Partial fractions sit at the intersection of algebra and calculus on the IB Mathematics Analysis & Approaches HL syllabus. The technique itself — splitting a rational function into simpler pieces with linear denominators — is short on theory but very long on procedural detail, and that is exactly why it is so heavily examined. The same handful of mistakes recur every year: missing the polynomial long-division step on improper fractions, leaving out the third constant in repeated-factor decompositions, dropping the minus sign in $\int \tfrac{1}{k-x}\,dx$, or forgetting the absolute-value bars inside $\ln$.
This cheatsheet condenses everything from AHL 2.14 (decomposition) and AHL 5.15 (integration via partial fractions, including the logistic DE) into one page you can revise from. If you want the printable PDF version, the full set of notes, the worked tutorials, and the marked-up solutions, scroll to the bottom for the download links and the gated full library.
§1 — What Are Partial Fractions? AHL 2.14
Partial fractions split a rational function into simpler parts with linear denominators.
Uses: integration, separable differential equations, the logistic model.
Proper: $\deg(\text{num}) < \deg(\text{den})$ — ready for PF decomposition.
Improper: $\deg(\text{num}) \geq \deg(\text{den})$ — do polynomial long division first.
§2 — Decomposition Forms AHL 2.14
Case 1 — Distinct linear factors
$$\frac{f(x)}{(ax+b)(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{cx+d}$$
Cover-up for $A$: set $ax+b = 0$, evaluate what remains. Cover-up for $B$: set $cx+d = 0$.
Example: $\dfrac{5x+1}{(x-1)(x+2)} = \dfrac{2}{x-1} + \dfrac{3}{x+2}$. Setting $x=1$: $6 = 3A \Rightarrow A=2$; setting $x=-2$: $-9 = -3B \Rightarrow B=3$.
Case 2 — Repeated linear factor
$$\frac{f}{(ax+b)(cx+d)^2} \equiv \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}$$
Cover-up gives $A$ and $C$ directly. Use $x = 0$ (or any third value) to find $B$.
Example: $\dfrac{1}{(x+1)^2(2x+1)} \equiv \dfrac{-2}{x+1} + \dfrac{-1}{(x+1)^2} + \dfrac{4}{2x+1}$.
Cover-Up Rule (Heaviside)
To find $A$ in $\dfrac{1}{(x-r)(x-s)\cdots}$:
- Set $x = r$ (root of that factor).
- "Cover" $(x-r)$ and evaluate what's left.
Always write the identity line before substituting.
Equating coefficients
After multiplying through: $1 \equiv A(cx+d) + B(ax+b)$.
Coefficient of $x$: $Ac + Ba = 0$. Constant: $Ad + Bb = 1$. Solve the $2\times 2$ system. Use this method when the roots are messy fractions.
§3 — 7-Step Strategy AHL 2.14
- Degree check: proper? If not, divide.
- Factorise the denominator fully.
- Write the PF form (repeated factor? Use 3 constants).
- Multiply both sides by the denominator.
- Cover-up to find $A$, $C$ (roots of linear factors).
- Equate coefficients or use $x = 0$ for $B$.
- Verify: add the fractions back — should equal the original.
Polynomial-division example
$\dfrac{3x^2 + 5x + 4}{x^2 + 3x + 2} = 3 + \dfrac{-4x - 2}{(x+1)(x+2)}$. Then decompose the remainder fraction. Key: check degrees before setting up PF.
§4 — Integration with Partial Fractions AHL 5.15
Integration formulas
Always write $|\;|$ inside $\ln$ — IB marks depend on it.
Full integration example
$\displaystyle\int \frac{1}{x^2 + 3x + 2}\,dx = \int \frac{1}{(x+1)(x+2)}\,dx$, with PF $\dfrac{1}{x+1} - \dfrac{1}{x+2}$.
$= \ln|x+1| - \ln|x+2| + C = \ln\!\left|\dfrac{x+1}{x+2}\right| + C$. (IB booklet example, AHL 5.15.)
Repeated-factor integration
$\displaystyle\int \frac{1}{(x+1)^2(2x+1)}\,dx = \int\!\left(\frac{4}{2x+1} - \frac{2}{x+1} - \frac{1}{(x+1)^2}\right)dx = 2\ln|2x+1| - 2\ln|x+1| + \frac{1}{x+1} + C$.
Note the $\dfrac{1}{x+1}$ term — not a log!
§5 — Logistic DE Connection AHL 5.18
The logistic DE via partial fractions
$$\frac{dP}{dt} = kP\!\left(1 - \frac{P}{N}\right)$$
Key PF step: $\dfrac{1}{P(N-P)} = \dfrac{1}{N}\!\left(\dfrac{1}{P} + \dfrac{1}{N-P}\right)$.
After integrating: $\ln\!\left|\dfrac{P}{N-P}\right| = Nkt + C$.
Closed-form solution: $P = \dfrac{P_0 N}{P_0 + (N - P_0)e^{-Nkt}}$. Maximum growth rate occurs when $P = N/2$.
Key identities
§6 — Exam Attack Plan All sections
| If you see this | Do this |
|---|---|
| $\displaystyle\int \dfrac{ax+b}{\text{quadratic}}\,dx$ | Factorise denominator, PF, then integrate |
| $\dfrac{dP}{dt} = kP(N-P)$ | PF of $\dfrac{1}{P(N-P)}$, separate and integrate |
| "Show that" PF identity | Write form, multiply through, solve |
| $(x+a)^2$ in denominator | 3 constants: $\dfrac{A}{x+a} + \dfrac{B}{(x+a)^2} + \dfrac{C}{\ldots}$ |
| "Hence integrate" | Must use PF from earlier part |
| $\deg(\text{num}) \geq \deg(\text{den})$ | Polynomial long division first |
Top exam traps — never make these
- Missing $|\;|$: write $\ln|ax+b|$, not $\ln(ax+b)$.
- Repeated factor: need $\dfrac{A}{x+a} + \dfrac{B}{(x+a)^2}$, NOT $\dfrac{2A}{x+a}$.
- Improper fraction: always divide first.
- Cover-up: must write the identity line before substituting.
- "Hence": must use PF from the earlier part — can't change method.
- $\displaystyle\int \frac{1}{k-x}\,dx = -\ln|k-x| + C$ — minus sign from chain rule.
- Missing $+ C$ in indefinite integrals.
- Assuming the denominator is factorised — always check.
Worked Example — IB-Style Partial Fractions Integration
Question (HL Paper 2 style — 8 marks)
(a) Express $\dfrac{4x - 1}{(x - 1)(x + 2)}$ in partial fractions. [3] (b) Hence find $\displaystyle\int \frac{4x - 1}{(x - 1)(x + 2)}\,dx$. [3] (c) Hence evaluate $\displaystyle\int_{2}^{4} \frac{4x - 1}{(x - 1)(x + 2)}\,dx$, giving your answer in the form $\ln k$. [2]
Solution
- Set up identity: $\dfrac{4x-1}{(x-1)(x+2)} \equiv \dfrac{A}{x-1} + \dfrac{B}{x+2} \;\Rightarrow\; 4x - 1 \equiv A(x+2) + B(x-1)$. (M1)
- Cover-up at $x = 1$: $4(1) - 1 = 3 = 3A \Rightarrow A = 1$. Cover-up at $x = -2$: $-9 = -3B \Rightarrow B = 3$. So $\dfrac{4x-1}{(x-1)(x+2)} = \dfrac{1}{x-1} + \dfrac{3}{x+2}$. (A1)(A1)
- Integrate term-by-term: $\displaystyle\int\!\left(\frac{1}{x-1} + \frac{3}{x+2}\right)dx = \ln|x-1| + 3\ln|x+2| + C$. (M1)(A1)(A1)
- Evaluate from 2 to 4: $\big[\ln|x-1| + 3\ln|x+2|\big]_2^4 = \big(\ln 3 + 3\ln 6\big) - \big(\ln 1 + 3\ln 4\big) = \ln 3 + 3\ln 6 - 3\ln 4$. (M1)
- Simplify to $\ln k$ form: $= \ln 3 + 3\ln\!\left(\tfrac{6}{4}\right) = \ln 3 + 3\ln\!\left(\tfrac{3}{2}\right) = \ln 3 + \ln\!\left(\tfrac{27}{8}\right) = \ln\!\left(\tfrac{81}{8}\right)$. So $k = \tfrac{81}{8}$. (A1)
Examiner's note: in part (b), every IB candidate who omits the absolute-value bars inside $\ln$ loses both A1 marks. In part (c), forgetting that $\ln 1 = 0$ produces an arithmetic error that propagates into a wrong final answer. Use log laws ($a\ln b = \ln b^a$ and $\ln a + \ln b = \ln ab$) to combine into a single $\ln k$ — IB demands the answer be in the requested form.
Common Student Questions
When do I need polynomial long division before partial fractions?
Why does my repeated factor need three constants instead of two?
How do I integrate $\dfrac{1}{k - x}$?
What is the cover-up rule and when can I use it?
What is the partial fraction identity for the logistic DE?
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