Aim for FiveCalculus BC

Unit 6 · Topic 6.11

BC only

6.11 Integrating Using Integration by Parts

BC only. Integration by parts reverses the product rule: ∫ u dv = uv − ∫ v du. It handles products like x·eˣ, x·sin x and x²·cos x, and single functions like ln x and arctan x.

Key terms

  • integration by parts
  • product rule
  • u and dv
  • uv − ∫ v du

Where the formula comes from

This topic is tested only on the BC exam.

The product rule says (uv)′ = u′v + uv′. Integrate both sides and rearrange: ∫ u v′ dx = uv − ∫ v u′ dx. In shorthand, with dv = v′ dx and du = u′ dx:

∫ u dv = uv − ∫ v du.

The goal is to trade the integral you can't do for one you can. You choose which part of the integrand is u (you'll differentiate it) and which part, along with dx, is dv (you'll integrate it).

Choosing u

Pick u to be the factor that gets simpler when you differentiate it, and dv to be something you can integrate. A common guide is LIATE: choose u from the first type that appears in this list.

  • L: logarithms (ln x)
  • I: inverse trig (arctan x, arcsin x)
  • A: algebraic (x, x², polynomials)
  • T: trig (sin x, cos x)
  • E: exponentials (eˣ, e²ˣ)

Repeated parts and the tabular shortcut

For x²·cos x, one round of parts leaves ∫ 2x sin x dx, which needs parts again. When u is a polynomial and dv is easy to integrate again and again (eˣ, sin, cos), you can use a table: list derivatives of u down one column until you hit 0, list repeated antiderivatives of dv down another, then multiply diagonally with alternating signs +, −, +, …

For ∫ x² cos x dx: derivatives x², 2x, 2, 0; antiderivatives cos x, sin x, −cos x, −sin x. Diagonal products: +x² sin x, −2x(−cos x), +2(−sin x). Result: x² sin x + 2x cos x − 2 sin x + C.

Special cases

  • A lone ln x or arctan x: let u = ln x and dv = dx. Then ∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − x + C.
  • Integrals like ∫ eˣ sin x dx cycle back to the original after two rounds. Call the original I, write the equation I = (stuff) − I, and solve for I. This shows up less often, but it's fair game.
  • For definite integrals: ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du. Evaluate the uv part at both limits too.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    One round of parts

    Find ∫ x e²ˣ dx.

    Show the solution
    1. Step 1: Choose u = x (algebraic; gets simpler) and dv = e²ˣ dx.
    2. Step 2: Then du = dx and v = ½e²ˣ.
    3. Step 3: ∫ u dv = uv − ∫ v du = ½x e²ˣ − ∫ ½e²ˣ dx = ½x e²ˣ − ¼e²ˣ + C.
    4. Step 4: Check: d/dx [½x e²ˣ − ¼e²ˣ] = ½e²ˣ + x e²ˣ − ½e²ˣ = x e²ˣ.

    Answer: ½x e²ˣ − ¼e²ˣ + C

  2. Example 2

    Definite integral of ln x

    Evaluate ∫₁ᵉ ln x dx.

    Show the solution
    1. Step 1: u = ln x, dv = dx, so du = (1/x) dx and v = x.
    2. Step 2: ∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − x.
    3. Step 3: At x = e: e·1 − e = 0. At x = 1: 1·0 − 1 = −1.
    4. Step 4: Subtract: 0 − (−1) = 1.

    Answer: 1

  3. Example 3

    Trap: a bad choice of u

    Find ∫ x² cos x dx, and explain why u = cos x is a bad choice.

    Show the solution
    1. Step 1: If u = cos x and dv = x² dx, then v = x³/3 and the new integral is ∫ (x³/3) sin x dx, which is worse: the power went up.
    2. Step 2: Choose u = x² (algebraic) and dv = cos x dx: du = 2x dx, v = sin x. So ∫ x² cos x dx = x² sin x − ∫ 2x sin x dx.
    3. Step 3: Parts again on ∫ 2x sin x dx with u = 2x, dv = sin x dx: = −2x cos x + ∫ 2 cos x dx = −2x cos x + 2 sin x.
    4. Step 4: Combine: x² sin x − (−2x cos x + 2 sin x) = x² sin x + 2x cos x − 2 sin x.

    Answer: x² sin x + 2x cos x − 2 sin x + C

Common mistakes

  • Losing a negative sign in the second round. Put the whole second integral's result in parentheses before subtracting.
  • Choosing u so that the new integral is harder. If the power of x goes up, switch your choice.
  • Forgetting to evaluate the uv part at the limits in a definite integral.
  • Using parts when a simple substitution works, as in ∫ x e^(x²) dx, where u = x² is all you need.

On the exam

  • BC multiple-choice questions often test one round of parts, such as ∫ x cos x dx or ∫ x eˣ dx, or a definite version. Differentiating each answer choice is a quick check.
  • On free response, write u, dv, du and v clearly. Integration by parts also appears inside improper integrals (6.13) and in finding antiderivatives for motion questions.

Connected topics

Videos

Check yourself

4 questions on 6.11 Integrating Using Integration by Parts. Pick an answer to see if you got it, and why.

Question 1 of 4

∫ from 1 to e of x ln x dx =

Question 2 of 4

∫ x cos(2x) dx =

Question 3 of 4

The function f is differentiable, with f′ continuous. It is known that f(0) = 4, f(2) = 3, and ∫₀² f(x) dx = 5. What is the value of ∫₀² x·f′(x) dx?

Let f be the function defined by f(x) = x·e^(−x/2) for x ≥ 0.

Described function

Question 4 of 4

Which of the following is equal to ∫ x·e^(−x/2) dx ?

0 of 4 answered