- Partial fractions split one awkward rational function into a sum of simple fractions that you can integrate one at a time.
- Each linear factor \((x-a)\) gives a term \(\frac{A}{x-a}\); a repeated factor \((x-a)^2\) needs two terms; an irreducible quadratic needs \(\frac{Bx+C}{x^2+k}\).
- In our case, a revision app’s weekly downloads \(\frac{1500}{(t+1)(t+4)}\) split into \(\frac{500}{t+1}-\frac{500}{t+4}\), so the first 8 weeks bring about 549 downloads.
- Even if the launch ran for ever, total downloads would never pass \(500\ln 4 \approx 693\): the integral converges.
- Exam point: if the top degree is not lower than the bottom degree, divide first. Splitting an improper fraction directly is the most common lost mark.
The case: how many downloads will a launch campaign bring?
A small team launches a free revision app for A Level students. They run a burst of adverts in the first week and then stop paying for promotion. Downloads are high at the start and then fall away as the advert effect fades. From the first fortnight of data they fit a simple model for the download rate, measured in downloads per week:
At launch the rate is \(R(0) = 1500/4 = 375\) downloads a week. One week later it has dropped to \(R(1) = 1500/10 = 150\). By week 4 it is \(37.5\), and by week 8 it is under 14 a week. (The figures are illustrative.)
The team’s questions are practical. How many downloads should they expect in the first 8 weeks? When will they reach half of everything the campaign will ever deliver? And is there a ceiling, or will the downloads keep adding up for ever if they simply wait?
Each question asks for a total, and a total of a rate is an integral. The trouble is the integrand. You cannot integrate \(\frac{1500}{(t+1)(t+4)}\) by any rule you learnt first. There is no obvious substitution, and integration by parts makes it worse. The tool that unlocks it is partial fractions.
The concept: undoing the common denominator
You already know how to add fractions. To add \(\frac{2}{x-1}\) and \(\frac{3}{x+2}\) you use the common denominator \((x-1)(x+2)\) and get \(\frac{5x+1}{(x-1)(x+2)}\). Partial fractions is that process run backwards. You start with the single combined fraction and recover the simple pieces it came from.
Why bother? Because the simple pieces are easy to integrate. Every term of the form \(\frac{A}{x-a}\) integrates to \(A\ln|x-a|\). Every term \(\frac{A}{(x-a)^2}\) integrates to \(-\frac{A}{x-a}\). A messy rational function becomes a short list of logarithms and reciprocals.
The method works only for a proper rational function, where the degree of the numerator is less than the degree of the denominator. If it is not proper, you first divide to get a polynomial plus a proper remainder, and then split the remainder.[3]
Three standard forms cover the questions you will meet in AP Calculus BC, IB higher level and Cambridge AS/A Level:
Here \(a\) and \(b\) are the roots of the linear factors and \(k>0\), so \(x^2+k\) cannot be factorised over the real numbers. The letters \(A\), \(B\) and \(C\) are constants you must find. The rule behind the forms is simple: the numerator over each factor is one degree lower than that factor. A linear factor gets a constant on top; a quadratic factor gets a linear expression \(Bx+C\) on top. A repeated factor \((x-b)^2\) needs a term for every power, \((x-b)\) and \((x-b)^2\), because the combined denominator could have come from either.
Mathematics Made Simple explains partial fractions and integration with worked examples and graphs.Get the textbook · $12
Why this method fits the case
The download rate is a rational function whose denominator already splits into two different linear factors, \((t+1)\) and \((t+4)\). That is the textbook situation for partial fractions. Once split, each piece is a constant over a linear factor, and the integral is a pair of logarithms.[3]
The decomposition also tells a story. The rate becomes \(\frac{500}{t+1} – \frac{500}{t+4}\): the difference of two decaying curves. Early on the first term is much larger than the second, so the rate is high. Later the two terms become almost equal and nearly cancel, which is why the rate falls so fast. The model has the shape of a launch: a strong start and a long tail.
Partial fractions is not the right tool in every case. If the numerator is the derivative of the denominator, as in \(\frac{2x+3}{x^2+3x+2}\), you can write the answer straight away as \(\ln|x^2+3x+2|\). If the denominator is an irreducible quadratic on its own, such as \(\frac{1}{x^2+9}\), you need the inverse tangent, not a split. And if the integrand is not a rational function at all, for example \(x e^{x}\), you need integration by parts. Check for these shortcuts before reaching for the full method.
Setting it up
Here is everything we know, and what each quantity means.
| Quantity | Symbol | Value or expression | Meaning |
|---|---|---|---|
| Time since launch | \(t\) | weeks, \(t \ge 0\) | The input variable |
| Download rate | \(R(t)\) | \(\frac{1500}{(t+1)(t+4)}\) | Downloads per week at time \(t\) |
| Rate at launch | \(R(0)\) | 375 per week | Peak of the campaign |
| Cumulative downloads | \(S(T)\) | \(\int_0^T R(t)\,dt\) | Total from launch to week \(T\) |
| Long-run total | \(S_\infty\) | \(\lim_{T\to\infty} S(T)\) | The ceiling, if there is one |
The plan has three stages. First split \(R(t)\) into partial fractions. Then integrate each term to find \(S(T)\). Finally, put in the numbers: \(T=8\), the half-way point and the limit as \(T\) grows.
Solving it
Split \(R(t) = \frac{1500}{(t+1)(t+4)}\) into partial fractions and find the number of downloads in the first 8 weeks.
The first 8 weeks bring about 549 downloads.
The same formula answers the team’s other questions. The long-run total is the limit of \(S(T)\). As \(T\to\infty\), \(\frac{T+1}{T+4}\to 1\), so
Half of that is \(250\ln 4 = 500\ln 2\). Setting \(500\ln\frac{4(T+1)}{T+4} = 500\ln 2\) gives \(\frac{4(T+1)}{T+4} = 2\), so \(4T+4 = 2T+8\) and \(T = 2\). Half of all downloads arrive in the first two weeks. To reach 90% of the ceiling, about 623.8 downloads, you need \(T=20\): \(S(19) \approx 623.3\) falls just short, while \(S(20) \approx 626.4\) clears it.
Repeated factor. Express \(\frac{7x+1}{(x-1)(x+1)^2}\) in partial fractions and hence find \(\int_2^3 \frac{7x+1}{(x-1)(x+1)^2}\,dx\) in exact form.
The integral equals \(2\ln\frac{3}{2} + \frac{1}{4} \approx 1.061\).
Improper fraction. Find \(\int_0^1 \frac{2x^2+5x+5}{(x+1)(x+2)}\,dx\) in the form \(p + q\ln 2 + r\ln 3\).
The integral is \(2 + 5\ln 2 – 3\ln 3 \approx 2.170\), so \(p=2\), \(q=5\), \(r=-3\).
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What the answer tells us
The partial fraction form turned a hard integral into a clean formula, \(S(T) = 500\ln\frac{4(T+1)}{T+4}\), and that formula gives the team a full picture of the campaign. Half of everything arrives in the first two weeks. About 549 arrive by week 8, which is already 79% of the ceiling. After that, waiting adds little: it takes until week 20 to reach 90%.
Figure 1 shows the cumulative total rising quickly and then flattening towards the dashed ceiling of about 693.
Reasonableness check. A quick estimate supports the answer. The rate starts at 375 a week and is 150 after one week, so the first week alone should bring roughly 200 to 250 downloads. The formula gives \(S(1) = 500\ln\frac{8}{5} \approx 235\), inside that range. Also, \(S(T)\) is increasing and can never pass 693, which fits a rate that is always positive but shrinking fast.
Why the total is finite. For large \(t\), the rate behaves like \(\frac{1500}{t^2}\). Rates that fall like \(1/t^2\) add up to a finite total, unlike rates that fall like \(1/t\), whose total grows without limit. The partial fractions show this directly: the two logarithms grow at the same speed and their difference settles down.
Sensitivity. The ceiling is \(\frac{1500}{3}\ln 4\), where 3 is the gap between the roots \(-1\) and \(-4\). If the team doubled the advertising effect so that the numerator became 3000, every figure would double: 1,099 in 8 weeks and a ceiling of 1,386. If instead the fade were slower, so the second factor were \((t+9)\), the split would be \(\frac{187.5}{t+1} – \frac{187.5}{t+9}\) and the ceiling would become \(187.5\ln 9 \approx 412\), lower overall but spread over a longer tail. The structure of the denominator drives the shape of the whole campaign.
To build fluency, work through the integration chapters of the Mathematics Made Simple textbook, which show each type of factor with a graph and a full worked solution, and then drill the method with the Mathematics Made Simple Practice Questions book, where every decomposition and integral is worked out step by step.
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Practice questions
Try each question before opening the worked answer.
EasyQ1. Express \(\frac{5}{(x-2)(x+3)}\) in partial fractions. Show worked answer
\(x=2\): \(5 = 5A\), so \(A=1\).
\(x=-3\): \(5 = -5B\), so \(B=-1\).
Answer: \(\frac{1}{x-2} – \frac{1}{x+3}\).
EasyQ2. Express \(\frac{3x}{(x-1)(x+2)}\) in partial fractions. Show worked answer
\(x=1\): \(3 = 3A\), so \(A=1\).
\(x=-2\): \(-6 = -3B\), so \(B=2\).
Answer: \(\frac{1}{x-1} + \frac{2}{x+2}\). Check at \(x=0\): \(-1 + 1 = 0\), matching \(\frac{0}{-2}=0\).
MediumQ3. Show that \(\int_0^2 \frac{10}{(2x+1)(x+3)}\,dx = 2\ln 3\). Show worked answer
\(x=-\tfrac12\): \(10 = 2.5A\), so \(A=4\). \(x=-3\): \(10 = -5B\), so \(B=-2\).
\(\int \Big(\frac{4}{2x+1} – \frac{2}{x+3}\Big)dx = 2\ln(2x+1) – 2\ln(x+3)\). Note the factor \(\tfrac12\) from the \(2x\).
Upper limit: \(2\ln 5 – 2\ln 5 = 0\). Lower limit: \(0 – 2\ln 3\).
Result: \(0 – (-2\ln 3) = 2\ln 3\) as required.
MediumQ4. Express \(\frac{x+3}{x^2(x+1)}\) in partial fractions and find \(\int_1^2 \frac{x+3}{x^2(x+1)}\,dx\) exactly. Show worked answer
\(x=0\): \(B=3\). \(x=-1\): \(2 = C\). Coefficients of \(x^2\): \(0 = A + C\), so \(A=-2\).
Integral: \(\Big[-2\ln x – \frac{3}{x} + 2\ln(x+1)\Big]_1^2\).
At 2: \(-2\ln 2 – 1.5 + 2\ln 3\). At 1: \(-3 + 2\ln 2\).
Answer: \(\frac{3}{2} + 2\ln 3 – 4\ln 2 \approx 0.925\).
ExamQ5. (a) Express \(\frac{2}{(x+1)(x^2+1)}\) in partial fractions. (b) Hence show that \(\int_0^1 \frac{2}{(x+1)(x^2+1)}\,dx = \frac{1}{2}\ln 2 + \frac{\pi}{4}\). [6 marks] Show worked answer
\(x=-1\): \(2 = 2A\), so \(A=1\). [M1]
\(x^2\): \(0 = A + B\), so \(B=-1\); constants: \(2 = A + C\), so \(C=1\). [A1 all three]
So the integrand is \(\frac{1}{x+1} + \frac{-x+1}{x^2+1} = \frac{1}{x+1} – \frac{x}{x^2+1} + \frac{1}{x^2+1}\).
(b) Integrate: \(\ln(x+1) – \tfrac12\ln(x^2+1) + \tan^{-1}x\). [M1 log terms, M1 inverse tangent]
Limits: \(\ln 2 – \tfrac12\ln 2 + \frac{\pi}{4} – 0 = \frac12\ln 2 + \frac{\pi}{4}\). [A1 exact answer shown]
Frequently asked questions
How do I know which partial fraction form to use?
Look at each factor of the denominator. A distinct linear factor gets a constant over it. A linear factor squared gets two terms, one over the factor and one over its square. An irreducible quadratic such as \(x^2+4\) gets a linear numerator \(Bx+C\). Add one term for each and you have the form.
Should I substitute values or compare coefficients?
Use both. Substituting the roots of the linear factors is fastest because each one removes all but one constant. When a constant cannot be isolated that way, as with \(B\) over a repeated factor or a quadratic factor, compare coefficients of the highest power or the constant term. Then check with a fresh value.
Why does the integral of \(\frac{1}{x+1}\) need a modulus?
The function \(\ln x\) is defined only for positive \(x\), but \(\frac{1}{x+1}\) has an antiderivative on both sides of \(x=-1\). Writing \(\ln|x+1|\) covers both. In a definite integral where \(x+1\) stays positive, as in the download model, you can drop the modulus.
Where else do partial fractions appear?
They appear in the logistic differential equation, where separating the variables gives \(\frac{1}{P(K-P)}\) to integrate. They also appear in binomial expansions of rational functions, in sums of series that telescope, and in Laplace transforms at university. Learning the method well pays off well beyond one topic.
References
- Cambridge Assessment International Education. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. cambridgeinternational.org
- College Board. AP Calculus AB and BC Course and Exam Description. New York: College Board.
- Strang, G. & Herman, E. (2016). Calculus Volume 2. Houston, TX: OpenStax. openstax.org
- International Baccalaureate Organization (2019). Mathematics: analysis and approaches guide. Cardiff: IBO.
