4A$ , so $A = 1$ .

$x = -1$ : $-2 + 6 + 2 - 2 = -2C$ , so $C = -2$ .

$x = 0$ : $-2 = 2(-1)(1) + A - B - C = -2 + 1 - B + 2$ , so $B = 3$ .

\frac{2x^3 + 6x^2 - 2x - 2}{(x - 1)(x + 1)^2} = 2 + \frac{1}{x - 1} + \frac{3}{x + 1} - \frac{2}{(x + 1)^2}.

An exam-style question with integration

Partial fractions, then a definite integral

Let $f(x) = \dfrac{8 + 6x - x^2}{(2 - x)(4 + x^2)}$ .

(a) Express $f(x)$ in partial fractions.

(b) Hence show that $\displaystyle\int_0^1 f(x)\,dx = \ln\left(\frac{5\sqrt{5}}{2}\right)$ .

Solution

(a) $f(x) \equiv \dfrac{A}{2 - x} + \dfrac{Bx + C}{4 + x^2}$ , so

8 + 6x - x^2 \equiv A(4 + x^2) + (Bx + C)(2 - x).

$x = 2$ : $8 + 12 - 4 = 8A$ , so $16 = 8A$ , $A = 2$ .

$x^2$ coefficients: $-1 = A - B$ , so $B = 3$ .

Constants: $8 = 4A + 2C = 8 + 2C$ , so $C = 0$ .

f(x) = \frac{2}{2 - x} + \frac{3x}{4 + x^2}.

(b) Integrate each term (see integration with partial fractions):

\int \frac{2}{2 - x}\,dx = -2\ln(2 - x), \qquad \int \frac{3x}{4 + x^2}\,dx = \tfrac{3}{2}\ln(4 + x^2).

\int_0^1 f(x)\,dx = \Big[-2\ln(2 - x) + \tfrac{3}{2}\ln(4 + x^2)\Big]_0^1 = \left(0 + \tfrac{3}{2}\ln 5\right) - \left(-2\ln 2 + \tfrac{3}{2}\ln 4\right).

Since $\tfrac{3}{2}\ln 4 = 3\ln 2$ , this is $\tfrac{3}{2}\ln 5 - \ln 2 = \ln\dfrac{5^{3/2}}{2} = \ln\left(\dfrac{5\sqrt{5}}{2}\right)$ .

Tip

A constant of zero, as with $C = 0$ above, is perfectly normal. It is the question's way of making the integral neat. If you get a zero, check it, but do not assume it is an error.

Common mistakes

Watch out

Wrong form for a repeated factor. Writing $\dfrac{A}{2x - 1} + \dfrac{C}{(x + 1)^2}$ without the $\dfrac{B}{x + 1}$ term leaves too few constants, and the coefficients will not match. Count: three constants for a cubic denominator.

Watch out

A constant numerator over a quadratic. Over $x^2 + 2$ the numerator must be $Bx + C$ . Writing only $\dfrac{B}{x^2 + 2}$ fails whenever the answer needs an $x$ term.

Watch out

Not spotting an improper fraction. If the top and bottom have the same degree and you omit the constant $K$ , the equations for $A$ , $B$ , $C$ become inconsistent. Always compare degrees first.

Watch out

Factorising the quadratic $cx^2 + d$ when it does not factorise. $x^2 + 4$ has no real factors. Only $cx^2 - d$ (with $c, d > 0$ ) splits, as a difference of squares, and then you have linear factors instead.

Exam tip

The form is often worth a mark in itself, so write it out with letters before substituting.
Show each substitution: " $x = -1$ : $9 = -3C$ , so $C = -3$ ". A list of answers with no working can lose method marks if any value is wrong.
Give the final answer as an explicit sum of fractions; do not leave it as " $A = 2$ , $B = 1$ , $C = -3$ ".
Check with a spare value of $x$ . Every later part of the question depends on these constants, so a 20-second check protects many marks.
Keep fractions exact: $\dfrac{9}{5(x + 1)}$ , not $\dfrac{1.8}{x + 1}$ .

Summary

Partial fractions reverse the addition of fractions; they make integration and series expansion possible.
Linear factor $(ax + b)$ : term $\dfrac{A}{ax + b}$ . Repeated $(cx + d)^2$ : terms $\dfrac{B}{cx + d} + \dfrac{C}{(cx + d)^2}$ . Quadratic $(cx^2 + d)$ : term $\dfrac{Bx + C}{cx^2 + d}$ .
The number of constants equals the degree of the denominator.
Multiply through, substitute the roots of the linear factors, then compare coefficients for whatever is left.
If numerator and denominator have equal degree, include a constant term (equal to the ratio of the leading coefficients) or divide first.
Check by substituting a spare value of $x$ .

Practice

Question

Express $\dfrac{5x + 5}{(x - 2)(x + 3)}$ in partial fractions.
Express $\dfrac{2x^2 + 6x - 2}{x(x - 1)(x + 2)}$ in partial fractions.
Express $\dfrac{4x^2 + 5x + 3}{(2x + 1)(x + 1)^2}$ in partial fractions.
Express $\dfrac{5x^2 - 3x + 1}{(2x + 1)(x^2 + 1)}$ in partial fractions.
Express $\dfrac{x^2 + 3x + 5}{(x + 1)(x + 2)}$ in partial fractions.
Express $\dfrac{2x^2 - x + 6}{(x + 1)(x^2 + 4)}$ in partial fractions.
Express $\dfrac{11 - x - x^2}{(2 + x)(1 - x)^2}$ in partial fractions.
Let $f(x) = \dfrac{2x^2 + 6x - 2}{x(x - 1)(x + 2)}$ . Using your answer to question 2, show that $\displaystyle\int_2^3 f(x)\,dx = \ln\frac{24}{5}$ .

Answers

$5x + 5 \equiv A(x + 3) + B(x - 2)$ . $x = 2$ : $15 = 5A$ , $A = 3$ . $x = -3$ : $-10 = -5B$ , $B = 2$ . Answer $\dfrac{3}{x - 2} + \dfrac{2}{x + 3}$ .
$2x^2 + 6x - 2 \equiv A(x - 1)(x + 2) + Bx(x + 2) + Cx(x - 1)$ . $x = 0$ : $-2 = -2A$ , $A = 1$ . $x = 1$ : $6 = 3B$ , $B = 2$ . $x = -2$ : $8 - 12 - 2 = -6 = 6C$ , $C = -1$ . Answer $\dfrac{1}{x} + \dfrac{2}{x - 1} - \dfrac{1}{x + 2}$ .
$4x^2 + 5x + 3 \equiv A(x + 1)^2 + B(2x + 1)(x + 1) + C(2x + 1)$ . $x = -1$ : $2 = -C$ , $C = -2$ . $x = -\tfrac{1}{2}$ : $1 - \tfrac{5}{2} + 3 = \tfrac{3}{2} = \tfrac{1}{4}A$ , $A = 6$ . $x^2$ : $4 = A + 2B$ , $B = -1$ . Answer $\dfrac{6}{2x + 1} - \dfrac{1}{x + 1} - \dfrac{2}{(x + 1)^2}$ .
$5x^2 - 3x + 1 \equiv A(x^2 + 1) + (Bx + C)(2x + 1)$ . $x = -\tfrac{1}{2}$ : $\tfrac{5}{4} + \tfrac{3}{2} + 1 = \tfrac{15}{4} = \tfrac{5}{4}A$ , $A = 3$ . $x^2$ : $5 = A + 2B$ , $B = 1$ . Constants: $1 = A + C$ , $C = -2$ . Answer $\dfrac{3}{2x + 1} + \dfrac{x - 2}{x^2 + 1}$ .
Equal degrees: $x^2 + 3x + 5 \equiv K(x + 1)(x + 2) + A(x + 2) + B(x + 1)$ . $K = 1$ . $x = -1$ : $3 = A$ . $x = -2$ : $3 = -B$ , $B = -3$ . Answer $1 + \dfrac{3}{x + 1} - \dfrac{3}{x + 2}$ .
$2x^2 - x + 6 \equiv A(x^2 + 4) + (Bx + C)(x + 1)$ . $x = -1$ : $9 = 5A$ , $A = \tfrac{9}{5}$ . $x^2$ : $2 = A + B$ , $B = \tfrac{1}{5}$ . Constants: $6 = 4A + C$ , $C = 6 - \tfrac{36}{5} = -\tfrac{6}{5}$ . Answer $\dfrac{9}{5(x + 1)} + \dfrac{x - 6}{5(x^2 + 4)}$ .
$11 - x - x^2 \equiv A(1 - x)^2 + B(2 + x)(1 - x) + C(2 + x)$ . $x = 1$ : $9 = 3C$ , $C = 3$ . $x = -2$ : $11 + 2 - 4 = 9 = 9A$ , $A = 1$ . $x^2$ : $-1 = A - B$ , $B = 2$ . Answer $\dfrac{1}{2 + x} + \dfrac{2}{1 - x} + \dfrac{3}{(1 - x)^2}$ .
$\displaystyle\int_2^3 \left(\frac{1}{x} + \frac{2}{x - 1} - \frac{1}{x + 2}\right)dx = \Big[\ln x + 2\ln(x - 1) - \ln(x + 2)\Big]_2^3$ $= (\ln 3 + 2\ln 2 - \ln 5) - (\ln 2 + 0 - \ln 4) = \ln 3 + 2\ln 2 - \ln 5 + \ln 2 = \ln 3 + 3\ln 2 - \ln 5 = \ln\dfrac{24}{5}$ .

The idea

The forms you must recall

Distinct linear factors

A repeated linear factor

A quadratic factor

Improper fractions: equal degrees

An exam-style question with integration

Common mistakes

Summary

Practice

Console