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SASpecialist MathematicsSyllabus dot point

How does splitting a rational function into simpler fractions make it integrable?

Decompose rational functions into partial fractions and use the decomposition to integrate them.

Decomposing a proper rational function into partial fractions over linear factors, finding the unknown constants, and integrating the pieces to logarithms.

Generated by Claude Opus 4.78 min answer

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  1. What this dot point is asking
  2. When and how to decompose
  3. Finding the constants
  4. Integrating the pieces
  5. Improper rational functions

What this dot point is asking

You need to decompose rational functions into partial fractions and use the decomposition to integrate them.

When and how to decompose

Partial fractions apply to a proper rational function (numerator degree less than denominator degree). If the function is improper, do polynomial division first and decompose the remaining proper part.

Finding the constants

Two reliable methods: substitute strategic xx-values (the cover-up idea), or expand and equate coefficients of like powers.

Integrating the pieces

Each Axβˆ’a\dfrac{A}{x - a} integrates to Aln⁑∣xβˆ’a∣A\ln|x - a|, using ∫1xβˆ’a dx=ln⁑∣xβˆ’a∣+C\int \dfrac{1}{x - a}\,dx = \ln|x - a| + C. So once decomposed, the integral is a sum of logarithms.

Improper rational functions

If the numerator degree is at least the denominator degree, divide first. For instance x2x2βˆ’1\dfrac{x^2}{x^2 - 1} is improper; dividing gives 1+1x2βˆ’11 + \dfrac{1}{x^2 - 1}, and only the proper remainder 1(xβˆ’1)(x+1)\dfrac{1}{(x-1)(x+1)} is decomposed.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2017 SACE Stage 21 marksShow that 1/(x-2) - 1/(x+3) = 5/((x-2)(x+3)).
Show worked answer β†’

Combine the two fractions on the left over the common denominator (x-2)(x+3):
1/(x-2) - 1/(x+3) = [(x+3) - (x-2)] / ((x-2)(x+3)).

Simplify the numerator: (x+3) - (x-2) = x + 3 - x + 2 = 5.

So 1/(x-2) - 1/(x+3) = 5/((x-2)(x+3)), as required. This is the partial-fraction decomposition of 5/((x-2)(x+3)) read in reverse. [1 mark]

2025 SACE Stage 22 marksShow that 1/((X-8)(X-4)) = (1/4)(1/(X-8) - 1/(X-4)).
Show worked answer β†’

Decompose 1/((X-8)(X-4)) = A/(X-8) + B/(X-4). Multiplying through by (X-8)(X-4) gives 1 = A(X-4) + B(X-8). [1 mark]

Use convenient values of X:
Let X = 8: 1 = A(8-4) + B(0) = 4A, so A = 1/4.
Let X = 4: 1 = A(0) + B(4-8) = -4B, so B = -1/4.

Therefore 1/((X-8)(X-4)) = (1/4)/(X-8) - (1/4)/(X-4) = (1/4)(1/(X-8) - 1/(X-4)), as required. [1 mark]