Integration by Parts (AA HL)
Integration by parts turns an integral of a product into an easier integral, at the cost of a minus sign and one differentiation. It rescues integrals that the reverse chain rule can't touch, such as \(\int x e^x\,dx\) or \(\int \ln x\,dx\). This page covers the formula, how to choose \(u\) and \(dv\), worked examples, and the mistakes that lose marks. It's part of the broader Integration topic.
24 questions on this sub-topic.
The formula
Covered under IB syllabus reference AHL5.16. The by-parts formula is given in the formula booklet, so the real skill is spotting when to use it and choosing \(u\) and \(dv\) sensibly.
Integration by parts
\[\int u\,\dfrac{dv}{dx}\,dx = uv - \int v\,\dfrac{du}{dx}\,dx\]
Used for products where one factor simplifies on differentiating, such as \(x\ln x\) or \(xe^{2x}\). Sometimes needs repeating.
✓ In the formula bookletRepeated by parts
Apply the formula more than once when \(u\) doesn't vanish after one differentiation, such as \(x^2 e^x\) - each pass hands you a simpler product until it's gone.
A special case is \(\int e^x\sin x\,dx\)-style integrals, where the original integral reappears and you solve for it algebraically.
Not a listed formula - techniqueNeed the full syllabus wording and formula-booklet reference table? See Integration.
Worked examples
Find \(\displaystyle\int x\sin x\,dx.\)
Worked solution
LIATE: let \(u = x,\ dv = \sin x\,dx.\) Then \(du = dx,\ v = -\cos x.\) M1
\(\int x\sin x\,dx = -x\cos x - \int(-\cos x)\,dx\) A1
\(= -x\cos x + \sin x + C.\) A1
Find \(\displaystyle\int x\ln x\,dx.\)
Worked solution
\(u = \ln x,\ dv = x\,dx.\) Then \(du = \tfrac1x dx,\ v = \tfrac{x^2}{2}.\) M1
\(\int x\ln x\,dx = \tfrac{x^2}{2}\ln x - \int \tfrac{x}{2}\,dx.\) A1
\(= \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C.\) A1
Find \(\displaystyle\int \ln x\,dx.\)
Worked solution
\(u = \ln x,\ dv = dx,\) so \(du = \tfrac{1}{x}dx,\ v\) M1
\(= x.\) A1
\(\int\ln x\,dx = x\ln x - \int 1\,dx\) M1
\(= x\ln x - x + C.\) A1
Common mistakes
- Choosing \(u\) and \(dv\) poorly. Pick \(u\) to be the factor that gets simpler when differentiated (like \(x\) or \(\ln x\)) - choosing the other way round usually makes the resulting integral harder, not easier.
- Stopping after one pass when a second is needed. For something like \(\int x^2 e^x\,dx\), one application of the formula still leaves a product to integrate - keep applying by parts until the polynomial factor has fully disappeared.
- Dropping the \(+C\). An indefinite integral by parts is still a family of antiderivatives - forgetting the constant costs an accuracy mark even when every other line is correct.
Ready to practise properly?
24 integration-by-parts questions, marked instantly like the real exam.
Quick answers
What is the formula for integration by parts?
\(\int u\,\dfrac{dv}{dx}\,dx = uv - \int v\,\dfrac{du}{dx}\,dx\). It's given in the formula booklet, so you only need to know when and how to apply it.
How do I choose u and dv in integration by parts?
Pick \(u\) to be the factor that gets simpler when differentiated, such as \(x\) or \(\ln x\), and \(dv\) to be the factor you can integrate directly, such as \(e^x\) or \(\sin x\).