Integration by Substitution (AI HL)
Some integrals aren't a straight power-rule job - the trick is spotting a hidden chain rule and undoing it with a substitution \(u=g(x)\). This page covers how to recognise the pattern, carry out the swap cleanly, and get back to the original variable at the end. It's part of the broader Integration topic.
11 questions on this sub-topic.
The two ideas you need
Covered under IB syllabus reference AHL5.11: "Integration by inspection, or substitution of the form \(\int f(g(x))g'(x)\,dx\)." Substitution is really just the power rule applied after a change of variable, so both ideas below work together.
Integration by substitution
\[\int f(g(x))\,g'(x)\,dx\]
If the integrand is a function of \(g(x)\) multiplied by \(g'(x)\), substitute \(u=g(x)\) to simplify it into a standard integral.
✓ Named in the formula bookletPower rule for integration
\[\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+c,\ n\neq-1\]
This is what you apply once the substitution has turned the integral into \(\int u^n\,du\) - it's the reason the substitution is worth doing in the first place.
✓ In the formula bookletFor evaluating the resulting integral numerically on a GDC, see the parent Integration page.
Worked examples
Find \(\displaystyle\int 2x(x^2+1)^3\,dx\) using the substitution \(u=x^2+1\).
Worked solution
\(u = x^2 + 1 \Rightarrow du\) M1
\(= 2x\,dx.\) A1
Integral \(= \int u^3\,du\) A1
\(= \tfrac{u^4}{4} + C\) A1
\(= \tfrac14(x^2+1)^4 + C.\) A1
Find \(\displaystyle\int \dfrac{6x^2}{x^3 + 5}\,dx.\)
Worked solution
the numerator is \(2\times\) the derivative of \(x^3 + 5.\) M1 A1
\(\int \dfrac{6x^2}{x^3+5}\,dx\) M1 \(= 2\ln|x^3 + 5| + C.\) A1
Find \(\displaystyle\int \dfrac{(\ln x)^2}{x}\,dx.\)
Worked solution
\(u = \ln x \Rightarrow du = \dfrac{1}{x}\,dx.\) M1
\(\int u^2\,du = \dfrac{u^3}{3}+C.\) A1
\(= \dfrac{(\ln x)^3}{3}+C.\) A1
Common mistakes
- Forgetting to back-substitute. An answer left in terms of \(u\) is incomplete - always replace \(u\) with \(g(x)\) again once you've integrated, so the final answer is in terms of the original variable.
- Missing a constant factor between \(du\) and the integrand. If \(du=2x\,dx\) but the integral only has \(x\,dx\) sitting in it, that missing factor of \(2\) needs to be accounted for (often by dividing outside the integral), not ignored.
- Trying to substitute when there's no matching \(g'(x)\) present. Substitution only simplifies things when the derivative of the inner function is genuinely a factor of the integrand - if it isn't, a different technique (or straight expansion) is needed instead.
Ready to practise properly?
11 substitution questions, marked instantly like the real exam.
Quick answers
When should I use substitution to integrate?
When the integrand looks like a function of \(g(x)\) multiplied by \(g'(x)\) - for example a chain-rule-style expression where an inner function's derivative is sitting alongside it.
Do I need to change du back to dx at the end?
Yes, for an indefinite integral - once you've integrated in terms of \(u\), always substitute \(u=g(x)\) back in so the final answer is written in terms of \(x\) again.