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.
Cross-reference with the official IB Mathematics subject page for the full syllabus.
§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.