IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~9 min read

Reverse Chain Rule

Reverse chain rule (RCR) is integration “by inspection”. You spot that the integrand has the form g′(x) · f′(g(x)) — the derivative of an inner function multiplied by an outer derivative — and undo the chain rule directly. The most common useful special case: when the numerator is the derivative of the denominator, the integral is ln of the denominator.

📘 What you need to know

Reverse chain rule (formal): ∫g′(x) f′(g(x)) dx = f(g(x)) + c.
The trigger: integrand contains a composite function whose inside’s derivative also appears (possibly with a coefficient adjustment).
The (ax + b)ⁿ special case: ∫(ax + b)ⁿ dx = 1a(n+1)(ax + b)ⁿ⁺¹ + c, for n ≠ −1.
The ln special case: ∫f′(x)f(x) dx = ln|f(x)| + c. The numerator must be EXACTLY the derivative of the denominator (or a constant multiple of it).
Adjust and compensate: if the integrand has 2x but you need 3x² to fit the pattern, multiply inside by 3 and put 1/3 outside. The two factors cancel — the integral is unchanged.
NOT in the formula booklet: the (ax + b)ⁿ result and the ln pattern aren’t given; you need to recognise them.
RCR vs substitution: every RCR problem can also be done by substitution (u = inside function). RCR is faster when you can spot the pattern; substitution is the safety net.
Always verify: differentiate your answer back to confirm you recover the original integrand.

The pattern — when can RCR be used?

Reverse chain rule ∫ g′(x) · f′(g(x)) dx = f(g(x)) + c

The integrand is the product of (derivative of inside) and (outside function applied to inside). When that pattern holds, the integral is just (outside antiderivative) of (inside). Two extremely common special cases:

(ax + b)ⁿ case

∫ (ax + b)ⁿ dx = (ax + b)ⁿ⁺¹a(n+1) + c

linear inside, power outside — raise the power, divide by new power AND by a

f′/f case → ln|f|

∫ f′(x)f(x) dx = ln|f(x)| + c

numerator IS derivative of denominator — answer is ln of denominator

How to spot it: look at the integrand. Identify the “inside” function (something like x² + 4, or 3x − 2). Compute its derivative. If you see that derivative (or a constant multiple of it) somewhere else in the integrand, RCR applies.

Adjust and compensate

Most RCR questions don’t hand you the exact pattern — you have to massage the integrand. If the integrand has 2x but you need 6x to match (derivative of inside), multiply the integrand by 3 to get 6x, then put 1/3 in front to cancel out. The integral is unchanged.

Example — adjust by a constant ∫ 10x cos(x²) dx = 5 ∫ 2x cos(x²) dx = 5 sin(x²) + c

🧭 Recipe — apply the reverse chain rule

Identify the inside function g(x) — usually the thing inside brackets, or in the denominator, or in the exponent.
Compute g′(x) mentally. Does the integrand contain g′(x) (possibly times a constant)?
Adjust and compensate if needed — multiply integrand by the constant you need, divide by the same constant outside.
Integrate the outside — power rule, sin → −cos, e^(…) → e^(…), 1/(…) → ln|…|.
Verify by differentiating; you should recover the original integrand.

Worked examples

WE 1

The (ax + b)ⁿ special case

Find ∫4(5x − 7)³ dx.

Spot the pattern: linear inside (5x − 7), power outside (n = 3) Use ∫(ax + b)^n dx = (ax + b)^(n+1) / (a(n+1)) + c Here a = 5, n = 3, so a(n + 1) = 5 · 4 = 20 ∫(5x − 7)³ dx = (5x − 7)⁴ / 20 + c Multiply by the coefficient 4 ∫4(5x − 7)³ dx = 4 · (5x − 7)⁴ / 20 + c = (1/5) (5x − 7)⁴ + c ∫4(5x − 7)³ dx = (1/5)(5x − 7)⁴ + c verify: d/dx[(1/5)(5x − 7)⁴] = (1/5) · 4(5x − 7)³ · 5 = 4(5x − 7)³ ✓

WE 2

Polynomial inside — adjust by a constant factor

Find ∫6x(x² + 4)⁵ dx.

Spot the inside: g(x) = x² + 4, so g'(x) = 2x Integrand has 6x, which is 3 × 2x — adjust by pulling out 3 ∫6x(x² + 4)⁵ dx = 3 ∫2x(x² + 4)⁵ dx Now in exact form: 2x is the derivative of x² + 4 ∫2x(x² + 4)⁵ dx = (x² + 4)⁶ / 6 (power up by 1, divide by 6) Multiply by the 3 from outside 3 · (x² + 4)⁶ / 6 + c = (x² + 4)⁶ / 2 + c ∫6x(x² + 4)⁵ dx = (1/2)(x² + 4)⁶ + c verify: d/dx[(1/2)(x² + 4)⁶] = (1/2) · 6(x² + 4)⁵ · 2x = 6x(x² + 4)⁵ ✓

WE 3

Trig with a cubic inside

Find ∫x² cos(x³ + 1) dx.

Spot the inside: g(x) = x³ + 1, so g'(x) = 3x² Integrand has x², which is (1/3) · 3x² — adjust and compensate ∫x² cos(x³ + 1) dx = (1/3) ∫3x² cos(x³ + 1) dx Now in exact form: 3x² is the derivative of x³ + 1, outside function cos → sin ∫3x² cos(x³ + 1) dx = sin(x³ + 1) Multiply by the 1/3 compensation (1/3) sin(x³ + 1) + c ∫x² cos(x³ + 1) dx = (1/3) sin(x³ + 1) + c verify: d/dx[(1/3) sin(x³ + 1)] = (1/3) · cos(x³ + 1) · 3x² = x² cos(x³ + 1) ✓

WE 4

f′/f pattern — exact match, no adjustment

Find ∫8x − 34x² − 3x + 7 dx.

Check the f’/f pattern denominator: f(x) = 4x² − 3x + 7 derivative: f'(x) = 8x − 3 ← exactly the numerator! Apply ∫(f’/f) dx = ln|f(x)| + c ∫(8x − 3)/(4x² − 3x + 7) dx = ln|4x² − 3x + 7| + c ∫(8x − 3)/(4x² − 3x + 7) dx = ln|4x² − 3x + 7| + c discriminant of 4x²−3x+7 is 9 − 112 = −103 < 0, so denominator > 0 for all x — the |·| isn’t strictly needed here, but always write it as a habit

WE 5

f′/f pattern with adjustment by 1/2

Find ∫xx² + 5 dx.

Check the f’/f pattern denominator: f(x) = x² + 5 derivative: f'(x) = 2x numerator: x = (1/2) · 2x — adjust and compensate Pull out 1/2 to get the exact form ∫ x/(x² + 5) dx = (1/2) ∫ 2x/(x² + 5) dx = (1/2) ln|x² + 5| + c ∫ x/(x² + 5) dx = (1/2) ln|x² + 5| + c (also written (1/2) ln(x² + 5) + c since x² + 5 > 0) x² + 5 ≥ 5 > 0 always, so the |·| is never engaged — still good practice to write it

WE 6

Exponential with quadratic inside

Find ∫x e^{x² + 2} dx.

Spot the inside: g(x) = x² + 2 (the exponent), so g'(x) = 2x Integrand has x, which is (1/2) · 2x — adjust and compensate ∫x e^(x² + 2) dx = (1/2) ∫2x e^(x² + 2) dx Now in exact form: 2x is derivative of x² + 2, outside function e^(…) → e^(…) ∫2x e^(x² + 2) dx = e^(x² + 2) Multiply by the 1/2 compensation (1/2) e^(x² + 2) + c ∫x e^(x² + 2) dx = (1/2) e^(x² + 2) + c verify: d/dx[(1/2) e^(x² + 2)] = (1/2) · e^(x² + 2) · 2x = x e^(x² + 2) ✓

💡 Top tips

Practice spotting: the more you do, the faster you recognise the patterns. Most exam-level RCR is one of the six WE archetypes above.
For f′/f, look for the derivative of the denominator in the numerator — that’s the giveaway.
The 1/a trick still works for linear inside: ∫(ax + b)ⁿ dx is the simplest RCR — just divide by a(n + 1).
If you can’t see the pattern, try substitution (next note) — substitution always works where RCR works, just slower.
Verify by differentiating — your answer should differentiate back to the integrand. If it doesn’t, check your “adjust and compensate” factors.

⚠ Common mistakes

Adjust without compensating — multiplying inside the integral by 3 changes the value. You MUST put 1/3 outside to cancel.
Wrong sign on the adjustment — careful when the inside derivative is negative. ∫sin(x²) · x dx: inside is x², derivative 2x. x = (1/2)·2x, so coefficient is +1/2, not −1/2.
Forgetting to divide by a on (ax + b)ⁿ: ∫(3x + 5)⁴ dx = (3x + 5)⁵/15, NOT (3x + 5)⁵/5.
Trying RCR when the pattern isn’t there: ∫x² cos(x) dx is NOT an RCR — the derivative of x isn’t x². (This needs integration by parts, an HL technique covered later.)
Confusing ln(f(x))′ with f′(x)/f(x) the other way round — the ln answer requires numerator = derivative of denominator, not the other way.

Up next: Integration by Substitution. When RCR is hard to spot, or the inside function isn’t quite simple, substitution gives you a step-by-step algorithm — let u = inside, replace dx with du/g'(x), and the integral often simplifies to something you already know how to do. It’s slower than RCR but always works.

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