Rational functions and partial fractions
Master rational functions and partial fractions in Year 12 VCE Specialist Mathematics. Partial fractions split a rational function whose denominator factorises into a sum of simpler fractions, turning a hard expression into standard pieces you can work with. It sits in the Functions, relations and graphs area of study of the VCE Mathematics Study Design (VCAA), within the Rational functions topic of Unit 3.
You will learn to set up the decomposition \(\dfrac{A}{x-a}+\dfrac{B}{x-b}\), find the constants by the cover-up rule, and rebuild a rational expression from its simpler fractions — an essential algebraic technique that also underpins integration and differential equations later in the course.
Theory
Partial fractions split a rational function with distinct linear factors in its denominator into a sum of simpler fractions, each of which integrates to a logarithm — a core technique of Year 12 Specialist Mathematics. This page shows how to decompose \(\dfrac{P(x)}{(x-a)(x-b)}\) as \(\dfrac{A}{x-a}+\dfrac{B}{x-b}\), find the constants by cover-up, and integrate to \(A\ln|x-a|+B\ln|x-b|+c\).
Partial fractions is a technique for rewriting a single rational function as a sum of simpler fractions. It is used here to integrate a proper rational function whose denominator is a product of two distinct linear factors, \(\dfrac{P(x)}{(x-a)(x-b)}\).
A fraction is proper when the degree of the numerator is less than the degree of the denominator. Such a fraction decomposes as \(\dfrac{P(x)}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}\), where \(A\) and \(B\) are constants (one over each factor).
The constants are found by the cover-up rule: multiply both sides by the denominator to get \(P(x)=A(x-b)+B(x-a)\), then substitute the value of \(x\) that makes one factor zero to isolate each constant. Equating coefficients gives the same result.
Each simple fraction integrates to a natural logarithm: \(\displaystyle\int\dfrac{A}{x-a}\,dx=A\ln|x-a|+c\). Two preliminary steps may be needed first — factorise the denominator if it is written as a quadratic, and divide first if the fraction is improper (numerator degree \(\ge\) denominator degree).
For a proper fraction with two distinct linear factors, the decomposition is
The single logarithm each term integrates to:
So, once decomposed, the whole integral is a sum of logarithms:
How to integrate by partial fractions
- Prepare: factorise the denominator into distinct linear factors; if the fraction is improper (numerator degree \(\ge\) denominator degree), divide first to get a whole part plus a proper fraction.
- Set up the form: write \(\dfrac{P(x)}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}\) with one unknown constant over each factor.
- Solve for the constants: multiply through to \(P(x)=A(x-b)+B(x-a)\), then cover-up — substitute \(x=a\) to find \(A\) and \(x=b\) to find \(B\).
- Integrate term by term: each fraction gives a logarithm, \(A\ln|x-a|+B\ln|x-b|+c\); for a definite integral, apply the limits and simplify to an exact value.
Write the form with a constant over each factor, then clear the denominator:
| \(\dfrac{2x+1}{(x-1)(x+2)}\) | \(=\) | \(\dfrac{A}{x-1}+\dfrac{B}{x+2}\) |
| \(2x+1\) | \(=\) | \(A(x+2)+B(x-1)\) |
Cover-up: substitute \(x=1\) to isolate \(A\):
| \(\text{let } x=1:\quad 2(1)+1\) | \(=\) | \(A(1+2)\) |
| \(3\) | \(=\) | \(3A\) |
| \(A\) | \(=\) | \(1\) |
Substitute \(x=-2\) to isolate \(B\):
| \(\text{let } x=-2:\quad 2(-2)+1\) | \(=\) | \(B(-2-1)\) |
| \(-3\) | \(=\) | \(-3B\) |
| \(B\) | \(=\) | \(1\) |
\(\dfrac{2x+1}{(x-1)(x+2)}=\dfrac{1}{x-1}+\dfrac{1}{x+2}\).
Decompose first; clear the denominator and cover-up for each constant:
| \(\dfrac{6}{(x-1)(x+5)}\) | \(=\) | \(\dfrac{A}{x-1}+\dfrac{B}{x+5}\) |
| \(6\) | \(=\) | \(A(x+5)+B(x-1)\) |
| \(\text{let } x=1:\quad 6\) | \(=\) | \(6A\) |
| \(A\) | \(=\) | \(1\) |
| \(\text{let } x=-5:\quad 6\) | \(=\) | \(-6B\) |
| \(B\) | \(=\) | \(-1\) |
Integrate each term to a logarithm:
| \(\int \dfrac{6}{(x-1)(x+5)}\,dx\) | \(=\) | \(\int \dfrac{1}{x-1}-\dfrac{1}{x+5}\,dx\) |
| \(=\) | \(\ln|x-1|-\ln|x+5|+c\) |
\(\displaystyle\int \dfrac{6}{(x-1)(x+5)}\,dx=\ln|x-1|-\ln|x+5|+c\).
Factorise the quadratic denominator into distinct linear factors:
| \(x^2-2x-8\) | \(=\) | \((x-4)(x+2)\) |
Decompose and cover-up for each constant:
| \(\dfrac{x+8}{(x-4)(x+2)}\) | \(=\) | \(\dfrac{A}{x-4}+\dfrac{B}{x+2}\) |
| \(x+8\) | \(=\) | \(A(x+2)+B(x-4)\) |
| \(\text{let } x=4:\quad 12\) | \(=\) | \(6A\) |
| \(A\) | \(=\) | \(2\) |
| \(\text{let } x=-2:\quad 6\) | \(=\) | \(-6B\) |
| \(B\) | \(=\) | \(-1\) |
Integrate term by term:
| \(\int \dfrac{x+8}{x^2-2x-8}\,dx\) | \(=\) | \(2\ln|x-4|-\ln|x+2|+c\) |
\(\displaystyle\int \dfrac{x+8}{x^2-2x-8}\,dx=2\ln|x-4|-\ln|x+2|+c\).
Decompose by cover-up:
| \(\dfrac{1}{(x-2)(x+1)}\) | \(=\) | \(\dfrac{A}{x-2}+\dfrac{B}{x+1}\) |
| \(1\) | \(=\) | \(A(x+1)+B(x-2)\) |
| \(\text{let } x=2:\quad 1\) | \(=\) | \(3A\) |
| \(A\) | \(=\) | \(\dfrac13\) |
| \(\text{let } x=-1:\quad 1\) | \(=\) | \(-3B\) |
| \(B\) | \(=\) | \(-\dfrac13\) |
Integrate to logarithms, then apply the limits \(3\) and \(4\):
| \(\int_3^4 \dfrac{1}{(x-2)(x+1)}\,dx\) | \(=\) | \(\dfrac13\Big[\ln|x-2|-\ln|x+1|\Big]_3^4\) |
| \(=\) | \(\dfrac13\big[(\ln2-\ln5)-(\ln1-\ln4)\big]\) | |
| \(=\) | \(\dfrac13(\ln2+\ln4-\ln5)\) | |
| \(=\) | \(\dfrac13\ln\dfrac{8}{5}\) |
\(\displaystyle\int_3^4 \dfrac{1}{(x-2)(x+1)}\,dx=\dfrac13\ln\dfrac{8}{5}\).
Common pitfalls
Frequently asked questions
What are partial fractions used for in integration?
They rewrite a proper rational function such as \(\dfrac{P(x)}{(x-a)(x-b)}\) as a sum \(\dfrac{A}{x-a}+\dfrac{B}{x-b}\), so each simple term can be integrated to a logarithm.
How do you find the constants A and B by cover-up?
Multiply both sides by the denominator to get \(P(x)=A(x-b)+B(x-a)\), then substitute the value of \(x\) that makes one factor zero: \(x=a\) gives \(A\) and \(x=b\) gives \(B\).
What does each partial fraction integrate to?
Each term \(\dfrac{A}{x-a}\) integrates to \(A\ln|x-a|+c\), so the whole integral is a sum of logarithms \(A\ln|x-a|+B\ln|x-b|+c\).
Do you have to factorise the denominator first?
Yes. The denominator must be written as a product of distinct linear factors, so factorise a quadratic like \(x^2-2x-8=(x-4)(x+2)\) before setting up the partial fractions.
What if the fraction is improper?
If the numerator degree is greater than or equal to the denominator degree, divide first to get a whole part plus a proper fraction, then decompose the proper part, e.g. \(\dfrac{x^2+1}{(x-1)(x-3)}=1+\dfrac{4x-2}{(x-1)(x-3)}\).
Why is there an absolute value in the logarithm?
The anti-derivative of \(\dfrac{1}{x-a}\) is \(\ln|x-a|+c\); the absolute value keeps the logarithm defined for \(x