IB Maths AA HL
Topic 5 — Calculus
Paper 1 & 2
~7 min read
Integrating Powers of x
The core integration rule: raise the power by 1, then divide by the new power. Works for any rational power except n = −1 (which would mean dividing by zero). Reciprocals and roots have to be rewritten as powers of x first. Sums and differences integrate term by term; products and quotients have to be expanded or simplified first.
📘 What you need to know
- Power rule: ∫xn dx = xn+1/(n+1) + c, for n ∈ ℚ, n ≠ −1 (formula booklet).
- With a constant coefficient: ∫axn dx = axn+1/(n+1) + c — the constant just rides along.
- Constant term: ∫a dx = ax + c (think of “a” as a·x⁰).
- Sums and differences: integrate term by term.
- Reciprocals → negative powers: 1/x² = x⁻², 5/x³ = 5x⁻³ before integrating.
- Roots → fractional powers: √x = x1/2, ∛x = x1/3, 1/√x = x−1/2.
- Products and quotients: expand or simplify BEFORE integrating — there’s no general “product rule for integration” at this level.
- The exception n = −1: ∫(1/x) dx ≠ x⁰/0. It’s actually ln|x| + c (covered later).
The power rule
Integration — power rule (formula booklet)
∫ xn dx = xn+1n + 1 + c (n ≠ −1)
Mantra: “Raise the power by one, divide by the new power.” Going up by 1 then dividing by the new power undoes the differentiation rule (which dropped the power by 1 and multiplied by the old power).
With a coefficient
∫ axn dx = axn+1n + 1 + c
the coefficient stays put — only the power changes
Constant term
∫ a dx = ax + c
a is a·x⁰, integrate to a·x¹/1 = ax
Rewriting before integrating
The power rule needs the integrand written as a sum of terms of the form axn. So before integrating: turn roots into fractional powers, reciprocals into negative powers, products into expanded polynomials, quotients into divided-out sums.
| Original form | Rewrite as power of x | Then integrate to… |
|---|
| √x | x1/2 | (2/3) x3/2 |
| 1/√x | x−1/2 | 2 x1/2 = 2√x |
| 1/x² | x−2 | −x−1 = −1/x |
| 1/x³ | x−3 | −x−2/2 = −1/(2x²) |
| ∛x | x1/3 | (3/4) x4/3 |
🧭 Recipe — integrate a sum of powers of x
- Rewrite each term as a power of x (roots → fractional, reciprocals → negative, constants → x⁰).
- Expand any products and simplify any quotients until you have a clean sum/difference of axn terms.
- Apply the power rule to each term: raise the power by 1, divide by the new power.
- Add the constant of integration “+ c” once at the end.
- Simplify the final answer — rewrite fractional/negative powers back as roots/reciprocals if the question asks for a specific form.
Worked examples
WE 1Basic polynomial integration
Find ∫(4x³ + 6x − 5) dx.
Integrate term by term, raising the power by 1 and dividing
∫ 4x³ dx = 4x⁴/4 = x⁴
∫ 6x dx = 6x²/2 = 3x²
∫ (−5) dx = −5x (special case for constants)
Combine and add + c
∫(4x³ + 6x − 5) dx = x⁴ + 3x² − 5x + c
∫(4x³ + 6x − 5) dx = x⁴ + 3x² − 5x + c
verify: d/dx(x⁴ + 3x² − 5x + c) = 4x³ + 6x − 5 ✓
WE 2Negative powers — reciprocals
Find ∫(2x + 5/x³) dx.
Rewrite the reciprocal as a negative power
5/x³ = 5x⁻³
→ integrand: 2x + 5x⁻³
Integrate term by term
∫ 2x dx = 2x²/2 = x²
∫ 5x⁻³ dx = 5x⁻²/(−2) = −(5/2)x⁻² = −5/(2x²)
Combine and add + c
= x² − 5/(2x²) + c
∫(2x + 5/x³) dx = x² − 5/(2x²) + c
when n = −3, new power n + 1 = −2 — careful with the sign of the divisor
WE 3Fractional powers — roots
Find ∫(6√x + 1/√x) dx.
Rewrite each root as a fractional power
6√x = 6x^(1/2)
1/√x = x^(−1/2)
Integrate term by term
∫ 6x^(1/2) dx = 6 · x^(3/2)/(3/2) = 6 · (2/3) x^(3/2) = 4x^(3/2)
∫ x^(−1/2) dx = x^(1/2)/(1/2) = 2x^(1/2) = 2√x
Combine and add + c
= 4x^(3/2) + 2√x + c
∫(6√x + 1/√x) dx = 4x^(3/2) + 2√x + c (also written 4x√x + 2√x + c)
dividing by a fraction = multiplying by its reciprocal: ÷(3/2) means ×(2/3)
WE 4Mixed: reciprocal + cube root + constant
Find ∫(8/x² − 3x2/3 + 7) dx.
Rewrite the reciprocal; root term and constant are already in usable form
8/x² = 8x⁻²
Integrate term by term
∫ 8x⁻² dx = 8x⁻¹/(−1) = −8x⁻¹ = −8/x
∫ −3x^(2/3) dx = −3 · x^(5/3)/(5/3) = −3 · (3/5) x^(5/3) = −(9/5) x^(5/3)
∫ 7 dx = 7x (special case)
Combine and add + c
= −8/x − (9/5) x^(5/3) + 7x + c
∫(8/x² − 3x^(2/3) + 7) dx = −8/x − (9/5)x^(5/3) + 7x + c
three different powers in one integral — the rule applies to each independently
WE 5Product — expand FIRST, then integrate
Find ∫(2x + 1)(3x − 4) dx.
Expand the product
(2x + 1)(3x − 4) = 6x² − 8x + 3x − 4 = 6x² − 5x − 4
Integrate term by term
∫ 6x² dx = 6x³/3 = 2x³
∫ −5x dx = −5x²/2 = −(5/2)x²
∫ −4 dx = −4x
Combine and add + c
= 2x³ − (5/2)x² − 4x + c
∫(2x + 1)(3x − 4) dx = 2x³ − (5/2)x² − 4x + c
NEVER integrate two factors separately. ∫(2x+1)(3x−4) dx ≠ [∫(2x+1)dx] · [∫(3x−4)dx]
WE 6Quotient — simplify FIRST, then integrate
Find ∫(x³ − 2x²)/√x dx.
Split the quotient and divide each term by √x = x^(1/2)
x³/x^(1/2) = x^(3 − 1/2) = x^(5/2)
2x²/x^(1/2) = 2x^(2 − 1/2) = 2x^(3/2)
→ integrand: x^(5/2) − 2x^(3/2)
Integrate term by term
∫ x^(5/2) dx = x^(7/2)/(7/2) = (2/7) x^(7/2)
∫ −2x^(3/2) dx = −2 · x^(5/2)/(5/2) = −2 · (2/5) x^(5/2) = −(4/5) x^(5/2)
Combine and add + c
= (2/7)x^(7/2) − (4/5)x^(5/2) + c
∫(x³ − 2x²)/√x dx = (2/7)x^(7/2) − (4/5)x^(5/2) + c
subtracting powers when dividing: x^a / x^b = x^(a−b). Don’t try to integrate the quotient as written
💡 Top tips
- Rewrite first, integrate second — every term should be axn before the power rule is applied.
- Add 1 to the power, then divide by the new power — this is the opposite of differentiation, where you multiplied by the old power and subtracted 1.
- Don’t forget the + c on indefinite integrals — it’s a common mark-loser.
- Verify by differentiating your answer — d/dx(answer) should give the integrand back.
- For n = −1: the power rule fails. ∫(1/x) dx = ln|x| + c — comes later in the chapter.
⚠ Common mistakes
- Sign errors with negative powers — when n = −3, new power is −2 (not 4). The “+1” makes negative numbers LESS negative.
- Integrating a product as a product of integrals — totally wrong. Always expand a product first.
- Integrating a quotient as a quotient of integrals — also wrong. Simplify or rewrite as a sum of powers first.
- Forgetting + c — indefinite integrals MUST end in “+ c“.
- Confusing the rule — integration RAISES the power; differentiation LOWERS it. Opposite operations, opposite shifts.
Up next: Finding the Constant of Integration. So far you’ve left + c as an unknown — but if the question gives you a point on the curve (or an initial condition like f(0) = 5), you can substitute and solve for c, pinning down the exact antiderivative. This is the difference between “an” antiderivative and “the” specific one.
Need help with Calculus?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
Book Free Session →