Integration by Substitution (AA HL)
When an integrand is a chain-rule derivative in disguise - something of the form \(kg'(x)f(g(x))\) - swapping in a new variable \(u = g(x)\) turns it into a standard integral. This page covers when to substitute, how to handle \(du\) and the limits, worked examples, and the mistakes that lose marks. It's part of the broader Integration topic.
17 questions on this sub-topic.
The technique
Covered under IB syllabus reference SL5.10: integration by inspection (reverse chain rule) or by substitution for expressions of the form \(\int kg'(x)f(g(x))\,dx\). Neither approach has a formula-booklet entry - both are techniques you apply, not results you look up.
Integration by substitution
Replace part of the integrand with a new variable \(u\) to turn an awkward integral into a standard one. On the exam, a substitution is provided unless the integral is already of the reverse-chain-rule form.
Not a listed formula - techniqueReverse chain rule (by inspection)
When \(g'(x)\) is already sitting next to \(f(g(x))\), you can often integrate directly without formally introducing \(u\) - useful for quicker questions like \(\int 2x(x^2+1)^4\,dx.\)
Not a listed formula - techniqueNeed the full syllabus wording and formula-booklet reference table? See Integration.
Worked examples
Find \(\displaystyle\int 2x(x^2+1)^4\,dx.\)
Worked solution
\(u = x^2+1,\ du = 2x\,dx.\) M1
\(\int u^4\,du = \dfrac{u^5}{5} + C.\) A1
\(\dfrac{(x^2+1)^5}{5} + C.\) A1
Evaluate \(\displaystyle\int_0^2 \dfrac{x}{x^2+1}\,dx.\)
Worked solution
\(u = x^2+1,\ du = 2x\,dx\): \(\dfrac12\int \dfrac{1}{u}\,du.\) M1
\(= \dfrac12\ln u.\) A1
\(x=0\Rightarrow u=1,\ x=2\Rightarrow u=5\): \(\dfrac12[\ln u]_1^5\) M1 \(= \dfrac12\ln 5 \approx 0.805.\) A1
Use the substitution \(u = 2x + 3\) to find \(\displaystyle\int x(2x+3)^4\,dx.\)
Worked solution
Let \(u = 2x + 3,\) so \(x = \dfrac{u-3}{2}\) and \(du = 2\,dx,\) i.e. \(dx\) M1
\(= \dfrac{1}{2}\,du.\) A1
Substituting gives
\(\int x(2x+3)^4\,dx = \int \dfrac{u-3}{2}\cdot u^4 \cdot \dfrac{1}{2}\,du = \dfrac{1}{4}\int (u^5 - 3u^4)\,du.\) M1
Integrating term by term,
\(= \dfrac{1}{4}\left(\dfrac{u^6}{6} - \dfrac{3u^5}{5}\right) + C = \dfrac{u^6}{24} - \dfrac{3u^5}{20} + C.\) A1
Now
back-substitute \(u = 2x+3.\) A1
Common mistakes
- Forgetting to convert \(dx\) into \(du\). Substituting \(u\) for part of the expression but leaving \(dx\) untouched breaks the integral - every \(x\) and \(dx\) in the original expression needs to be replaced in terms of \(u\) and \(du\).
- Not changing the limits on a definite integral. When you substitute \(u=g(x)\) in a definite integral, the limits must be converted to \(u\)-values too, rather than integrating in \(u\) and switching back to the original \(x\)-limits at the end.
- Reaching for a full substitution when inspection would do. If \(g'(x)\) is already sitting next to \(f(g(x))\), such as \(2x(x^2+1)^4\), you can often integrate directly by the reverse chain rule - writing out a full substitution just adds unnecessary steps.
Ready to practise properly?
16 integration-by-substitution questions, marked instantly like the real exam.
Quick answers
When should I use substitution instead of the reverse chain rule?
Use substitution when the reverse chain rule isn't obvious by inspection, especially on unfamiliar or messy integrands - it turns the awkward expression into a standard integral in a new variable \(u\).
Do I need to change the limits when substituting in a definite integral?
Yes - once you substitute \(u\) for an expression in \(x\), the limits of integration must be converted into \(u\)-values too, rather than converting back to \(x\) at the end.