Integration by parts and cyclic integrals (IB Math AA HL)
Updated October 2026
Quick answer: ∫u (dv/dx) dx = uv − ∫v (du/dx) dx. Choose u to be the factor that gets simpler when you differentiate it (a logarithm, then a power of x), and dv/dx to be the factor you can integrate.
∫u·dv = uv - ∫v·du, with the LIATE rule for picking u. HL AA Paper 3 loves cyclic integrals like ∫ex·sin(x) dx where you apply parts twice.
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Choose u to be the factor that becomes simpler when you differentiate it: a logarithm first, then a power of x. The other factor, dv/dx, must be one you can integrate.
Where can I practise integration by parts and cyclic integrals questions?
On the Advanced Integration questions page: 52 exam-style HL AA questions with full mark schemes, plus free step-by-step worked examples.
Can I use my calculator for integration by parts and cyclic integrals?
In HL AA, Paper 1 is non-calculator and the other papers allow a GDC, so learn the exact method as well as the calculator one.