Free Cheatsheet · AHL 2.14 · AHL 5.15

IB Math AA HL Partial Fractions — Complete Cheatsheet

Cover-up rule, repeated factors, integration formulas, and the logistic differential equation. Every Partial Fractions trick and trap for IB Mathematics Analysis & Approaches HL, hand-built by an IBO-certified Singapore tutor.

Topic: Partial Fractions (Algebra & Calculus) Syllabus: AHL 2.14, AHL 5.15 Read time: ~10 minutes Last updated: Apr 2026

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 $\dfrac{1}{(x-r)(x-s)\cdots}$:

  1. Set $x = r$ (root of that factor).
  2. "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.

TrickCover-up is the fastest method for distinct linear factors — sub the root of each factor and read off the constant in one step. No simultaneous equations needed.
TrapIf $\deg(\text{num}) \geq \deg(\text{den})$, you must do polynomial long division first before setting up the PF decomposition. Students who skip this step get the wrong form entirely.

§3 — 7-Step Strategy AHL 2.14

  1. Degree check: proper? If not, divide.
  2. Factorise the denominator fully.
  3. Write the PF form (repeated factor? Use 3 constants).
  4. Multiply both sides by the denominator.
  5. Cover-up to find $A$, $C$ (roots of linear factors).
  6. Equate coefficients or use $x = 0$ for $B$.
  7. 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

$\displaystyle\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + C$
$\displaystyle\int \frac{1}{(ax+b)^2}\,dx = \frac{-1}{a(ax+b)} + C$
$\displaystyle\int \frac{1}{(ax+b)^n}\,dx = \frac{-1}{a(n-1)(ax+b)^{n-1}} + C$

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!

Trap$\displaystyle\int \frac{1}{k-x}\,dx = -\ln|k-x| + C$ — the minus sign from the chain rule is the most commonly dropped mark in PF integration.
TrickAfter decomposing, add the fractions back to verify. Takes 30 seconds and catches all sign errors before you integrate.

§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

$\dfrac{1}{P(N-P)} = \dfrac{1}{N}\!\left(\dfrac{1}{P} + \dfrac{1}{N-P}\right)$
$\dfrac{1}{(x-a)(x-b)} = \dfrac{1}{a-b}\!\left(\dfrac{1}{x-a} - \dfrac{1}{x-b}\right)$
$\dfrac{1}{(ax+b)^2}$ — never a $\ln$ term
$\displaystyle\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C$   (not a PF integral)
NoteThe logistic DE is the most common exam application of partial fractions. Know the identity $\dfrac{1}{P(N-P)} = \dfrac{1}{N}\!\left(\dfrac{1}{P} + \dfrac{1}{N-P}\right)$ by heart.

§6 — Exam Attack Plan All sections

If you see thisDo 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 identityWrite form, multiply through, solve
$(x+a)^2$ in denominator3 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

  1. Missing $|\;|$: write $\ln|ax+b|$, not $\ln(ax+b)$.
  2. Repeated factor: need $\dfrac{A}{x+a} + \dfrac{B}{(x+a)^2}$, NOT $\dfrac{2A}{x+a}$.
  3. Improper fraction: always divide first.
  4. Cover-up: must write the identity line before substituting.
  5. "Hence": must use PF from the earlier part — can't change method.
  6. $\displaystyle\int \frac{1}{k-x}\,dx = -\ln|k-x| + C$ — minus sign from chain rule.
  7. Missing $+ C$ in indefinite integrals.
  8. 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

  1. 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)
  2. 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)
  3. 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)
  4. 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)
  5. 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?
Whenever the fraction is improper — that is, the degree of the numerator is greater than or equal to the degree of the denominator. Long division converts it to a polynomial plus a proper fraction; only the proper fraction gets decomposed. Skipping this step gives you the wrong PF form entirely and zero marks for the decomposition.
Why does my repeated factor need three constants instead of two?
For a denominator like $(x+1)^2(2x+1)$ the decomposition is $\dfrac{A}{x+1} + \dfrac{B}{(x+1)^2} + \dfrac{C}{2x+1}$ — three terms, three constants. The $(x+1)^2$ splits into both a $(x+1)$ and a $(x+1)^2$ term because together they span the same algebraic space the original repeated factor covers. Using only two constants gives an under-determined system and the wrong answer.
How do I integrate $\dfrac{1}{k - x}$?
The integral is $-\ln|k - x| + C$. The minus sign comes from the chain rule because the inner function $k - x$ has derivative $-1$. Forgetting this minus is the single most commonly dropped mark in PF integration — always pause when you see $(k - x)$ in the denominator and write the minus sign.
What is the cover-up rule and when can I use it?
Cover-up (Heaviside's method) is the fastest way to find constants over distinct linear factors. To find $A$ in $\dfrac{f(x)}{(x-r)(x-s)\ldots}$, substitute $x = r$ into what remains after covering $(x-r)$. It works directly only for distinct linear factors and for the highest-power term of a repeated factor; for the middle constants, substitute a third value of $x$ or equate coefficients.
What is the partial fraction identity for the logistic DE?
For $\dfrac{dP}{dt} = kP(1 - P/N)$, separating variables gives $\displaystyle\int \dfrac{1}{P(N-P)}\,dP = \int k\,dt$. The PF identity is $\dfrac{1}{P(N-P)} = \dfrac{1}{N}\!\left(\dfrac{1}{P} + \dfrac{1}{N-P}\right)$. After integrating: $\ln\!\left|\dfrac{P}{N-P}\right| = Nkt + C$, leading to the closed-form logistic solution. This identity appears every other year on AHL 5.18.

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