Integration by Parts

The formula and its origin

The formula:

∫ u dv = uv − ∫ v du

This comes directly from the product rule for derivatives. Differentiate uv:

d(uv)/dx = u' v + u v'

Integrate both sides with respect to x:

uv = ∫u'v dx + ∫uv' dx
   = ∫v du + ∫u dv

Rearrange — ∫u dv = uv − ∫v du. The formula is the product rule, integrated. We use it when the original integral ∫u dv is hard but ∫v du is easier.

How to apply it

Given an integral, pick parts of the integrand to label u and dv. Then:

1. Write u — what you'll differentiate.
2. Write dv — what you'll integrate.
3. Compute du = (du/dx) dx.
4. Compute v by integrating dv.
5. Apply the formula — ∫u dv = uv − ∫v du.

The tricky step is the choice. LIATE (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) suggests which to pick as u — earlier in the list = better choice. The idea — pick u so that du is simpler than u.

Worked examples

Example 1 — ∫ x · e^x dx

Algebraic (x) before Exponential (e^x), so u = x. Then dv = e^x dx.

u = x        →  du = dx
dv = e^x dx  →  v = e^x

∫x e^x dx = uv − ∫v du
          = x e^x − ∫e^x dx
          = x e^x − e^x + C
          = e^x(x − 1) + C

Example 2 — ∫ ln(x) dx

This looks tricky because there's no obvious "product." Trick — let u = ln(x) and dv = dx.

u = ln(x)  →  du = (1/x) dx
dv = dx    →  v = x

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

One application of integration by parts converts the seemingly impossible ∫ln(x) dx into the trivial ∫1 dx. This is the technique's elegance.

Example 3 — repeated application: ∫ x²·sin(x) dx

Algebraic before Trig — u = x², dv = sin(x) dx. First pass:

u = x²        →  du = 2x dx
dv = sin x dx →  v = −cos x

∫x² sin x dx = −x² cos x + ∫2x cos x dx

The new integral ∫2x cos x dx is easier (lower degree). Apply IBP again — u = 2x, dv = cos x dx.

u = 2x       →  du = 2 dx
dv = cos x   →  v = sin x

∫2x cos x dx = 2x sin x − ∫2 sin x dx
            = 2x sin x + 2 cos x + C

Combining:

∫x² sin x dx = −x² cos x + 2x sin x + 2 cos x + C
            = (2 − x²) cos x + 2x sin x + C

Tabular integration

For polynomial × (sin, cos, e^x), repeated IBP becomes tedious. Tabular integration shortcuts the bookkeeping. For ∫ x³ e^x dx:

Sign	D (derivatives of u = x³)	I (antiderivatives of dv = e^x)
+	x³	e^x
−	3x²	e^x
+	6x	e^x
−	6	e^x
+	0	e^x

Multiply diagonally with the signs:

∫x³ e^x dx = +x³·e^x − 3x²·e^x + 6x·e^x − 6·e^x + C
           = e^x (x³ − 3x² + 6x − 6) + C

Five lines of work for what would take four IBP iterations. Tabular integration is just bookkeeping — same theorem, less notation.

LIATE vs other strategies

Integrand	u choice	Why
x · e^x	u = x (Algebraic)	du = dx is simpler
x · ln(x)	u = ln(x) (Logarithmic)	L beats A in LIATE; du = dx/x
x · arctan(x)	u = arctan(x) (Inverse trig)	du = dx/(1+x²) is simpler
x² · sin(x)	u = x² (Algebraic)	A beats T; reduces by IBP twice
e^x · sin(x)	Either — both reduce to themselves after two IBPs	Cyclic; solve algebraically for the integral

JavaScript — symbolic IBP demo

// This is conceptual — real symbolic integration uses CAS like math.js.
// Here we numerically verify by integrating both forms.

function trapezoidIntegrate(f, a, b, n = 10000) {
  const h = (b - a) / n;
  let sum = 0.5 * (f(a) + f(b));
  for (let i = 1; i < n; i++) sum += f(a + i * h);
  return sum * h;
}

// ∫_0^1 x e^x dx — direct numerical
const direct = trapezoidIntegrate(x => x * Math.exp(x), 0, 1);

// IBP transformed — should equal e − 2·∫_0^1 e^x dx... let's verify the closed form
// ∫_0^1 x e^x dx = [x e^x]_0^1 − ∫_0^1 e^x dx = e − (e − 1) = 1
console.log(direct);  // ≈ 1.0
console.log(Math.E - (Math.E - 1));  // exactly 1.0

When integration by parts is the right tool

• Polynomial × (sin / cos / e^x). Tabular integration handles these mechanically.
• Functions that are "lone" — log, arctan, arcsin. Pick u = the function, dv = dx. ln(x), arctan(x) — all integrate this way.
• Cyclic problems — ∫e^x sin(x) dx. Two IBPs return the original; solve algebraically for the unknown integral.
• Reduction formulas. ∫sin^n(x) dx = (recurrence in n) — IBP gives the recurrence relation that lets you compute integrals of any power.
• Probability and statistics. Many expected-value calculations involve ∫x · density(x) dx, naturally suited to IBP.

Common mistakes

• Wrong choice of u and dv. Forgetting LIATE and picking the wrong one makes the new integral harder. If your IBP makes things worse, swap u and dv.
• Forgetting the constant of integration. Standard for indefinite integrals; some textbooks omit; production calculus must include + C.
• Sign errors in the formula. ∫u dv = uv − ∫v du. The minus sign in front of ∫v du is critical and easy to drop.
• Computing v wrong. v = ∫ dv is an antiderivative — pick any one (often the simplest, with constant 0). Errors in v propagate to the final answer.
• Not noticing cyclic patterns. ∫e^x sin x dx returns the original after two IBPs; you have to solve the resulting equation algebraically rather than recursing forever.
• Forgetting limits in definite integrals. The definite version is ∫_a^b u dv = [uv]_a^b − ∫_a^b v du. The boundary term [uv]_a^b is often forgotten; it changes the answer.