IB Maths AI SLIntegrationPaper 1 & 2∫xn dx~6 min read
Integrating Powers of x
Integration reverses the power rule for differentiation. Where differentiating drops the power by 1, integrating raises it by 1 — then divides by the new power. This note covers that rule, the special cases, and how to handle sums, fractions and products.
📘 What you need to know
The integration power rule: ∫xn dx = xn+1/(n+1) + c — raise the power by 1, then divide by the new power.
A constant multiplier rides along: ∫axn dx = axn+1/(n+1) + c.
Special case: ∫a dx = ax + c — a constant integrates to that constant times x.
The rule fails when n = −1 — you cannot integrate 1/x this way.
Rewrite first: turn fractions like a/xn into ax−n, and expand any products.
Integrate a sum or difference of powers term by term, and never forget the + c.
The integration power rule
Integration reverses the power rule for differentiation. Differentiating drops the power by 1; integrating does the opposite — it raises the power by 1, then divides by that new power.
The integration power rule
∫ axn dx = axn+1n + 1 + craise the power by 1, then divide by the new power · n ≠ −1 · given in the formula booklet
To integrate axn: raise the power by 1, then divide by the new power. Below, 4x3 becomes x4 + c.
A constant multiplier simply rides along, and — because this is an indefinite integral — the result always ends in + c. The rule is given in the formula booklet.
Special cases and negative powers
A plain constant has its own short rule: integrating a number a gives ax.
Integrating a constant
∫ a dx = ax + ca constant integrates to that constant times x — e.g. ∫4 dx = 4x + c
The power rule also handles negative powers — but a fraction has to be rewritten first. A term like 6/x3 becomes 6x−3, which integrates as normal. The one case the rule cannot handle is n = −1: that would mean dividing by zero, so 1/x cannot be integrated this way.
Sums, products and quotients
An expression that is a sum or difference of powers is integrated term by term — integrate each piece separately and add the results, with a single + c at the end.
Products and quotients cannot be integrated term by term as they stand. Expand a product of brackets, or simplify a quotient by dividing through, so that every term is a single power of x — only then does the power rule apply.
🧠Recipe — integrating an expression
Rewrite every term as a power of x: turn a/xn into ax−n, and expand any brackets.
Take one term at a time — a sum or difference is integrated piece by piece.
Apply the power rule: for each axn, raise the power by 1, then divide by the new power.
Integrate any constant term using ∫a dx = ax.
Add a single + c — and convert negative powers back to fractions if asked.
Worked examples
WE 1
A single power-rule term
Find ∫6x2 dx.
raise the power by 1, then divide by the new power 3∫6x2 dx = 6x3/3 + c∫6x2 dx = 2x3 + cthe coefficient 6 rides along; 6 ÷ 3 = 2, and don’t forget the + c.
WE 2
A polynomial, term by term
Find ∫(4x3 − 9x2 + 2x − 5) dx.
integrate each term separately4x3 → 4x4/4 = x4−9x2 → −9x3/3 = −3x32x → 2x2/2 = x2 · −5 → −5x∫(…) dx = x4 − 3x3 + x2 − 5x + cthe constant −5 integrates to −5x; one + c covers the whole expression.
WE 3
A negative power from a fraction
Find ∫(x2 + 6/x3) dx.
rewrite the fraction as a negative power: 6/x3 = 6x−3x2 → x3/36x−3 → 6x−2/(−2) = −3x−2∫(…) dx = x3/3 − 3x−2 + c = x3/3 − 3/x2 + craising the power −3 by 1 gives −2; dividing by −2 flips the sign.
WE 4
A constant and a negative power
Find ∫(8 − 2/x2) dx.
rewrite: 8 − 2x−28 → 8x (the constant rule)−2x−2 → −2x−1/(−1) = 2x−1∫(…) dx = 8x + 2x−1 + c = 8x + 2/x + c−2 ÷ (−1) = +2, so the sign flips — a common slip with negative powers.
WE 5
A product — expand first
Find ∫3x(x − 4) dx.
a product — expand before integrating3x(x − 4) = 3x2 − 12x3x2 → 3x3/3 = x3−12x → −12x2/2 = −6x2∫3x(x − 4) dx = x3 − 6x2 + cyou cannot integrate a product bracket by bracket — expand to a sum of powers first.
WE 6
Full question: find y, then check it
It is given that dy/dx = 12x3 − 6x + 3/x2. (a) Find an expression for y. (b) Verify your answer by differentiating it.
(a) rewrite: dy/dx = 12x3 − 6x + 3x−2, then integrate12x3 → 12x4/4 = 3x4−6x → −6x2/2 = −3x23x−2 → 3x−1/(−1) = −3x−1y = 3x4 − 3x2 − 3/x + c(b) differentiate y to checkd/dx(3x4 − 3x2 − 3x−1 + c) = 12x3 − 6x + 3x−2(a) y = 3x4 − 3x2 − 3/x + c · (b) differentiating gives back 12x3 − 6x + 3/x2integrating a derivative recovers the original function — differentiating your answer is the perfect check.
💡 Top tips
The integration power rule has two moves: raise the power by 1, then divide by the new power.
Never forget the + c — an indefinite integral without it loses a mark.
Rewrite fractions as negative powers before integrating: a/xn becomes ax−n.
The rule fails for n = −1 — 1/x cannot be integrated with the power rule.
Check any integration by differentiating the answer — you should get back what you started with.
âš Common mistakes
Forgetting the constant of integration “+ c“.
Raising the power but not dividing by the new power — or dividing by the old one.
Integrating a product or quotient term by term — expand or simplify it first.
Mishandling negative powers: raising −3 by 1 gives −2, and dividing by −2 flips the sign.
Forgetting that a constant integrates to ax, not to 0.
Next up: Finding the Constant of Integration — using a known point to pin down that “+ c” and turn a whole family of curves into one definite answer. The integration rule itself never changes: rewrite every term as a power of x, raise the power, divide by the new power, + c.
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