IB Maths AI HL
Further Integration
Paper 1 & 2
~6 min read
Integration by Substitution
Substitution is the systematic version of the reverse chain rule. You let u stand for the “messy inside”, swap every x and dx for u and du, and an awkward integral collapses into a simple one. Get the du step right and the rest is routine integration.
📘 What you need to know
- The idea: choose u = g(x) (usually the inside of a composite) and rewrite the whole integral in terms of u.
- The swap: from u = g(x), get dudx = g′(x), then treat it as du = g′(x) dx.
- Goal: the new integral must be entirely in u — no stray x left.
- Indefinite: integrate in u, then substitute back x and add “+ c“.
- Definite: either change the limits to u-values (cleaner) or convert back to x first.
- Choosing u: pick the inside whose derivative also appears — that’s what makes the dx disappear.
The method
Integration by substitution
∫ f(g(x)) g′(x) dx = ∫ f(u) du, where u = g(x)
✓ substitution result is in the formula booklet
🤔 Why can we treat du = g′(x) dx?
The chain rule says dudx = g′(x). Although dudx is a single derivative, in integration it behaves like a ratio, so multiplying across by dx gives du = g′(x) dx. That tidy swap is exactly what replaces the x-part of the integral with a u-part.
🧠 “Let u, find du, swap, integrate, swap back”
Five beats: let u = inside, find du, swap everything into u, integrate, then swap back to x (or change limits for a definite integral).
Carrying it out
🧭 Recipe — substitution
- Let u = g(x) — the inside whose derivative is also present.
- Differentiate: dudx = g′(x), so du = g′(x) dx (rearrange for dx if needed).
- Substitute to get an integral purely in u.
- Integrate in u.
- Finish: substitute x back (indefinite, + c) or change limits to u and evaluate (definite).
Definite integrals — change the limits. If u = g(x), then the lower limit x = a becomes u = g(a) and the upper x = b becomes u = g(b). Then you never need to convert back to x.
Worked examples
WE 1Find ∫ 2x(x2 + 1)4 dx
Let u = x²+1, so du = 2x dx — the 2x dx is already there.
u = x²+1, du = 2x dx
∫ u4 du = u55
(x²+1)55 + c
Let u = x²+3, du = 2x dx. We only have x dx, so x dx = 12 du.
u = x²+3, du = 2x dx → x dx = ½ du
12 ∫ u5 du = 12 · u66
(x²+3)612 + c
WE 3Find ∫ (2x + 1)(x2 + x + 4)3 dx
Let u = x²+x+4, so du = (2x+1) dx — matched exactly.
u = x²+x+4, du = (2x+1) dx
∫ u3 du = u44
(x²+x+4)44 + c
WE 4Evaluate ∫01 2x(x2 + 1)4 dx by changing the limits
Same substitution as WE 1, but convert the limits: x=0 → u=1, x=1 → u=2.
u = x²+1; x:0→1 gives u:1→2
∫12 u4 du = u55 from 1 to 2
= 325 − 15
315 (= 6.2)
WE 5Find ∫ cos x (sin x)3 dx
Let u = sin x, so du = cos x dx — the cos x dx is present.
u = sin x, du = cos x dx
∫ u3 du = u44
(sin x)44 + c
💡 Top tips
- Pick the inside of the composite as u — the one whose derivative appears.
- Rearrange for dx and substitute so no x remains before integrating.
- Definite integrals: change the limits to u — it’s cleaner than converting back.
- Indefinite: don’t forget to swap back to x and add “+ c“.
- Constant out by a number? Factor it out — substitution handles that fine.
- Differentiate to check your final answer.
⚠ Common mistakes
- Leaving an x in the u-integral — you can’t integrate a mix of variables.
- Forgetting to change the dx into du.
- Not changing the limits (or changing them but still substituting back) in a definite integral.
- Forgetting to substitute back to x in an indefinite integral.
- Mishandling the constant when du = k dx — divide, don’t multiply.
That wraps up Further Integration. The unit built one idea in layers: you started with the standard integrals for the special functions (1x, ex, sin, cos), met the reverse chain rule for spotting a composite alongside its inside’s derivative, and finished with substitution — the systematic tool that makes that reversal work every time by switching to a new variable u. Differentiation and integration stay mirror images: the chain rule going forwards, substitution coming back.
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