IB Maths AA SLTopic 5 — CalculusPaper 1 & 2~8 min read
Reverse Chain Rule
Reverse chain rule is integration by spotting. If you can recognise that an expression looks like the result of chain rule (a composite function multiplied by the derivative of its inside), you can integrate it instantly. The key skill is pattern recognition — once you see it, the answer just falls out.
📘 What you need to know
Reverse chain rule (RCR) is the inverse of the chain rule — it’s used to integrate composite functions by inspection.
Spot the pattern: g'(x) × f'(g(x)) — a composite function with the derivative of its inside sitting next to it.
The general result: ∫g'(x) f'(g(x)) dx = f(g(x)) + c.
Special case: ∫f'(x)/f(x) dx = ln|f(x)| + c (numerator is the derivative of the denominator).
If the coefficient isn’t perfect, adjust and compensate — multiply and divide by the same number.
This extends the (ax+b) results from special functions to any “inside” function.
The pattern to spot
Differentiating a composite like sin(x²) using chain rule gives:
d/dx [sin(x²)] = 2x · cos(x²)
Reverse it: if you see 2x · cos(x²) in an integral, the answer is sin(x²) + c.
Spotting the RCR pattern
Read the integrand and ask: “is something here the derivative of something else?” If yes, you’ve spotted RCR. The answer is whatever differentiates back to the outer function — no chain rule needed in reverse, because it’s already there.
The two patterns
Main RCR pattern
∫ g‘(x) · f‘(g(x)) dx = f(g(x)) + c
Special case — fraction with derivative on top
∫ f‘(x)f(x) dx = ln|f(x)| + c
✓ derivable from formula booklet
The second pattern is just the first one applied to f(x) → ln|f(x)|. If the top of a fraction is exactly the derivative of the bottom, the integral is ln|denominator|.
📍
Example: ∫(3x² + 1)/(x³ + x) dx
Here f(x) = x³ + x, so f'(x) = 3x² + 1 — exactly the numerator. So the integral is just ln|x³ + x| + c. No working needed.
The (ax + b)n rule
Here’s a useful extension to your special-functions toolkit:
Power of a linear function
∫(ax + b)n dx = 1a(n + 1) (ax + b)n + 1 + c
where n ∈ ℚ, n ≠ −1
This is just “raise the power and divide” — the same as ∫xn dx — but you also divide by a because of the linear inside. It’s RCR baked into a formula.
Adjust and compensate
Sometimes the integrand almost fits the RCR pattern but the coefficient is wrong. The fix: multiply by what you need, then divide by the same thing outside the integral. The integral stays equal — you’ve just made it look right.
Adjust & compensate trick
∫ 10x · cos(x²) dx=5 · ∫ 2x · cos(x²) dx
pull out a 5, leaves “2x” inside — now it fits the pattern perfectly
→5 sin(x²) + c
🧠
“Multiply and divide by the same thing”
If you multiply by k inside, divide by k outside (or vice versa). Like adjusting both sides of a balance — the value doesn’t change, but the shape becomes the RCR pattern you want.
Three-step method
Spot · adjust · integrate
Spot the main function. Find the composite (the “outer” piece). Ask: what would I differentiate to get this?
Adjust and compensate. Check the coefficient. If it’s not a perfect match for chain rule’s “× derivative of inside”, multiply and divide to fix it.
Integrate and simplify. Write down the antiderivative of the outer function with the inside unchanged. Add + c.
After some practice, step 2 happens in your head. For trickier expressions (negatives, fractions, weird coefficients) write it out — it’ll save you sign errors.
Worked examples
WE 1
Power of a linear function
Find ∫ 3(7 − 2x)5/3 dx.
spot the standardIt’s (ax + b)^n with a = −2, b = 7, n = 5/3apply formulaI = 3 × [1/(−2 × 8/3) (7 − 2x)^(8/3)] + csimplify: 1/(−2 × 8/3) = −3/16I = −9/16 (7 − 2x)^(8/3) + cremember: divide by both a (= −2) AND (n+1) (= 8/3)!
WE 2
Linear cos with a constant out front
Find ∫12 cos(3x − 2) dx.
step 1 — pull constant outI = ½ ∫cos(3x − 2) dxstep 2 — apply standard, a = 3I = ½ × ⅓ sin(3x − 2) + cI = ⅙ sin(3x − 2) + calways pull constants out first — keeps the working clean!
WE 3
Spot the pattern by inspection
Find ∫ 2x cos(x²) dx.
spot the patterncos(x²) is the composite, x² is the inside2x is the derivative of x² ✓ — perfect RCR!integratecos integrates to sin, inside stays as x²I = sin(x²) + ccheck by differentiating: chain rule gives 2x cos(x²) ✓
WE 4
The f'(x)/f(x) special case
Find ∫3x² + 1x³ + x dx.
check the patterndenominator: f(x) = x³ + xderivative: f'(x) = 3x² + 1 ← matches numerator!apply ∫f’/f = ln|f|I = ln|x³ + x| + cno working needed — once you spot the pattern, just write the answer down!
WE 5
Full adjust & compensate — find f(x) from f'(x)
A curve has gradient function f'(x) = 5x² sin(2x³). Find an expression for f(x).
step 1 — pull constant outf(x) = 5 ∫x² sin(2x³) dxstep 2 — spot the patternmain function: sin(2x³), comes from −cos(2x³)inside: 2x³, derivative: 6x²we have x² but need 6x² → adjust!step 3 — adjust and compensatef(x) = 5 × ⅙ ∫6x² sin(2x³) dxnow it fits: 6x² is exactly the derivative of 2x³step 4 — integrate= 5/6 × [−cos(2x³)] + cf(x) = −5/6 cos(2x³) + ccheck by differentiating: 5/6 × 6x² sin(2x³) = 5x² sin(2x³) ✓
💡 Top tips
Look for products and fractions. RCR shows up when there’s a composite next to its inside derivative — or as f'(x)/f(x).
Differentiate to check. Your answer should differentiate back to the original integrand.
Use the (ax + b)n formula when the inside is linear — much faster than substitution.
Adjust and compensate when the coefficient is off — multiply and divide by the same number.
If you can’t spot the pattern, switch to integration by substitution (next note).
⚠ Common mistakes
Forgetting to divide by a in the (ax + b)n rule. The answer needs 1/(a(n+1)), not just 1/(n+1).
Confusing RCR with chain rule. RCR integrates; chain rule differentiates. The factor of “derivative of inside” is already given in the integrand for RCR.
Bad adjust & compensate. If you multiply by 6 inside, you must divide by 6 outside (not multiply!).
Missing the f’/f pattern. If the top is the derivative of the bottom, it’s ln — not a power rule.
Forgetting + c on indefinite integrals. Easy mark to lose.
Using RCR when the pattern doesn’t fit. If the “inside derivative” isn’t there (and can’t be made to appear with a constant), use substitution instead.
Next up: Integration by Substitution — what to do when RCR is too messy or the pattern is hard to spot. Substitution always works; RCR is just the fast version when you can see it.
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