Free Cheatsheet · SL 5.5, 5.10–5.11 · AHL 5.15–5.17

IB Math AA HL Integration — Complete Cheatsheet

Every standard integral, technique, area formula, and volume formula you need for IB Mathematics Analysis & Approaches HL Integration. Inspection, substitution, integration by parts, partial fractions, areas, volumes of revolution, kinematics — all in one revisable page.

Topic: Integration (Calculus) Syllabus: SL 5.5, 5.10–5.11, AHL 5.15–5.17 Read time: ~16 minutes Last updated: Apr 2026

Integration is the highest-density topic on IB Mathematics Analysis & Approaches HL. It carries the largest mark allocation in the calculus strand, it is examined on every Paper 1, Paper 2 and Paper 3, and it gates almost all of the modelling and applied questions later in the syllabus (kinematics, differential equations, statistics, Maclaurin series). The HL extension over SL is huge: integration by substitution, integration by parts (including cyclic IBP), partial fractions, and volumes of revolution are all AHL content that students see for the first time at HL.

This cheatsheet condenses every formula, technique, and exam trap from SL 5.5, 5.10–5.11 and AHL 5.15–5.17 into one revisable page. The hardest part of HL Integration is not memorising rules — it is choosing the right technique fast under exam pressure. The Exam Attack Plan in §10 is built specifically for that decision. Scroll to the bottom for the printable PDF and the full notes booklet.

§1 — Standard Integrals SL 5.5, 5.10, AHL 5.15

Core table

$f(x)$$\int f(x)\,dx$
$x^n$   $(n \neq -1)$$\dfrac{x^{n+1}}{n+1} + C$
$\dfrac{1}{x}$$\ln|x| + C$
$e^x$$e^x + C$
$a^x$$\dfrac{a^x}{\ln a} + C$
$\sin x$$-\cos x + C$
$\cos x$$\sin x + C$
$\tan x$$\ln|\sec x| + C$
$\sec^2 x$$\tan x + C$
$\csc^2 x$$-\cot x + C$
$\sec x \tan x$$\sec x + C$
$\sec x$$\ln|\sec x + \tan x| + C$

Inverse trig & AHL

$f(x)$$\int f(x)\,dx$
$\dfrac{1}{1+x^2}$$\arctan x + C$
$\dfrac{1}{a^2 + x^2}$$\dfrac{1}{a}\arctan\dfrac{x}{a} + C$
$\dfrac{1}{\sqrt{1 - x^2}}$$\arcsin x + C$
$\dfrac{1}{\sqrt{a^2 - x^2}}$$\arcsin\dfrac{x}{a} + C$

Linear composite rule

$$\int f(ax+b)\,dx = \frac{F(ax+b)}{a} + C$$

e.g. $\int e^{3x}\,dx = \dfrac{e^{3x}}{3} + C$,   $\int\sin(2x)\,dx = -\dfrac{\cos 2x}{2} + C$.

TrickFor $\int\dfrac{1}{x^2 + bx + c}\,dx$: complete the square first to get $(x+p)^2 + q^2$, then use $\dfrac{1}{q}\arctan\dfrac{x+p}{q}$.
Trap$\int\dfrac{1}{x}\,dx = \ln|x| + C$ — the absolute value bars are required and cost a mark every time they are omitted.

§2 — Integration by Inspection (Reverse Chain Rule) SL 5.10

Three-step method

  1. Spot the inner function $g(x)$.
  2. Compute $g'(x)$.
  3. Check whether $g'(x)$ appears in the integrand (possibly times a constant $k$). If yes, integrate directly.

Core rule: $\int k \cdot g'(x) \cdot f(g(x))\,dx = k \cdot F(g(x)) + C$.

Pattern table

PatternAntiderivative ($+C$)
$\int [g]^n \cdot g'\,dx$$\dfrac{[g]^{n+1}}{n+1}$
$\int \dfrac{g'}{g}\,dx$$\ln|g|$
$\int e^g \cdot g'\,dx$$e^g$
$\int \sin(g)\cdot g'\,dx$$-\cos(g)$
$\int \cos(g)\cdot g'\,dx$$\sin(g)$
$\int \sec^2(g)\cdot g'\,dx$$\tan(g)$
$\int \dfrac{g'}{\sqrt{1 - g^2}}\,dx$$\arcsin(g)$
$\int \dfrac{g'}{1 + g^2}\,dx$$\arctan(g)$

Key examples

  • $\int 2x(x^2 + 1)^4\,dx = \dfrac{(x^2+1)^5}{5} + C$
  • $\int \dfrac{3x^2}{x^3 + 7}\,dx = \ln|x^3 + 7| + C$
  • $\int 4x\sin(x^2)\,dx = -2\cos(x^2) + C$
  • $\int \tan x\,dx = \ln|\sec x| + C$
  • $\int \dfrac{e^x}{1 + e^{2x}}\,dx = \arctan(e^x) + C$
  • $\int \dfrac{2x}{1 + x^4}\,dx = \arctan(x^2) + C$

Verify: differentiate your answer — you should get the integrand back.

TrickFor $\int \dfrac{f'(x)}{f(x)}\,dx$: any fraction where the top is (a multiple of) the derivative of the bottom integrates to $\ln|$bottom$|$. Check this pattern first before reaching for substitution.
Trap$\int 6x(x^2 + 1)^4\,dx$: don't multiply $6x$ into the answer. Factor as $3 \cdot 2x$ first, then apply the power rule: answer is $\dfrac{3(x^2 + 1)^5}{5} + C$.
NoteIf $g'(x)$ is NOT present in the integrand (not even as a scaled multiple), inspection will fail — switch to formal substitution instead.

§3 — Integration by Substitution AHL 5.16

Method — Indefinite

  1. Choose $u = g(x)$ (the inner / messy part).
  2. $du = g'(x)\,dx \Rightarrow dx = \dfrac{du}{g'(x)}$.
  3. Eliminate every $x$ from the integrand.
  4. Integrate in $u$.
  5. Back-substitute $u \to g(x)$, add $+C$.

Method — Definite

Same steps 1–4, plus: change the limits $x = a \Rightarrow u = g(a)$, $x = b \Rightarrow u = g(b)$. With converted limits, no back-substitution is needed.

Half-angle substitution (given in exam): $t = \tan(x/2) \Rightarrow \sin x = \dfrac{2t}{1 + t^2}$, $dx = \dfrac{2}{1 + t^2}\,dt$.

TrickAfter substituting, check no $x$ remains. If an $x$ is left over, rearrange $u = g(x)$ to express $x$ in terms of $u$.
TrapForgetting to change the limits in a definite integral after substitution. Continuing with the original $x$-values is wrong and loses every accuracy mark in the question.

§4 — Integration by Parts AHL 5.16

Formula & LIATE

$$\int u\,dv = uv - \int v\,du$$

LIATE priority for choosing $u$: Log > Inverse trig > Algebraic > Trig > Exponential.

Key results

  • $\int \ln x\,dx = x\ln x - x + C$
  • $\int x e^x\,dx = (x - 1)e^x + C$
  • $\int x \sin x\,dx = \sin x - x\cos x + C$
  • $\int \arcsin x\,dx = x\arcsin x + \sqrt{1 - x^2} + C$
  • $\int \arctan x\,dx = x\arctan x - \tfrac{1}{2}\ln(1 + x^2) + C$

Cyclic IBP & reduction

Cyclic IBP ($\int e^x\sin x$, $\int e^x\cos x$): apply IBP twice, keeping the same $u$ each time. Name the integral $I$, solve $2I = \ldots$:

  • $\int e^x \sin x\,dx = \dfrac{e^x(\sin x - \cos x)}{2} + C$
  • $\int e^x \cos x\,dx = \dfrac{e^x(\sin x + \cos x)}{2} + C$

Tabular method (polynomial $\times$ exp/trig): differentiate the polynomial column, integrate the exp/trig column, alternate $+/-$ signs, read diagonally.

Reduction formula: $n I_n = (n-1) I_{n-2}$ for $I_n = \int_0^{\pi/2} \cos^n x\,dx$.

Trick$\int \ln x\,dx$: treat as $\int \ln x \cdot 1\,dx$ with $u = \ln x$, $dv = dx$. The $1$ is the invisible $dv$.
TrapCyclic IBP — swapping your choice of $u$ in round 2 collapses to the trivial $0 = 0$. Always keep the same function as $u$ in both rounds.

§5 — Partial Fractions AHL 5.15

Decomposition forms

  • Distinct linear: $\dfrac{1}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$
  • Repeated linear: $\dfrac{1}{(x-a)^2} = \dfrac{A}{(x-a)^2} + \dfrac{B}{x-a}$
  • Irreducible quadratic: $\dfrac{1}{(x-a)(x^2+b)} = \dfrac{A}{x-a} + \dfrac{Bx + C}{x^2 + b}$

Each term integrates to $\ln|\cdot|$ or $\arctan$.

5-step method

  1. Factorise the denominator.
  2. Write the decomposition with unknowns $A$, $B$, $C$.
  3. Multiply through by the denominator.
  4. Substitute strategic $x$-values or compare coefficients to solve for $A$, $B$, $C$.
  5. Integrate each term.

Complete the square for $\int \dfrac{1}{x^2 + bx + c}\,dx$: $(x+p)^2 + q^2 \to \dfrac{1}{q}\arctan\dfrac{x+p}{q}$.

TrapIf $\deg(P) \geq \deg(Q)$, do polynomial long division first before decomposing. Skipping this step gives an incorrect partial-fraction setup.
TrickSubstitute $x = a$ (a root of a linear factor) to find the corresponding constant in one step — no simultaneous equations required.

§6 — Areas SL 5.11

Five area cases

CaseSetupFormula
1$f(x) \geq 0$ on $[a,b]$$A = \int_a^b f\,dx$
2$f(x) \leq 0$ on $[a,b]$$A = -\int_a^b f\,dx$
3$f$ changes sign, root at $c$$A = \left|\int_a^c f\,dx\right| + \left|\int_c^b f\,dx\right|$
4Between curves $f \geq g$ on $[a,b]$$A = \int_a^b (f - g)\,dx$
5Curves cross inside $[a,b]$Find crossing, split, $\int |f-g|$

Area with $y$-axis: $A = \int_c^d g(y)\,dy$ where $x = g(y)$ (invert the curve).

Sign-change procedure

  1. Find all roots of $f$ on $[a,b]$.
  2. Sketch — identify sign on each interval.
  3. Split the integral at every root.
  4. Take $|\ldots|$ of each piece.
  5. Sum all pieces.

Even function: $\int_{-a}^a f = 2\int_0^a f$.   Odd function: $\int_{-a}^a f = 0$.

Trap$\int_a^b f\,dx$ gives signed area — it can be negative. "Find the area" always means geometric (positive). Never integrate directly when $f$ dips below the axis.
TrickOdd function on a symmetric interval: $\int_{-a}^a f = 0$ instantly. Even function: $\int_{-a}^a f = 2\int_0^a f$. Check this before starting any integral on symmetric limits.

§7 — Volumes of Revolution AHL 5.17

All four formulas

SituationAxisFormula
Single curve $y = f(x)$$x$-axis$V = \pi\int_a^b [f(x)]^2\,dx$
Single curve (invert to $x = g(y)$)$y$-axis$V = \pi\int_c^d [g(y)]^2\,dy$
Outer $f$, inner $g$ (washer)$x$-axis$V = \pi\int_a^b ([f]^2 - [g]^2)\,dx$
Outer $g_1$, inner $g_2$$y$-axis$V = \pi\int_c^d ([g_1]^2 - [g_2]^2)\,dy$

Worked examples

  • $y = \sqrt{x}$, $x \in [0, 4]$, about $x$-axis: $V = \pi\int_0^4 x\,dx = 8\pi$.
  • $y = \sin x$, $x \in [0, \pi]$, about $x$-axis: use $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ to get $V = \dfrac{\pi^2}{2}$.
  • $y = e^x$, $x \in [0, 2]$, about $x$-axis: $V = \pi\int_0^2 e^{2x}\,dx = \dfrac{\pi}{2}(e^4 - 1)$.
  • Washer between $y = \sqrt{x}$ and $y = x^2$: $V = \pi\int_0^1 (x - x^4)\,dx = \dfrac{3\pi}{10}$.

Useful IBP result for $y$-axis rotations: $\int (\ln y)^2\,dy = y[(\ln y)^2 - 2\ln y + 2] + C$.

Trap$(f - g)^2 \neq f^2 - g^2$. Square each radius separately: $\pi\int ([f]^2 - [g]^2)\,dx$. Squaring the difference is wrong — a very common exam mistake.
TrickFor trig rotations: $\int \sin^2 x\,dx$ and $\int \cos^2 x\,dx$ always need the double-angle identity first — $\sin^2 x = \dfrac{1 - \cos 2x}{2}$, $\cos^2 x = \dfrac{1 + \cos 2x}{2}$ — before integrating.
NoteFor $y$-axis rotation, the limits must be $y$-values. Substitute the $x$-limits into $y = f(x)$ to convert them.

§8 — Kinematics (Integration View) SL 5.9

$v = \dfrac{ds}{dt}$,   $a = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2}$.   $s = \int v\,dt$,   $v = \int a\,dt$.

Displacement vs total distance

Displacement (signed): $\Delta s = \int_{t_1}^{t_2} v(t)\,dt$.   Total distance (always non-negative): $d = \int_{t_1}^{t_2} |v(t)|\,dt$. Split at zeros of $v$ (direction changes).

Distance procedure

  1. Solve $v(t) = 0$ to find the times $t_1, t_2, \ldots$ at which the particle reverses.
  2. Check the sign of $v$ on each sub-interval.
  3. Integrate $v$ on each sub-interval.
  4. Sum the absolute values.

Speed $= |v|$ (always $\geq 0$). Particle at rest: $v = 0$.

Trap$\int v\,dt$ gives displacement, not distance. "How far does it travel?" $\Rightarrow$ total distance $= \int |v|\,dt$.
TrickIf $v(t)$ on Paper 2 is messy (e.g. $e^{-\sin t}\cos(2t)$), use the GDC for numerical integration: nInt(abs(v(t)),t,a,b).

§9 — Maclaurin Series & Integration AHL 5.19

Standard series (memorise)

  • $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$
  • $\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots$
  • $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots$
  • $\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$

Key substitution: replace $x$ with the relevant expression (e.g. $e^{-x^2}$: substitute $x \to -x^2$ in the series for $e^x$).

Integrating series term by term

Some integrals have no elementary closed form — e.g. $\int e^{-x^2}\,dx$, $\int \dfrac{\sin x}{x}\,dx$. Expand as a Maclaurin series and integrate each polynomial term separately:

  • $\int e^{-x^2}\,dx \approx x - \dfrac{x^3}{3} + \dfrac{x^5}{10} - \cdots$
  • $\dfrac{1}{1 + x^2} = 1 - x^2 + x^4 - \cdots \Rightarrow \arctan x = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots$
TrickSetting $x = 1$ in the $\arctan x$ series gives the Leibniz formula $\dfrac{\pi}{4} = 1 - \dfrac{1}{3} + \dfrac{1}{5} - \cdots$. A favourite IB "show that" question.
TrapThe series for $\ln(1+x)$ only converges for $-1 < x \leq 1$. Integrating outside this interval gives wrong answers.

§10 — Exam Attack Plan All sections

When you see this in the question — reach for that:

TriggerMethod
Numerator $=$ derivative of denominator$\int \dfrac{g'}{g}\,dx = \ln|g|$ — inspection
Product: one factor is the derivative of the otherInspection (reverse chain rule)
Power $\times$ composite functionInspection or substitution
Product: two different function typesIntegration by parts (LIATE)
$\int \ln x$, $\int \arctan x$, $\int \arcsin x$IBP with $u = \ln / \arctan / \arcsin$, $dv = dx$
$\int e^x \sin x$ or $\int e^x \cos x$Cyclic IBP, name it $I$
Rational function (poly / poly)Partial fractions (long divide if needed)
$\int \dfrac{1}{x^2 + bx + c}\,dx$Complete the square $\to \arctan$
"Use substitution $u = \ldots$"Formal substitution, change limits if definite
Area "enclosed by" curvesFind intersections, set up $\int (f - g)$
Area where $f$ changes signSplit at roots, sum $|\ldots|$ of pieces
"Rotate about $x$-axis"$V = \pi \int [f(x)]^2\,dx$
"Rotate about $y$-axis"Invert to $x = g(y)$, $V = \pi\int [g(y)]^2\,dy$
"Total distance travelled"$\int |v|\,dt$ — split at $v = 0$
"Find displacement"$\int v\,dt$ — signed, can be negative
Integral with no closed formMaclaurin series, integrate term by term

Top 5 marks lost — based on past papers

  1. Missing $|x|$ in $\ln$: writing $\ln x$ instead of $\ln|x|$ in $\int \dfrac{1}{x}\,dx$ loses a mark every time. Build $|x|$ into muscle memory.
  2. Signed area instead of geometric area: "Find the area" always means take $|\ldots|$ and split at roots. Computing $\int_a^b f\,dx$ directly when $f$ dips below the axis loses every accuracy mark.
  3. Wrong constant in inspection: $\int 6x(x^2 + 1)^4\,dx \neq \dfrac{(x^2+1)^5}{5} \times 6x$. Always isolate the constant before applying the pattern.
  4. Cyclic IBP — swapping $u$: choosing $u = e^x$ in round 1, then $u = \sin x$ in round 2, gives the trivial $0 = 0$. Keep the same $u$ throughout.
  5. Volume with wrong variable: rotating about the $y$-axis means integrating with respect to $y$. Writing $\pi \int [f(x)]^2\,dx$ for a $y$-axis rotation scores zero marks for method.

Worked Example — IB-Style Integration by Parts (Cyclic)

Question (HL Paper 1 style — 8 marks)

Use integration by parts twice to show that $\int e^{2x}\sin(3x)\,dx = \dfrac{e^{2x}}{13}\bigl(2\sin 3x - 3\cos 3x\bigr) + C$.

Solution

  1. Let $I = \int e^{2x}\sin(3x)\,dx$. Apply IBP with $u = \sin(3x)$, $dv = e^{2x}\,dx$  (or equivalently $u = e^{2x}$, $dv = \sin(3x)\,dx$ — choice is fixed for both rounds). Take $u = e^{2x}$, $dv = \sin(3x)\,dx$, so $du = 2e^{2x}\,dx$ and $v = -\tfrac{1}{3}\cos(3x)$. Then $I = -\tfrac{1}{3}e^{2x}\cos(3x) + \tfrac{2}{3}\int e^{2x}\cos(3x)\,dx$.  (M1)(A1)
  2. Apply IBP again to $J = \int e^{2x}\cos(3x)\,dx$, keeping $u = e^{2x}$, $dv = \cos(3x)\,dx$, so $du = 2e^{2x}\,dx$ and $v = \tfrac{1}{3}\sin(3x)$. Then $J = \tfrac{1}{3}e^{2x}\sin(3x) - \tfrac{2}{3}\int e^{2x}\sin(3x)\,dx = \tfrac{1}{3}e^{2x}\sin(3x) - \tfrac{2}{3}I$.  (M1)(A1)
  3. Substitute back into the expression for $I$: $I = -\tfrac{1}{3}e^{2x}\cos(3x) + \tfrac{2}{3}\!\left[\tfrac{1}{3}e^{2x}\sin(3x) - \tfrac{2}{3}I\right]$.  (M1)
  4. Expand: $I = -\tfrac{1}{3}e^{2x}\cos(3x) + \tfrac{2}{9}e^{2x}\sin(3x) - \tfrac{4}{9}I$. Collect $I$ on the LHS: $\tfrac{13}{9}I = e^{2x}\!\left(\tfrac{2}{9}\sin 3x - \tfrac{1}{3}\cos 3x\right)$.  (A1)
  5. Multiply both sides by $\tfrac{9}{13}$: $I = \dfrac{e^{2x}}{13}\bigl(2\sin 3x - 3\cos 3x\bigr) + C$.  (A1)(AG)

Examiner's note: The fatal error in cyclic IBP is swapping the choice of $u$ in round 2. If you chose $u = e^{2x}$ in round 1, you must keep $u = e^{2x}$ in round 2 — otherwise the second IBP undoes the first and you get $I = I$ (no progress). Also remember to write "$+ C$" on indefinite integrals, even on "show that" questions.

Common Student Questions

How do I decide which integration technique to use?
Check in this order: (1) Is the numerator a (multiple of the) derivative of the denominator? Use $\int \dfrac{g'}{g} = \ln|g|$. (2) Is one factor the derivative of the inside of the other? Use inspection / reverse chain rule. (3) Two different function types (poly $\times$ exp, log $\times$ poly)? Use integration by parts with LIATE. (4) Rational function? Try partial fractions. (5) "Forced substitution $u = \ldots$"? Apply formal substitution and remember to change limits if the integral is definite.
What does LIATE mean and how do I apply it for integration by parts?
LIATE is the priority order for choosing $u$ in $\int u\,dv = uv - \int v\,du$. Pick $u = $ whichever appears first in: Logarithm > Inverse trig > Algebraic > Trig > Exponential. So for $\int x \ln x\,dx$ choose $u = \ln x$ (L beats A); for $\int x \sin x\,dx$ choose $u = x$ (A beats T). The remaining factor (with $dx$) becomes $dv$. Following LIATE almost always leaves a simpler integral on the RHS.
Why is "find the area" different from computing $\int f\,dx$ directly?
$\int f\,dx$ gives signed area — it can be negative when $f$ dips below the $x$-axis. "Find the area" always means geometric (positive) area. If $f$ changes sign on $[a, b]$, find every root of $f$ in the interval, split the integral at each root, take the absolute value of each piece, then add them. Computing $\int_a^b f\,dx$ directly when $f$ changes sign cancels positive and negative regions and gives the wrong answer.
What's the difference between rotating about the $x$-axis and the $y$-axis?
For rotation about the $x$-axis: $V = \pi \int_a^b [f(x)]^2\,dx$, integrating in $x$ with the original function. For rotation about the $y$-axis: you must invert the curve to $x = g(y)$, find the corresponding $y$-limits, and integrate $V = \pi \int_c^d [g(y)]^2\,dy$ with respect to $y$. Writing $\pi \int [f(x)]^2\,dx$ for a $y$-axis rotation is a method-zero error — always check that the variable matches the axis of rotation.
When do I need partial fractions instead of just substitution?
Partial fractions are needed for rational functions $P(x)/Q(x)$ where $Q$ has multiple distinct factors and the numerator is NOT a (multiple of the) derivative of the denominator. If $\deg(P) \geq \deg(Q)$, do polynomial long division first. Then decompose: distinct linear factors give $\dfrac{A}{x-a} + \dfrac{B}{x-b}$; repeated linear give $\dfrac{A}{(x-a)^2} + \dfrac{B}{x-a}$; irreducible quadratic give $\dfrac{Ax + B}{x^2 + c}$. Each piece integrates to $\ln|\cdot|$ or $\arctan$.

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