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Integration by Parts

Key Terms

Formula
∫u dv = uv − ∫v du. This reverses the product rule for differentiation.
When to use
When the integrand is a product of two different function types that cannot be integrated by substitution.
LIATE rule for choosing u
Choose u from the first applicable category:
   Logarithms   •   Inverse trig   •   Algebraic (polynomials)   •   Trigonometric   •   Exponential
   The function higher in the list is usually the best choice for u (easier to differentiate), while the other becomes dv (integrated).
Setting out
Identify u and dv. Find du by differentiating u. Find v by integrating dv. Then apply the formula.
Repeated application
Sometimes the resulting ∫v du still requires integration by parts. Apply the formula again. This is common for ∫x²ex dx and similar.
Cyclic case
For ∫excos x dx, applying by parts twice returns the original integral. Collect it on one side and solve algebraically.
Key Formula:
  • ∫u dv = uv − ∫v du
  • Equivalently: ∫f(x)g′(x) dx = f(x)g(x) − ∫g(x)f′(x) dx
  • LIATE: Logarithms → Inverse trig → Algebraic → Trig → Exponential
Hot Tip Always write your choices explicitly: “Let u = ..., dv = ... dx, so du = ... dx, v = ...”. Then write the formula and substitute. The most common mistake is choosing u and dv the wrong way around — if your integral becomes harder after one step, swap them. Logarithms are almost always the best choice for u because ln x differentiates to a simple rational function.

Worked Example 1 — Standard application

Find ∫xex dx.

Let u = x (Algebraic), dv = ex dx (Exponential).

Then du = dx and v = ex.

∫xex dx = xex − ∫ex dx = xex − ex + C = ex(x − 1) + C

Worked Example 2 — Logarithm as u

Find ∫ln(x) dx.

Let u = ln x (Logarithm), dv = dx (Algebraic: 1).

Then du = (1/x) dx and 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

Worked Example 3 — Cyclic integrals

Find ∫excos(x) dx.

Let u = cos x, dv = ex dx.   ⇒ du = −sin x dx, v = ex.

∫excos x dx = excos x − ∫ex(−sin x) dx = excos x + ∫exsin x dx

Apply by parts again to ∫exsin x dx: Let u = sin x, dv = ex dx. ⇒ du = cos x dx, v = ex.

∫exsin x dx = exsin x − ∫excos x dx

Substituting back: ∫excos x dx = excos x + exsin x − ∫excos x dx

2∫excos x dx = ex(cos x + sin x)

∫excos x dx = (ex/2)(cos x + sin x) + C

The Product Rule in Reverse

The product rule states that d/dx[f(x)g(x)] = f′(x)g(x) + f(x)g′(x). Integrating both sides over x gives f(x)g(x) = ∫f′(x)g(x) dx + ∫f(x)g′(x) dx. Rearranging: ∫f(x)g′(x) dx = f(x)g(x) − ∫g(x)f′(x) dx. In Leibniz notation with u = f(x) and v = g(x), this is the familiar ∫u dv = uv − ∫v du.

The idea is to trade one integral for another that is hopefully simpler. Success depends on choosing u and dv so that ∫v du is easier than the original ∫u dv.

Using LIATE to Choose u

LIATE provides a hierarchy of function types. The function higher on the list is usually chosen as u, because it either simplifies or disappears when differentiated:

  • Logarithms (ln x, log x): d/dx[ln x] = 1/x — algebraic, much simpler.
  • Inverse trig (arctan x, arcsin x): derivatives are rational — simplify nicely.
  • Algebraic (x², x, etc.): polynomials reduce their degree when differentiated.
  • Trigonometric (sin x, cos x): derivatives cycle but do not simplify.
  • Exponential (ex): does not change, always easy to integrate as dv.

LIATE is a guideline, not a rule. Sometimes the cyclic technique (for ∫exsin x dx) or direct substitution is more efficient.

Repeated Integration by Parts

When the integrand involves a polynomial multiplied by ex, sin x, or cos x, each application of by parts reduces the power of the polynomial by one. For ∫x²ex dx, two applications are needed; for ∫x³ex dx, three. Keep u as the polynomial each time.

The Cyclic Technique

For ∫exsin x dx or ∫excos x dx, two applications of integration by parts return the original integral with a negative sign. Let I denote the integral. After two steps you obtain I = [terms] − I, so 2I = [terms], giving I = [terms]/2 + C. This “solving for the integral” technique is an important exam skill.

Exam Tip: For ∫ln(x) dx, write it as ∫ln(x) × 1 dx. Choose u = ln x and dv = 1 dx. This gives v = x, du = dx/x, and the integral reduces to x ln x − ∫1 dx = x ln x − x + C. This same technique works for ∫arctan(x) dx and any other “lonely” function that cannot otherwise be integrated easily.
Exam Tip: For definite integrals using by parts, apply the formula as ∫ab u dv = [uv]ab − ∫ab v du. Evaluate [uv] at the limits and then deal with the remaining definite integral.

Mastery Practice

See Answers ➔

  1. Find ∫xex dx. Fluency

  2. Find ∫x cos(x) dx. Fluency

  3. Find ∫ln(x) dx. (Hint: write as ∫ln(x) × 1 dx, let u = ln x, dv = dx.) Fluency

  4. Find ∫x²ex dx. (Apply integration by parts twice.) Fluency

  5. Find ∫x² sin(x) dx. Understanding

  6. Find ∫excos(x) dx. (Apply by parts twice and solve for the integral.) Understanding

  7. Find ∫x ln(x) dx. Understanding

  8. Evaluate ∫1e x ln(x) dx. Understanding

  9. Find ∫arctan(x) dx. (Let u = arctan x, dv = dx.) Problem Solving

  10. Find the area between y = xe−x and the x-axis for 0 ≤ x ≤ 2. Problem Solving