Partial Fractions Revision Notes for AQA A-Level Further Maths

Partial Fractions

Introduction to partial fractions

Partial fractions are a technique that allows you to break down complex rational functions into simpler fractions. This is particularly useful when you need to integrate rational functions, as the simpler forms are much easier to work with.

A rational function is a fraction where both the numerator and denominator are polynomials. When the denominator can be factorised into linear factors, you can split the fraction into partial fractions to make integration straightforward.

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Breaking complex fractions into simpler partial fractions transforms difficult integration problems into manageable ones. Each simpler fraction can be integrated using standard integration formulas you already know.

Improper fractions

Before you can work with partial fractions, you need to identify whether your fraction is proper or improper.

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Definition: An improper fraction is a fraction $\frac{f(x)}{g(x)}$ where the degree of $f(x)$ is greater than or equal to the degree of $g(x)$ .

When you have an improper fraction, it can be written in the form:

$P(x) + \frac{Q(x)}{g(x)}$

where $P(x)$ is the quotient and $\frac{Q(x)}{g(x)}$ is the proper fraction remainder.

Converting improper fractions

You have two methods to convert an improper fraction:

Long division method: Divide the numerator by the denominator using algebraic long division to find $P(x)$ and $Q(x)$ directly.
Degree comparison method: Look at the degrees of the polynomials to determine the general form of the quotient. For example, if $f(x)$ is a quartic (degree 4) and $g(x)$ is a quadratic (degree 2), then the quotient $P(x)$ must be a quadratic of the form $Ax^2 + Bx + C$ .

Once you have the proper fraction $\frac{Q(x)}{g(x)}$ , you can then split it into partial fractions.

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Tip: The degree comparison method is often faster than long division when you can quickly identify the form of the quotient by comparing polynomial degrees.

Worked example 1: improper fraction with linear factors

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Worked Example: Converting and Integrating an Improper Fraction

Question: Write $\frac{30x^3-13x^2+6x+6}{15x^2+x-6}$ in partial fractions, then find its integral.

Solution:

Since we're dividing a cubic by a quadratic, the quotient will be linear (of the form $Ax + B$ ).

We assume:

$\frac{30x^3-13x^2+6x+6}{15x^2+x-6} = Ax + B + \frac{C}{3x+2} + \frac{D}{5x-3}$

Note that $15x^2+x-6 = (3x+2)(5x-3)$ after factorising.

Multiplying both sides by $(15x^2+x-6)$ :

$30x^3-13x^2+6x+6 = (Ax+B)(15x^2+x-6) + C(5x-3) + D(3x+2)$

Expanding and equating coefficients:

Coefficient of $x^3$ : $30 = 15A \Rightarrow A = 2$
Coefficient of $x^2$ : $-13 = A + 15B \Rightarrow B = -1$
Coefficient of $x$ : $6 = -6A + B + 5C + 3D \Rightarrow 5C + 3D = 19$
Constant term: $6 = -6B - 3C + 2D \Rightarrow 3C - 2D = 0$

Solving simultaneously gives $C = 2$ and $D = 3$ .

Therefore:

$\frac{30x^3-13x^2+6x+6}{15x^2+x-6} = 2x - 1 + \frac{2}{3x+2} + \frac{3}{5x-3}$

Integration:

$\int \frac{30x^3-13x^2+6x+6}{15x^2+x-6} dx = \int \left(2x - 1 + \frac{2}{3x+2} + \frac{3}{5x-3}\right) dx$

$= x^2 - x + \frac{2}{3}\ln|3x+2| + \frac{3}{5}\ln|5x-3| + c$

Partial fractions with irreducible quadratic factors

Sometimes the denominator includes a quadratic factor that cannot be factorised further, such as $x^2+5$ (this has no real roots). In these cases, you need to use a linear expression $Ax + B$ as the numerator for the quadratic term.

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Key Principle for Irreducible Quadratics:

$\frac{f(x)}{(\alpha x^2+\beta)(\gamma x+\delta)} = \frac{Ax+B}{\alpha x^2+\beta} + \frac{C}{\gamma x+\delta}$

The key difference is that the irreducible quadratic factor gets a numerator of the form $Ax + B$ rather than just a constant.

Common mistake: Students often forget to use $Ax + B$ for irreducible quadratics and incorrectly use just a constant $A$ . This will not work!

Worked example 2: partial fractions with quadratic factors

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Worked Example: Integration with Irreducible Quadratic Factors

Question: Work out $\int \frac{2x+12}{(x+1)(x^2+9)} dx$

Solution:

First, note that $x^2+9$ cannot be factorised (it has no real roots), so we use: $\frac{2x+12}{(x+1)(x^2+9)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+9}$

Multiplying both sides by $(x+1)(x^2+9)$ :

$2x+12 = A(x^2+9) + (Bx+C)(x+1)$

$= Ax^2 + 9A + Bx^2 + Bx + Cx + C$

Equating coefficients:

$x^2$ : $0 = A + B$ , so $A = -B$
$x$ : $2 = B + C$ (equation 1)
Constant: $12 = 9A + C$ (equation 2)

From equation 1: $C = 2 - B$

Substituting into equation 2: $12 = 9(-B) + (2-B)$

$12 = -9B + 2 - B$

$10 = -10B$

$B = -1$

Therefore $A = 1$ and $C = 3$ .

So: $\frac{2x+12}{(x+1)(x^2+9)} = \frac{1}{x+1} + \frac{3-x}{x^2+9}$

Integration:

Split the second fraction:

$\int \frac{2x+12}{(x+1)(x^2+9)} dx = \int \left(\frac{1}{x+1} + \frac{3}{x^2+9} - \frac{x}{x^2+9}\right) dx$

For the first term: $\int \frac{1}{x+1} dx = \ln|x+1| + c_1$

For the second term: $\int \frac{3}{x^2+9} dx = 3 \times \frac{1}{3}\arctan\left(\frac{x}{3}\right) + c_2 = \arctan\left(\frac{x}{3}\right) + c_2$

For the third term, use substitution $u = x^2+9$ , $du = 2x dx$ :

$\int \frac{x}{x^2+9} dx = \frac{1}{2}\ln|x^2+9| + c_3$

Combining all parts:

$\int \frac{2x+12}{(x+1)(x^2+9)} dx = \ln|x+1| + \arctan\left(\frac{x}{3}\right) - \frac{1}{2}\ln(x^2+9) + c$

This can be written more elegantly as:

$= \ln\left(\frac{x+1}{\sqrt{x^2+9}}\right) + \arctan\left(\frac{x}{3}\right) + c$

Improper integrals and convergence

Partial fractions are also useful when evaluating improper integrals - integrals with infinite limits or discontinuities in the integration range. However, you must check whether the integral converges (has a finite value) or diverges (does not have a finite value).

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An improper integral will only converge if all parts of it converge. If any single part diverges, the entire integral diverges.

Strategy for testing convergence

When working with improper integrals involving rational functions:

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Three-Step Strategy for Testing Convergence:

Factorise the denominator as far as possible.
Split into partial fractions and establish any points of discontinuity for each fraction. A point of discontinuity occurs where the denominator equals zero.
Integrate and consider the behaviour of the integral as the variable tends towards the original limit. Evaluate limits carefully at both points of discontinuity and at infinity.

Worked example 3: convergence of improper integrals

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Worked Example: Testing Convergence of Improper Integrals

Question:

(a) Show algebraically that $\int_2^{\infty} \frac{x-9}{2x^3-x^2+8x-4} dx$ does not converge.
(b) Explain whether $\int_1^{\infty} \frac{x-9}{2x^3-x^2+8x-4} dx$ converges.

Solution:

First, factorise the denominator:

$2x^3-x^2+8x-4 = (2x-1)(x^2+4)$

Note that $x^2+4$ cannot be factorised further (it's an irreducible quadratic).

Set up partial fractions:

$\frac{x-9}{(2x-1)(x^2+4)} = \frac{A}{2x-1} + \frac{Bx+C}{x^2+4}$

Multiplying through by $(2x-1)(x^2+4)$ :

$x-9 = A(x^2+4) + (Bx+C)(2x-1)$

Equating coefficients:

$x^2$ : $0 = A + 2B$
$x$ : $1 = -B + 2C$
Constant: $-9 = 4A - C$

Solving simultaneously: $A = -2$ , $B = 1$ , $C = 1$

Therefore:

$\frac{x-9}{(2x-1)(x^2+4)} = \frac{-2}{2x-1} + \frac{x+1}{x^2+4}$

(a) Testing convergence from 2 to ∞:

The integral $\int_2^{\infty} \frac{x+1}{x^2+4} dx$ is a proper integral (no discontinuities in the range).

However, we need to check the behaviour at infinity:

$\int_2^{\infty} \frac{2}{2x-1} dx = \lim_{a \to \infty} \int_2^a \frac{2}{2x-1} dx$

$= \lim_{a \to \infty} [\ln|2x-1|]_2^a$

$= \lim_{a \to \infty} (\ln|2a-1| - \ln|3|)$

As $a \to \infty$ , $\ln|2a-1| \to \infty$ , so this integral diverges.

Therefore, the overall integral does not converge.

(b) Testing convergence from 1 to ∞:

For the range starting at 1, we still need to examine both parts:

$\int_1^{\infty} \frac{x-9}{2x^3-x^2+8x-4} dx = \int_1^{\infty} \left(\frac{x+1}{x^2+4} - \frac{2}{2x-1}\right) dx$

For $\int_1^{\infty} \frac{x+1}{x^2+4} dx$ :

Split this as: $\int_1^{\infty} \frac{x}{x^2+4} dx + \int_1^{\infty} \frac{1}{x^2+4} dx$

First part: $\int_1^{\infty} \frac{x}{x^2+4} dx = \lim_{a \to \infty} \left[\frac{1}{2}\ln(x^2+4)\right]_1^{\infty}$

As $a \to \infty$ , $\frac{1}{2}\ln(a^2+4) \to \infty$ so we need to check more carefully.

Actually, $\frac{a^2+4}{(2a-1)^2} = \frac{a^2+4}{4a^2-4a+1} \to \frac{1}{4}$ as $a \to \infty$

So $\frac{1}{2}\ln(a^2+4) - \ln(2a-1) = \ln\left(\frac{\sqrt{a^2+4}}{2a-1}\right) \to \ln\left(\frac{1}{2}\right) = -\ln 2$

The second part: $\int_1^{\infty} \frac{1}{x^2+4} dx = \lim_{a \to \infty} \left[\frac{1}{2}\arctan\left(\frac{x}{2}\right)\right]_1^{\infty}$

As $a \to \infty$ , $\arctan\left(\frac{a}{2}\right) \to \frac{\pi}{2}$

This part converges to: $\frac{1}{2}\left(\frac{\pi}{2}\right) - \frac{1}{2}\arctan\left(\frac{1}{2}\right) = \frac{\pi}{4} - \frac{1}{2}\arctan\left(\frac{1}{2}\right)$

After careful limit analysis (combining all terms), the integral converges to:

$\frac{\pi}{4} - \frac{1}{2}\arctan\left(\frac{1}{2}\right) - \frac{1}{2}\ln 20$

Exam tips

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Essential Tips for Success:

Always check if the fraction is improper first. If the degree of the numerator is greater than or equal to the degree of the denominator, you must divide first.
Factorise the denominator completely before setting up partial fractions. Sometimes you may need to factorise the denominator yourself - don't assume it's already factorised.
Watch out for irreducible quadratics. If you have a quadratic factor like $x^2+9$ that cannot be factorised, remember to use $Ax+B$ in the numerator, not just $A$ .
For integration involving $\frac{x}{x^2+a^2}$ , use substitution $u = x^2 + a^2$ to get $\frac{1}{2}\ln|x^2+a^2| + c$ .
When testing convergence of improper integrals, check each partial fraction separately. The integral only converges if all parts converge.
Points of discontinuity occur where any denominator equals zero. Always identify these before integrating.

Remember!

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Key Points to Remember:

An improper fraction has numerator degree ≥ denominator degree and must be converted to $P(x) + \frac{Q(x)}{g(x)}$ form before finding partial fractions.
For irreducible quadratic factors (like $x^2+5$ ), use $\frac{Ax+B}{x^2+5}$ rather than just $\frac{A}{x^2+5}$ .
Use the equating coefficients method to find unknown constants in partial fractions by multiplying through by the common denominator and comparing coefficients of like powers.
For improper integrals, the integral only converges if all individual parts converge. Check limits at both points of discontinuity and at infinity.
Standard integrals to remember:
- $\int \frac{1}{ax+b} dx = \frac{1}{a}\ln|ax+b| + c$
- $\int \frac{1}{x^2+a^2} dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + c$

Partial Fractions (AQA A-Level Further Maths): Revision Notes