A-Level Maths integration revision: every method, worked
A-Level integration covers five techniques: recognition (reverse chain rule), substitution, integration by parts, partial fractions leading to logs, and separating variables in differential equations — plus definite integration for areas under and between curves and the trapezium rule. The revision task is learning to choose the method fast: the exam does not tell you which to use, and that decision is where most marks are actually won or lost.
Integration is the topic where A-Level students most reliably lose marks, not because the techniques are hard but because the exam never labels them. A question that needs substitution looks identical to one that needs parts until you have tried to classify a few hundred of them. Here is the full toolkit, in the order you will meet it, and how to choose fast.
The specification, in plain English
| Method | When to reach for it | Typical look |
|---|---|---|
| Standard results | Anything in the formula booklet — learn them cold anyway | ∫e^(ax), ∫sin(ax), ∫xⁿ |
| Reverse chain rule | A function of a function with its derivative nearby | ∫2x(x²+1)⁵ dx, ∫cos x·sin³x dx |
| Substitution | A composite the reverse-chain shortcut won't reach | ∫x√(x+1) dx |
| By parts | A product where one factor simplifies on differentiating | ∫x e^x dx, ∫x cos x dx, ∫ln x dx |
| Partial fractions | A rational function the others won't touch | ∫(5x+1)/(x²−x−6) dx |
| Separation of variables | dy/dx given as a product or quotient of x and y | dy/dx = ky, dy/dx = xy + y |
The recognition patterns worth memorising
- ∫ f′(x)/f(x) dx = ln|f(x)| + c — numerator is the denominator's derivative. Solves ∫tan x dx, ∫cot x dx and half of all substitution questions in disguise.
- ∫ f′(x)·[f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/(n+1) + c — the 'spot the derivative' pattern.
- ∫ f′(x)·e^(f(x)) dx = e^(f(x)) + c.
- If you see √(a² − x²), think trig substitution x = a sin θ (yes, it's on the spec).
Worked questions
Use the substitution u = 1 + x² to find ∫ 4x(1 + x²)³ dx.
- du/dx = 2x, so du = 2x dx and 4x dx = 2 du.
- Substitute: ∫4x(1+x²)³ dx = ∫2u³ du.
- Integrate: 2u³ → u⁴/2.
- Back-substitute: u⁴/2 = (1 + x²)⁴/2 + c.
- Check by differentiating — the chain rule returns 4x(1+x²)³. Always available, always worth twenty seconds.
(1 + x²)⁴/2 + c. The giveaway was the 4x outside being (a multiple of) the inside's derivative.
Find ∫ x·sin x dx.
- Choose u = x (simplifies on differentiating) and dv/dx = sin x.
- Then du/dx = 1 and v = −cos x.
- Apply ∫u dv = uv − ∫v du: ∫x sin x dx = −x cos x − ∫(−cos x) dx.
- ∫cos x dx = sin x, so −x cos x + sin x.
−x cos x + sin x + c. Choosing u = sin x instead gives a worse integral — the u choice is the whole skill.
Solve dy/dx = y·x with y = 2 when x = 0, giving y in terms of x.
- Separate: dy/y = x dx.
- Integrate both sides: ln|y| = x²/2 + c.
- Use the initial condition: y = 2 at x = 0 gives ln 2 = c.
- So ln y = x²/2 + ln 2, and exponentiating: y = 2e^(x²/2).
- Marks here are lost by students who forget the +c until the end and then can't find it — apply the condition as soon as it exists.
y = 2e^(x²/2). Separate, integrate, then apply the condition — in that order.
Areas and the trapezium rule
Area questions have their own traps. The area between a curve and the x-axis is a definite integral only when the curve stays on one side — if it crosses, integrate each part separately or the regions cancel. For the area between two curves, find the intersection points first; they are your limits, and getting them is often the hard half of the question. The trapezium rule is a Year 1 survival: it approximates, and the follow-up question — 'is your answer an over- or under-estimate?' — is answered by the curve's shape, not by redoing it with more strips.
Where marks get dropped
- The missing +c — worth a mark on indefinite integrals and a whole deduction in differential equations.
- Not changing the limits on a definite-integral substitution — either convert limits to u or convert back to x.
- By parts with the wrong u: if the leftover integral is harder, swap the choice.
- ln|f(x)| written without the modulus — required when the argument could be negative.
- In DEs, integrating one side only, or treating k as the constant of integration.
- ∫e^(2x) dx written as 2e^(2x) — the derivative of the inside divides, not multiplies.
Your integration checklist
- I can classify any integral into a method within fifteen seconds of reading it.
- I know the three recognition patterns (ln, power, exponential) without the formula booklet.
- I can run a full substitution including changing the limits.
- I can integrate by parts twice in one question — ∫x²e^x and friends.
- I can split into partial fractions and integrate to logs.
- I can separate variables, apply an initial condition and give y explicitly.
- I can find the area between two curves, including finding the intersections first.
The moment integration clicks is when the method choice becomes automatic — and that only comes from doing classified, then unclassified, then mixed sets. If a mark scheme's substitution looks like magic, ask Lumi to show you what gave it away. It draws the reasoning, not just the algebra.
What to take from this
- The hard part is choosing the method, not executing it — classify every integral before starting.
- ∫ f′(x)/f(x) dx = ln|f(x)| and ∫ f′(x)·f(x)ⁿ dx cover a surprising share of exam questions.
- By parts: choose u as the part that simplifies when differentiated (usually x or ln x).
- Differential equations: separate variables, integrate both sides, and the +c becomes a real constant via the initial condition.
- The +c, the limits and the modulus signs are where examiners set traps — check them first.
Questions people also ask
Look at the structure: if the integrand contains a function and (roughly) its derivative, it is substitution or recognition. If it is a product of two unrelated types — polynomial times exponential or trig — it is parts. If you can differentiate one factor down to a constant, parts is the answer.
Yes — it is the classic parts question with u = ln x and dv/dx = 1, giving x·ln x − x + c. It appears constantly because it tests whether you can manufacture a product from a single function.
The booklet covers the basics but not the patterns, and checking it costs time on every question. e^(ax), sin(ax), cos(ax), 1/x and xⁿ should be instant — the booklet is for the ones you meet rarely.
Approximating an integral you cannot do exactly — the exam pairs it with a follow-up asking whether your estimate is an over- or underestimate, which depends on whether the curve is concave or convex over the interval.