IB Maths AI HL
Integration
Paper 1 & 2
~6 min read
Integrating Powers of x
The power rule for differentiation ran in reverse gives the rule for integration: raise the power by one and divide by the new power — then add “+ c“. It’s the mirror image of “multiply by the power and drop it by one”, and it handles any rational power except n = −1.
📘 What you need to know
- The rule: ∫xn dx = xn+1n + 1 + c, for n ≠ −1 — it’s in the booklet.
- Constant multiple: ∫axn dx = axn+1n + 1 + c.
- Special case: ∫a dx = ax + c (integrating a constant).
- The exception: the rule fails for n = −1, so you can’t integrate 1x this way.
- Rewrite first: roots → fractional powers, fractions → negative powers.
- Term by term: integrate a sum/difference one term at a time (expand products first).
The power rule for integration
Integrating a power of x
∫xn dx = xn+1n + 1 + c (n ≠ −1)
✓ given in the formula booklet
🧠 “Add one, divide by it”
Raise the power by one, then divide by that new power. The exact reverse of differentiating, where you’d multiply down and subtract one. And never forget the “+ c“.
⚠ The n = −1 exception
- The rule would divide by n + 1 = 0, so it can’t integrate 1x = x−1. Every other rational power — positive, negative or fractional — is fine.
Rewriting and integrating term by term
Roots and fractions must become powers of x before you integrate; products must be expanded. Then integrate each term separately.
| Original | Rewrite as |
|---|
| 5∛x | 5x1/3 |
| 4x2 + x2 | 4x−2 + x2 |
| 8x2(2x − 3) | 16x3 − 24x2 (expand) |
🧭 Recipe — integrating powers of x
- Rewrite roots as fractional powers, fractions as negative powers, products expanded.
- Integrate each term: add one to the power, divide by the new power.
- Add “+ c“ once, for the whole expression.
- Tidy up — convert negative/fractional powers back to roots and fractions if needed.
Products and quotients can’t be integrated term by term as they stand — expand or simplify them into a sum/difference of powers first.
Worked examples
The main example: given dydx = 3x4 − 2x2 + 3 − 1√x, find y.
WE 1Rewrite as powers of x
The last term is both fractional and negative.
1√x = x^(−1/2)
dy/dx = 3x⁴ − 2x² + 3 − x^(−1/2)
WE 2Integrate term by term
Add one to each power, divide by the new power; the constant 3 → 3x.
3x⁴ → 3x⁵5
−2x² → −2x³3
3 → 3x
−x^(−1/2) → −x^(1/2)1/2 = −2x^(1/2) (since −½+1 = ½)
y = 35x⁵ − 23x³ + 3x − 2√x + c
WE 3Integrate f(x) = 8x3 − 2x + 4
Sum/difference — straight term by term.
8x⁴4 − 2x²2 + 4x + c
2x⁴ − x² + 4x + c
WE 4Integrate f(x) = 8x2(2x − 3)
A product — expand first, then integrate.
expand: 16x³ − 24x²
16x⁴4 − 24x³3 + c
4x⁴ − 8x³ + c
WE 5Integrate f(x) = 4x2 + 5√x
Rewrite the fraction and the root as powers, then integrate.
= 4x^(−2) + 5x^(1/2)
4x^(−2) → 4x^(−1)−1 = −4x^(−1)
5x^(1/2) → 5x^(3/2)3/2 = 103x^(3/2)
−4x + 103√x³ + c
💡 Top tips
- Rewrite first — every term a power of x before integrating.
- “Add one, divide by it” — commit the pattern to memory for speed.
- One “+ c“ for the whole expression, not per term.
- Expand products — they can’t be integrated as a product.
- Watch fractional powers — e.g. −½ + 1 = ½, and dividing by ½ means × 2.
- Tidy the form — convert back to roots/fractions if the question expects it.
⚠ Common mistakes
- Forgetting “+ c“ on an indefinite integral.
- Trying to integrate 1x with the power rule — the n = −1 case doesn’t work.
- Integrating a product term by term — expand it first.
- Power/fraction slips — dividing by a fractional new power trips people up.
- Differentiating by habit — going down a power instead of up.
Next up — Finding the Constant of Integration. So far every answer has ended in an unknown “+ c“. The next topic shows how a single known point on the curve lets you pin down c exactly — turning a whole family of antiderivatives into one specific function.
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