IB Maths AI HL
Further Integration
Paper 1 & 2
~6 min read
The Reverse Chain Rule
The chain rule differentiates a function of a function; running it backwards integrates one. The trick is to spot a composite together with (a multiple of) the derivative of its inside — integrate the outer function and divide by that inside derivative. When the inside is just linear, this is exactly the “divide by a” you’ve already met.
📘 What you need to know
- The pattern: ∫ g′(x) · f′(g(x)) dx = f(g(x)) + c — the inside’s derivative must be there (up to a constant).
- Linear inside: ∫ (ax+b)n dx = (ax+b)n+1a(n+1) + c — divide by both a and the new index.
- Spotting it: look for a “messy inside” with its derivative sitting beside it as a factor.
- The log case: ∫ g′(x)g(x) dx = ln|g(x)| + c — derivative on top, function on the bottom.
- Fix the constant: if the g′ factor is out by a number, multiply by the reciprocal to balance.
- Check by differentiating — the chain rule should bring you straight back.
Recognising the pattern
Reverse chain rule
∫ g′(x) f′(g(x)) dx = f(g(x)) + c
✗ not stated as a formula — you recognise the pattern
🤔 Why does the inside’s derivative have to be there?
Differentiating f(g(x)) by the chain rule gives f′(g(x)) · g′(x) — the extra g′ factor is the “tail” the chain rule leaves behind. To reverse it, that tail must already be in the integrand (or you supply the missing constant). No g′, no clean reverse — you’d need substitution or another method.
🧠 “Inside and its derivative — a matching pair”
Scan for a composite (something raised to a power, e to something, ln of something) and ask: is the derivative of the inside also a factor? If yes, integrate the outer layer and you’re done. If it’s off by a number, fix it with a constant.
Using it
🧭 Recipe — reverse chain rule
- Identify the inside g(x) and find g′(x).
- Check that g′(x) appears as a factor (allowing for a constant multiple).
- Integrate the outer function, treating g(x) as a single block.
- Balance the constant — multiply by the reciprocal of any number g′ is out by — and add “+ c“.
- Differentiate to check you land back on the integrand.
The three forms you’ll see most: a bracket to a power (ax+b)n, an exponential eg(x) with g′ outside, and a fraction g′g that integrates to ln|g|.
Reverses cleanly
∫ 2x ex² dx
Inside x², derivative 2x is right there → ex² + c.
Doesn’t (yet)
∫ ex² dx
No 2x factor — the reverse chain rule can’t touch it.
Worked examples
Linear inside, a = 2. Raise the index, then divide by a and the new index.
inside = 2x+3, a = 2; index 5 → 6
(2x+3)6 ÷ (2 × 6) = ÷ 12
(2x+3)612 + c
WE 2Find ∫ 2x(x2 + 1)4 dx
Inside x²+1 has derivative 2x — already present. Integrate the outer power.
g = x²+1, g′ = 2x ✓ present
∫ (x²+1)4 · g′ dx → (x²+1)55
(x²+1)55 + c
Inside x², derivative 2x. We only have x, so we’re out by a factor of 2 — balance with 12.
g = x², g′ = 2x; have x → out by ½
∫ x ex² dx = 12 ∫ 2x ex² dx
12 ex² + c
The g′g form: bottom is 2x³+1, its derivative 6x² is on top exactly.
g = 2x³+1, g′ = 6x² ✓ on top
∫ g′g dx = ln|g|
ln|2x³ + 1| + c
WE 5Find ∫ cos x · esin x dx
Inside sin x, derivative cos x — sitting right outside. Integrate e to itself.
g = sin x, g′ = cos x ✓ present
∫ g′ eg dx = eg
esin x + c
💡 Top tips
- Always identify g and g′ first — it tells you instantly if the pattern fits.
- Out by a number? Pull out the reciprocal constant; the reverse chain rule only fixes constant mismatches, not variable ones.
- Differentiate to check every answer — the surest safeguard.
- Linear inside is the easy case: divide by a and the new index.
- Top-over-bottom with matching derivative → think ln.
- Keep the modulus in any ln answer.
⚠ Common mistakes
- Forcing it without g′ — e.g. ∫ ex² dx has no 2x, so the reverse chain rule fails.
- Multiplying instead of dividing by the constant when balancing.
- Trying to cancel a variable (not a constant) mismatch — that needs substitution.
- Forgetting to divide by the new index in the bracket-power case.
- Dropping the modulus in ln|g(x)|.
Next up — Integration by Substitution. The reverse chain rule works when the inside’s derivative is already sitting in the integrand; substitution is the systematic version for when it’s almost there or the algebra is messy. You’ll let u = g(x), swap dx for du using dudx, and turn an awkward integral into a clean one in u.
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