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Integration by parts: ∫ u dv = uv − ∫ v du

The integration counterpart of the product rule — the formula that handles x·eˣ, x·sin(x), x²·ln(x), and a hundred other classroom integrals.

∫ x · eˣ dx (integration by parts)u = xdv = eˣ dxdu = dxv = eˣ∫ u dv = u·v − ∫ v du∫ x · eˣ dx = x · eˣ − ∫ eˣ dx = x · eˣ − eˣ + C(pick u = x because differentiating it simplifies the integrand)

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What this shows

The product rule for differentiation says d(uv) = u dv + v du. Integrate both sides:

    uv = ∫ u dv + ∫ v du
    ⟹  ∫ u dv = uv − ∫ v du

That is integration by parts. The skill is in choosing which factor of the integrand to call u and which to call dv: pick u to be something that becomes simpler when differentiated, and dv to be something easy to integrate. A common mnemonic for the priority order is LIATE — Logarithmic, Inverse trig, Algebraic, Trig, Exponential — choose u as the leftmost type present.

For ∫ x · eˣ dx (algebraic times exponential, so u is the algebraic part):

    u  = x           →  du = dx
    dv = eˣ dx       →  v  = eˣ
    ∫ x·eˣ dx = u·v − ∫ v du
              = x·eˣ − ∫ eˣ dx
              = x·eˣ − eˣ + C

The single application reduced the original integral (with an x factor) to a simpler integral (without one). That's the point of the technique.

Where it shows up

Integration by parts is the standard tool whenever the integrand is a product where one factor's derivative is simpler than itself: log(x) times anything, polynomials times sin/cos, polynomials times exponentials. Repeated application eventually reduces a polynomial-times-exponential integral to a pure exponential, which integrates trivially.

In probability theory, the expectation of a non-negative random variable can be written as E[X] = ∫₀^∞ P(X > t) dt — a result that comes from one careful application of integration by parts on the definition of expectation. In Fourier analysis, it gives the smoothness-decay relationship between a function and its Fourier transform.

Frequently asked questions

How do I choose u and dv?

Use LIATE as a starting guide: among the factors of the integrand, choose u from the highest-priority type listed (Log, Inv-trig, Alg, Trig, Exp) and dv as the rest. This usually but not always works; if the resulting integral is harder than the original, swap the choices.

What if it needs more than one application?

Apply the rule again. For ∫ x²·eˣ dx, first take u = x², dv = eˣdx → x²eˣ − ∫ 2x·eˣ dx; the new integral is then handled by another integration by parts. A tabular method (sometimes called the DI method) speeds this up by columnising u's derivatives next to v's antiderivatives.

Why does the formula not contain an integration constant?

It's hidden inside the indefinite integrals on both sides. Whatever constant you add on the right cancels the constant on the left. In practice, write +C only at the very end.

What's the definite-integral version?

∫_a^b u dv = [u·v]_a^b − ∫_a^b v du. Same formula, with the boundary term evaluated at the limits. The +C constant disappears because every antiderivative is evaluated between fixed endpoints.

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