How to Know Which Integration Method to Use (Decision Guide)

Quick answer: To decide which integration method to use, run a fixed order: simplify or rewrite first, then check whether it's a standard result you know on sight, then look for an inner function whose derivative is present (substitution), then a product of two unrelated function types (parts), then a fraction with a factorisable denominator (partial fractions).

For most of last year I did integration by guessing. See an integral, try substitution because that's what we'd just been taught, get stuck, switch to parts, get stuck again, burn eleven minutes on a six-mark question. What fixed it wasn't more methods — it was the order to test them in, which takes twenty seconds and rules out the wrong ones before you write a line. I'm doing IGCSE and A-Levels, and history is my favourite subject, so this is written by someone who needed the system.

The order to test methods in

Run these five checks in order, stopping at the first that fires. Most integrals in a Cambridge, Edexcel, AQA, OCR, AP or CBSE Class 12 paper are decided by check 1, 2 or 3.

1. Can I simplify or rewrite it? Expand brackets, split a fraction term by term, turn roots into fractional powers, or use a trig identity.
2. Is it a standard result? Powers, e^(ax+b), sin and cos of a linear function, 1/(ax+b).
3. Is there an inner function with its derivative sitting there as a factor? Use substitution.
4. Is it a product of two unrelated function types? Use integration by parts.
5. Is it a fraction whose denominator factorises, with the top of lower degree? Use partial fractions.

The order matters because a rewrite often turns a scary integral into check 2, and anyone jumping straight to substitution misses it.

Step one: rewriting solves more integrals than any method

Before choosing a method, ask whether the integrand can be simplified. Three rewrites I use constantly:

• ∫ (x^3 + 2x)/x dx looks like a quotient, so people reach for partial fractions. Split it term by term: x^2 + 2, giving x^3/3 + 2x + C.
• ∫ 1/(x^2 sqrt(x)) dx looks like nothing you know. Rewrite as x^(-5/2) and the power rule gives -2/(3 x^(3/2)) + C.
• ∫ sin^2(x) dx can't be done directly. Use cos(2x) = 1 - 2 sin^2(x), so sin^2(x) = (1 - cos(2x))/2, and it becomes x/2 - sin(2x)/4 + C.

That last one is the classic "there's no method for this" integral. There isn't one — you rewrite instead.

Substitution or parts? One question separates them

Ask whether the two pieces of the integrand are related by differentiation. Related means substitution, unrelated means parts. Compare these, which look almost identical:

• ∫ x e^(x^2) dx — the inner function is x^2 and its derivative 2x is present, up to a constant. Related, so substitution.
• ∫ x e^x dx — x and e^x have nothing to do with each other. Unrelated, so parts.

For parts, pick u using LIATE: Logarithms, Inverse trig, Algebraic, Trig, Exponential. Whichever comes first becomes u, because it either simplifies when differentiated or has no easy integral. That's why ∫ ln(x) dx works at all — take u = ln(x) and dv = 1 dx.

Worked example: the decision run twice

Evaluate ∫ from 0 to 2 of x(x^2 + 1)^3 dx. Rewriting means expanding a degree-7 polynomial, so skip it. Not a standard result either. Check 3 fires: the inner function is x^2 + 1, its derivative is 2x, and an x factor is sitting right there.

Let u = x^2 + 1, so du = 2x dx, meaning x dx = du/2. Change the limits: x = 0 gives u = 1, and x = 2 gives u = 5. The integral becomes (1/2) ∫ from 1 to 5 of u^3 du = (1/2)[u^4/4] from 1 to 5 = (1/8)(625 - 1) = 78.

Now the near-identical twin: ∫ from 0 to 1 of x e^(2x) dx. Nothing to rewrite, not standard, and check 3 fails — the inner function is 2x, whose derivative is the constant 2, so that x out front didn't come from the chain rule. Check 4 fires: algebraic times exponential, unrelated.

By LIATE, algebraic beats exponential, so u = x and dv = e^(2x) dx, giving du = dx and v = e^(2x)/2. Using ∫ u dv = uv - ∫ v du: x e^(2x)/2 - ∫ e^(2x)/2 dx = x e^(2x)/2 - e^(2x)/4. With the limits: (e^2/2 - e^2/4) - (0 - 1/4) = (e^2 + 1)/4, about 2.10.

Two integrals that look the same on the page, separated in seconds by one question about the derivative. To check your intermediate lines rather than just the answer, run the integral through the math solver and compare.

The shapes you should recognise on sight

Some integrals should never reach check 3. (I write mod(...) for modulus signs because the bars don't display reliably here — use them in your exam.)

• ∫ f'(x)/f(x) dx = ln mod(f(x)) + C. So ∫ 6x/(3x^2 + 5) dx = ln mod(3x^2 + 5) + C, because 6x is exactly the derivative of the bottom.
• ∫ f'(x)[f(x)]^n dx = [f(x)]^(n+1)/(n+1) + C. Substitution done in your head.
• ∫ e^(ax+b) dx = e^(ax+b)/a + C, and ∫ 1/(ax+b) dx = (1/a) ln mod(ax+b) + C.

The first is worth real marks. Whenever you see a fraction, differentiate the denominator in your head — if it matches the numerator up to a constant, you're one line from done.

Partial fractions: when it's the only route

Reach for partial fractions when the denominator factorises into linear brackets and the numerator is of lower degree. If the numerator's degree is equal or higher, divide first.

Worked: ∫ (4x + 5)/((x - 1)(x + 2)) dx. Set it equal to A/(x - 1) + B/(x + 2), so 4x + 5 = A(x + 2) + B(x - 1). Put x = 1: 9 = 3A, so A = 3. Put x = -2: -3 = -3B, so B = 1. That leaves two standard logs: 3 ln mod(x - 1) + ln mod(x + 2) + C.

Check the f'(x)/f(x) shape before committing — plenty of questions that look like partial fractions are one-line logs. If the topic still feels blurry, the calculus hub or a fresh framing from explain beats re-reading the same solution.

The mark-scheme details that cost me marks

• Write +C on every indefinite integral. It's an independent mark in A-Level and Cambridge schemes, and I've lost it twice.
• State the substitution explicitly. "Let u = x^2 + 1, du = 2x dx" earns the method mark even if the arithmetic later goes wrong. Silent working earns nothing.
• Change the limits, or convert back to x before substituting numbers. Pick one and be consistent — never evaluate a u-limit into an x-expression.
• Keep the modulus signs in logarithms.
• Differentiate your answer to check it. Twenty seconds, and it catches every sign error.

Test yourself

1. Which method for ∫ x^2 sqrt(x^3 + 4) dx, and why?
2. Rewrite ∫ (2x^4 - x)/x^2 dx so that no method is needed, then integrate it.
3. Which method for ∫ (2x + 1)/((x + 3)(x - 1)) dx, and what should you check before starting?

Answers: 1. Substitution — the derivative of x^3 + 4 is 3x^2 and an x^2 is present. 2. Split into 2x^2 - 1/x, giving 2x^3/3 - ln mod(x) + C. 3. Partial fractions, but check the log shape first: the denominator expands to x^2 + 2x - 3, whose derivative is 2x + 2.

FAQ

How do I know when substitution definitely won't work?

If the extra factor isn't the derivative of the inner function up to a constant, substitution won't clear the x terms. Missing a number is fine, since constants can be juggled. Missing a variable is not, and that's your cue to try parts instead.

Is LIATE always the right way to choose u?

It's a reliable default for Cambridge, Edexcel, AQA, OCR and AP Calculus questions. It works because you want u to simplify when differentiated and dv to be easy to integrate, so if LIATE ever hands you something horrible, swap them and see.

What if I pick a method and get stuck halfway?

Stop after about ninety seconds and go back to the checklist rather than pushing on. Getting stuck usually means step one was skipped, so try rewriting. Timed practice on mock exams trains that stop-and-switch instinct better than untimed homework does.

In short: integration isn't a memory test with dozens of methods, it's a short decision procedure. Simplify, check for a standard result, look for an inner function with its derivative, then an unrelated product, then a factorisable fraction. Run those five checks in order and choosing a method stops being a guess.