Integration by parts
Objective. Derive and apply the integration-by-parts identity.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Integration by parts reverses the product rule, using the identity ∫u dv = uv − ∫v du, where u and dv are chosen from the integrand so that ∫v du is simpler than the original integral. A common guide is LIATE (logarithmic, inverse trig, algebraic, trig, exponential) for choosing which factor is u. For ∫x·eˣ dx, let u=x (so du=dx) and dv=eˣdx (so v=eˣ), giving x·eˣ − ∫eˣdx = x·eˣ − eˣ + C. Repeated application is sometimes needed, as with ∫x²eˣdx, where parts must be applied twice to eliminate the polynomial factor entirely.
Worked example
Evaluate ∫x·eˣ dx using integration by parts.
- 1Choose u=x (so du=dx) and dv=eˣdx (so v=eˣ).
- 2Apply the formula: ∫x·eˣdx = x·eˣ − ∫eˣdx.
- 3Integrate the remaining term: ∫eˣdx = eˣ.
- 4Combine: x·eˣ − eˣ + C, which is fully simplified.
Answer: x·eˣ − eˣ + C
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Practice
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Start practisingCommon mistake. Students pick u and dv so that ∫v du is harder than the original integral, for example choosing u=eˣ and dv=x dx in ∫x·eˣdx, which produces a messier integral than the sensible choice u=x.
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