\int u\,dv=uv-\int v\,duVariables
- u: selected function to differentiate
- dv: selected differential to integrate
- du: derivative of u
- v: antiderivative of dv
How to use this formula
Transforms an integral of a product using the product rule in reverse.
Important notes
- Choose u so that differentiating it simplifies the expression.
- Remember the minus sign before the remaining integral.
Quick example
∫x eˣ dx = xeˣ − eˣ + C.
Applicability, worked calculation, and verification
Domain and applicability
Applies to integrals where the integrand can be expressed as a product u·dv and both required derivatives and antiderivatives exist.
Units
- the antiderivative has integrand units multiplied by the input unit
Assumptions and domain checks
- Choose u so that differentiating it simplifies the expression.
- Integration limits, the differential, and any constant of integration must be included where required.
- The function must satisfy the differentiability, continuity, or integrability conditions required by the operation being used.
Do not use this formula when
- Do not use Integration by Parts without checking differentiability, integrability, convergence, interval, and transform-convention requirements.
Boundary and special cases
- For Integration by Parts, check zero, negative, and extreme input values before relying on the result.
- When using Integration by Parts, confirm that denominators, radicals, logarithms, and domain restrictions remain valid for the chosen values.
Equivalent and alternative forms
- Keep the canonical LaTeX form \int u\,dv=uv-\int v\,du for copying; rearrange only after preserving equivalence and domain restrictions.
Worked example
Evaluate ∫x eˣ dx. Choose u = x and dv = eˣdx, giving du = dx and v = eˣ. The result is xeˣ − eˣ + C = eˣ(x − 1) + C.
- Choose u = x because it simplifies when differentiated, and choose dv = eˣ dx.
- Compute du = dx and integrate dv to obtain v = eˣ.
- Apply ∫u dv = uv − ∫v du to get xeˣ − ∫eˣ dx.
- Integrate the remaining term and include the constant: eˣ(x − 1) + C.
Independent verification
Differentiating eˣ(x − 1) gives eˣ(x − 1) + eˣ = xeˣ, which recovers the original integrand.
Common mistakes
- Keep the integration bounds and differential attached to Integration by Parts, and include a constant of integration for an indefinite integral.
- Do not apply a differentiation or integration rule outside its domain or omit endpoint, continuity, or constant-of-integration checks.
Continue the workflow
Use Integration by Parts in your own work
- Check the domainMatch the variables and assumptions to the problem before substituting values.
- Copy the exact notationPreserve grouping, signs, and exponents in
\int u\,dv=uv-\int v\,du. - Edit or convertOpen the expression in the LaTeX editor, then export it for your document or web page.
Review and verification
Last reviewed: 2026-07-26
Review method: Reviewed the notation, variable definitions, applicability conditions, worked workflow, and verification method for Integration by Parts.
Verified against: OpenStax Calculus Volume 1
Automated quality check: Passed the core formula indexing gate.
Formula references
- Calculus Volume 1OpenStax, Rice University — Reviewed limits, derivatives, integrals, and foundational calculus formulas.
Frequently asked questions
What is the Integration by Parts used for?
Transforms an integral of a product using the product rule in reverse.
Can I copy this formula as LaTeX?
Yes. Copy \int u\,dv=uv-\int v\,du or open it in the LaTeX editor.
What should I check before using it?
Confirm that each variable, unit, domain restriction, and assumption matches the problem.