Basic Integration (AI HL)
Integration undoes differentiation - and the power rule is the workhorse behind almost every antiderivative you'll find at this level. This page covers integrating polynomials and index expressions term by term, and the constant of integration that goes with every indefinite integral. It's part of the broader Integration topic.
44 questions on this sub-topic.
The core formula
Covered under IB syllabus reference SL5.5. The power rule handles any term of the form \(x^n\) - the main job is rewriting roots and fractions in index form first, so it clearly applies.
Power rule for integration
\[\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+c,\ n\neq-1\]
Add one to the power, then divide by the new power. Integrate a sum of terms one term at a time, and add a single \(+c\) at the end.
✓ In the formula bookletIntegration by substitution
\[\int f(g(x))\,g'(x)\,dx\]
If the integrand is a function of \(g(x)\) multiplied by \(g'(x)\), the power rule alone isn't enough - substitute \(u=g(x)\) first, covered on its own page.
✓ Named in the formula bookletOnce you're integrating between limits, see the parent Integration page for GDC methods.
Worked examples
Find \(\displaystyle\int (6x^2 - 4x + 5)\,dx\).
Worked solution
Integrate term by term. \(\int(6x^2-4x+5)\,dx=2x^3-2x^2+5x+C.\) M1 A1
Find.
(a) \(\displaystyle\int \dfrac{2x}{x^2+5}\,dx\)
(b) \(\displaystyle\int \cos(2x+1)\,dx\)
Worked solution
(a) Numerator is the derivative of the denominator: M1
\(\ln|x^2 + 5| + C.\) A1
(b) Reverse chain on \(\cos(2x+1)\): M1
\(\tfrac12\sin(2x+1)\) A1
\(+ C.\) A1
A curve has gradient \(\dfrac{dy}{dx} = 3x^2 - 4\) and passes through \((2, 5).\) Find the equation of the curve.
Worked solution
\(y = x^3 - 4x + C.\) M1 A1
At \((2,5)\): \(5 = 8 - 8 + C\) M1
\(\Rightarrow C = 5.\) A1
So \(y = x^3 - 4x + 5.\) A1
A curve has gradient \(\dfrac{dy}{dx}=6x-2\) and passes through \((1,4)\).
(a) Find \(y\) in terms of \(x\).
(b) Find \(y\) when \(x=3\).
Worked solution
(a) \(y = 3x^2 - 2x + C.\) At \((1,4)\): \(C\) M1
\(= 3.\) A1
So \(y = 3x^2 - 2x + 3.\) A1
(b) \(y = 27 - 6 + 3\) M1
\(= 24.\) A1
Common mistakes
- Forgetting the \(+C\). It's an easy mark to lose - every indefinite integral needs the constant of integration, even in a multi-part question where it seems like an afterthought.
- Dividing by the old power instead of the new one. The rule is "add one, then divide by the new power" - dividing by \(n\) instead of \(n+1\) is a common slip, especially when the working is rushed.
- Applying the power rule when \(n=-1\). \(\int x^{-1}\,dx\) is \(\ln|x|+C\), not \(\dfrac{x^0}{0}\) - the power rule explicitly breaks down at this one value, and it's easy to apply it anyway out of habit.
Ready to practise properly?
44 basic-integration questions, marked instantly like the real exam.
Quick answers
What is the power rule for integration?
\(\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+c,\) for \(n\neq-1\): add one to the power, then divide by the new power.
Why does every indefinite integral need a +c?
Because any constant disappears when you differentiate, so infinitely many antiderivatives differ only by a constant - the \(+c\) represents that whole family, and it's a mark on its own on IB mark schemes.