A-Level Maths Partial Fractions Before Integration Explained
Partial fractions break a rational expression into simpler terms that can be integrated separately. They are particularly useful when the denominator factorises into linear factors and the numerator has lower degree. The method is algebra first and calculus second: factorise, choose the correct fraction structure, solve for coefficients and only then integrate.
When are partial fractions appropriate?
Consider (5x + 1)/((x + 1)(x + 2)). The denominator already contains two distinct linear factors, and the numerator has lower degree. Write the expression as A/(x + 1) + B/(x + 2). If the numerator degree were equal to or greater than the denominator degree, polynomial division would usually be required first. Recognising the appropriate form is more important than immediately integrating a complicated fraction.
Find the unknown constants
Multiply through by (x + 1)(x + 2) to obtain 5x + 1 = A(x + 2) + B(x + 1). Set x = −1 to get −4 = A, and x = −2 to get −9 = −B, so B = 9. Thus the decomposition is −4/(x + 1) + 9/(x + 2). Check with x = 0: the original gives 1/2, while the decomposition gives −4 + 9/2 = 1/2.
Integrate the decomposed expression
The integral becomes −4∫1/(x + 1) dx + 9∫1/(x + 2) dx. The logarithm rule gives −4ln|x + 1| + 9ln|x + 2| + C on any interval avoiding the poles. Absolute value bars are important because the linear factors can be negative even though their logarithmic antiderivatives are defined for negative arguments through ln|.|. Differentiate the result to check.
Handle a repeated linear factor
For a denominator (x + 1)², include terms A/(x + 1) and B/(x + 1)². For example (3x + 5)/(x + 1)² = 3/(x + 1) + 2/(x + 1)² because 3(x + 1) + 2 = 3x + 5. Integrating gives 3ln|x + 1| − 2/(x + 1) + C. Omitting the second power term is a common error when setting up the decomposition.
Avoid algebraic setup mistakes
Do not put only a single constant over a repeated denominator without checking the required structure. Multiply by the full denominator carefully and use substitution or coefficient comparison to solve for constants. Check signs when substituting roots of factors, and verify the resulting identity at an easy ordinary value. Partial fraction identities hold only where the original rational expression is defined, so never substitute directly into its poles when checking.
Choose the next calculus rule correctly
A term of the form 1/(x + a) integrates to ln|x + a|, while 1/(x + a)² integrates to −1/(x + a). More complicated irreducible quadratic factors can require different integration methods depending on the course. Keep the algebra and calculus stages separate in written work; this makes it easier to identify whether a wrong answer arose from the decomposition or integration.
Frequently asked questions
Why factorise the denominator first?
Its factorisation determines which simpler rational terms should appear in the decomposition.
Do all partial fractions integrate to logarithms?
No. Repeated powers and quadratic factors may lead to power-rule or other antiderivatives.
Explore related online maths resources
These existing MasterMaths articles cover connected skills. Choose the explanations relevant to your specific problem.
Book a free online maths introduction
If you are struggling with these ideas, MasterMaths Tutoring offers one-to-one online mathematics sessions. In a free twenty-minute introductory conversation, discuss the course, sticking points and study goals with Ryan. A suitable starting point can be agreed, without promising a particular grade or result.
Talk to Ryan: Book a free maths introduction session
Comments