Partial fractions
Master partial fractions for Year 12 Specialist Mathematics in Queensland (QCAA). This technique splits a rational function whose denominator is a product of two distinct linear factors into a sum of simpler fractions, turning a hard integral into two standard logarithm integrals.
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 integrate to \(A\ln|x-a|+B\ln|x-b|+c\) — an essential integration technique that also underpins 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 (QCAA, Queensland). 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