04 // Interactive Formula Compendium // CALCULUS-ALGEBRA∫ u dv = u · v − ∫ v du

Integration by Parts Formula — LIATE Rule & Definite Integral Form

Integration by parts states that ∫ u dv = u v − ∫ v du, or for definite integrals ∫_a^b u(x)v'(x)dx = [u(x)v(x)]_a^b − ∫_a^b v(x)u'(x)dx, serving as the integral counterpart to the product rule.

Formula Reference

Main Formula

∫ u dv = u · v − ∫ v du

Rearranged Forms

∫_a^b u(x) v'(x) dx = [u(x) v(x)]_a^b − ∫_a^b v(x) u'(x) dx

d(u · v) = u dv + v du (Product Rule Differential Form)

∫ x e^(kx) dx = (x/k − 1/k²) e^(kx) + C

Variables & Units

Variables & Units — Integration by Parts Formula — LIATE Rule & Definite Integral Form
Symbol	Name	Unit
I	Definite Integral ∫₀ˣ t·e^(k·t) dt via Parts	dimensionless
x	Upper Integration Limit (x)	1
k	Exponential Rate Constant (k ≠ 0)	1

Inputs & Parameters

xUpper Integration Limit (x)[1]

kExponential Rate Constant (k ≠ 0)[1]

Calculated Result (Definite Integral ∫₀ˣ t·e^(k·t) dt via Parts)

I =8.389056dimensionless

Formula Notes & Background

Start with the product rule for differentiation: d/dx [u(x)v(x)] = u(x)v'(x) + v(x)u'(x).
Integrate both sides with respect to x: u(x)v(x) = ∫ u(x)v'(x)dx + ∫ v(x)u'(x)dx.
Rearrange terms and substitute dv = v'(x)dx and du = u'(x)dx to obtain ∫ u dv = u v − ∫ v du.

Step-by-Step Examples

#1: Evaluate the definite integral ∫₀² x eˣ dx using integration by parts.

u = x ⇒ du = dxdv = eˣ dx ⇒ v = eˣLimits: a = 0, b = 2

[x eˣ]₀² − ∫₀² eˣ dx = (2e² − 0) − (e² − e⁰) = e² + 1

→ e² + 1 ≈ 8.38906

#2: Find the indefinite integral ∫ ln(x) dx.

u = ln(x) ⇒ du = (1/x) dxdv = 1 dx ⇒ v = x

∫ ln(x) dx = x ln(x) − ∫ x · (1/x) dx = x ln(x) − ∫ 1 dx

→ x ln(x) − x + C

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Frequently Asked Questions

What is the LIATE mnemonic for choosing u in integration by parts?+

Choose u in priority order: Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential—because functions earlier in LIATE simplify when differentiated.

What is tabular integration (the DI method)?+

When integrating a polynomial times an exponential or sine/cosine, repeatedly differentiate u in column D and integrate dv in column I with alternating signs (+ − + −) to write out multi-step integration by parts in seconds.

Why does ∫ eˣ cos(x) dx require integrating by parts twice?+

Neither factor vanishes upon differentiation, but after two passes the original integral ∫ eˣ cos(x) dx reappears with a minus sign and can be solved algebraically.