Integrating Using Integration by Parts: Study Guide (Calculus BC 6.11)

BC-only material. Integration by parts is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

You need four earlier ideas. If any is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Product rule	2.8	The formula comes straight from it
Fundamental Theorem of Calculus and definite integrals	6.7	Evaluating [uv] between limits
Basic antiderivatives	6.8	Finding v from dv
Substitution	6.9	Some v’s and some leftover integrals need it; it is also the method to try first

Integral notation on this page: ∫ from a to b of f(x) dx is the definite integral with lower limit a and upper limit b. [F(x)] from a to b means F(b) − F(a).

Why we need a new technique

Substitution undoes the chain rule. It works when the integrand contains a function and (a multiple of) its derivative, such as 2x cos(x²).

But many integrands are a product of two unrelated functions: x cos x, x eˣ, x² sin x. There is no inner function whose derivative is sitting outside, so substitution does not help. These products come from the product rule, so we undo the product rule instead.

Deriving the formula from the product rule

Let u and v be differentiable functions of x. The product rule says

d/dx [u v] = u′ v + u v′

Rearrange to isolate one product:

u v′ = d/dx [u v] − v u′

Integrate both sides with respect to x. The integral of a derivative gives back the function (up to a constant), so

∫ u v′ dx = u v − ∫ v u′ dx

In differential shorthand, write dv = v′ dx and du = u′ dx:

∫ u dv = u v − ∫ v du

The formula does not finish the integral. It trades one integral for another. The trade is worth it only if ∫ v du is easier than the original.

Definite version. Using the Fundamental Theorem on each part:

∫ from a to b of u dv = [u v] from a to b − ∫ from a to b of v du

The limits apply to both pieces.

Choosing u and dv

You split the integrand into two parts: u (which you will differentiate) and dv (which you will integrate). Two conditions decide a good choice:

1. dv must be something you can integrate. If you cannot find v, you are stuck immediately.
2. du should be simpler than u, so that v du is easier than u dv.

LIATE as a guide. Many teachers suggest choosing u from whichever factor appears first in this list: Logarithmic, Inverse trig, Algebraic (powers of x), Trigonometric, Exponential. It works often because logs and inverse trig functions get simpler when differentiated (ln x → 1/x), while exponentials and sin/cos are easy to integrate.

But LIATE is a habit, not a theorem. It can point you the wrong way. Consider ∫ x³ e^(x²) dx:

• LIATE says u = x³ and dv = e^(x²) dx. But e^(x²) has no elementary antiderivative, so condition 1 fails.
• A better split is u = x² and dv = x e^(x²) dx. Then v = ½ e^(x²) (by substitution) and du = 2x dx: ∫ x³ e^(x²) dx = ½ x² e^(x²) − ∫ x e^(x²) dx = ½ x² e^(x²) − ½ e^(x²) + C.

Always test a choice against the two conditions, not against a mnemonic.

When dv = dx. If the integrand is a single function that is hard to integrate but easy to differentiate, take dv = dx. For example, ∫ ln x dx: u = ln x, dv = dx, so du = (1/x) dx and v = x.

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

Worked example 1: ∫ x cos x dx, and a bad choice

Question. Find ∫ x cos x dx without a calculator. Check your answer.

1. Is substitution possible? No: x is not the derivative of anything inside cos x. The integrand is a product of an algebraic and a trig function, so try parts.
2. Choose. u = x (its derivative, 1, is simpler). dv = cos x dx (easy to integrate).
3. Find du and v. du = 1 dx. v = sin x (no + C needed here).
4. Apply the formula. ∫ x cos x dx = x sin x − ∫ sin x dx
5. Finish the easier integral. ∫ sin x dx = −cos x, so ∫ x cos x dx = x sin x − (−cos x) + C = x sin x + cos x + C

Check by differentiating. d/dx [x sin x + cos x] = (1 · sin x + x cos x) − sin x = x cos x. ✓

What if you choose the other way? u = cos x and dv = x dx give du = −sin x dx and v = ½x². Then

∫ x cos x dx = ½x² cos x + ½ ∫ x² sin x dx.

The new integral has x², which is harder than the original x. The formula is still true, but it has moved you backwards. Swap the choice.

Worked example 2: accumulation in context (definite integral)

Context. Rain-water flows into a storage tank at a rate of r(t) = 20 t e^(−t/2) litres per hour, where t is the time in hours after a storm starts. The tank holds 30 litres at t = 0. How much water is in the tank at t = 4 hours? (Assume no water leaves the tank.)

1. Set up. Volume at t = 4 = starting volume + volume added: V(4) = 30 + ∫ from 0 to 4 of 20 t e^(−t/2) dt
2. Choose u and dv. u = t (simpler when differentiated). dv = e^(−t/2) dt (easy to integrate).
3. Find du and v. du = dt. v = −2 e^(−t/2), because d/dt [−2 e^(−t/2)] = −2 · (−½) e^(−t/2) = e^(−t/2).
4. Apply the definite formula to ∫ from 0 to 4 of t e^(−t/2) dt: = [−2t e^(−t/2)] from 0 to 4 − ∫ from 0 to 4 of (−2 e^(−t/2)) dt = [−2t e^(−t/2)] from 0 to 4 + 2 ∫ from 0 to 4 of e^(−t/2) dt
5. Evaluate each piece.
   • [−2t e^(−t/2)] from 0 to 4 = (−8 e^(−2)) − (0) = −8e⁻².
   • 2 ∫ from 0 to 4 of e^(−t/2) dt = 2 [−2 e^(−t/2)] from 0 to 4 = 2(−2e⁻² + 2) = 4 − 4e⁻².
   • Total: −8e⁻² + 4 − 4e⁻² = 4 − 12e⁻².
6. Multiply by 20. ∫ from 0 to 4 of 20 t e^(−t/2) dt = 20(4 − 12e⁻²) = 80 − 240e⁻² ≈ 47.52 L.
7. Add the starting volume. V(4) = 30 + 80 − 240e⁻² = 110 − 240e⁻² ≈ 77.5 L.

Answer. About 77.5 litres (exact value 110 − 240e⁻² L).

Checks.

• Units: litres per hour × hours = litres. ✓
• Size: the rate stays below about 14.7 L/h for 4 hours, so the added volume must be below 4 × 14.7 ≈ 59 L. 47.5 L is sensible.
• Antiderivative: d/dt [−2t e^(−t/2) − 4 e^(−t/2)] = −2e^(−t/2) + t e^(−t/2) + 2e^(−t/2) = t e^(−t/2). ✓
• Calculator (where allowed): a numerical integral of r(t) from 0 to 4 gives 47.52. ✓

Interpretation. In the first 4 hours after the storm starts, about 47.5 litres flow in, so the tank holds about 77.5 litres at t = 4.

Using parts more than once

Repeated parts. For ∫ x² eˣ dx, one application lowers the power of x by one:

• u = x², dv = eˣ dx → ∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx.
• Apply parts again to ∫ 2x eˣ dx with u = 2x, dv = eˣ dx: it equals 2x eˣ − 2eˣ.
• So ∫ x² eˣ dx = x² eˣ − 2x eˣ + 2eˣ + C = eˣ(x² − 2x + 2) + C.

Check: d/dx [eˣ(x² − 2x + 2)] = eˣ(x² − 2x + 2) + eˣ(2x − 2) = x² eˣ. ✓

Integrals that come back: solve for the integral. Let I = ∫ eˣ sin x dx. Neither factor ever disappears when differentiated, but after two steps the original integral returns.

1. u = sin x, dv = eˣ dx: I = eˣ sin x − ∫ eˣ cos x dx.
2. u = cos x, dv = eˣ dx (keep the trig function as u; swapping would undo step 1): ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I.
3. Substitute: I = eˣ sin x − eˣ cos x − I.
4. Solve: 2I = eˣ(sin x − cos x), so I = ½ eˣ(sin x − cos x) + C.

Check: d/dx [½ eˣ(sin x − cos x)] = ½ eˣ(sin x − cos x) + ½ eˣ(cos x + sin x) = eˣ sin x. ✓

Optional shortcut: the tabular method. When u is a polynomial (so repeated differentiation reaches 0) and dv is easy to integrate repeatedly, list derivatives of u and antiderivatives of dv side by side, then multiply along the diagonals with alternating signs + − + − …

Sign	Differentiate u	Integrate dv
+	x²	eˣ
−	2x	eˣ
+	2	eˣ
	0	eˣ

Diagonal products: +x² eˣ − 2x eˣ + 2eˣ, the same answer as above. The table is just repeated parts written compactly. It is not required, and it does not suit integrals like eˣ sin x, where nothing differentiates to 0.

Common misconceptions

• “The integral of a product is the product of the integrals.” ∫ x eˣ dx is not ½x² · eˣ. Differentiate ½x² eˣ and you get x eˣ + ½x² eˣ, not x eˣ.
• “LIATE always gives the right u.” It fails for ∫ x³ e^(x²) dx. Check that dv can actually be integrated.
• Forgetting the minus sign in uv − ∫ v du, or losing a minus inside v (for example v = −cos x from dv = sin x dx).
• Applying the limits only to the integral part in a definite integral. [uv] from a to b must be evaluated too.
• Forgetting the chain-rule factor in v. From dv = e^(−t/2) dt, v = −2e^(−t/2), not e^(−t/2).
• Swapping the choice of u halfway through a self-returning integral. You end up with 0 = 0 or back at the start.
• Using parts when substitution works. ∫ x e^(x²) dx is a substitution integral (u = x²). Try substitution first.
• Dropping + C on an indefinite integral, or adding + C to a definite one.

Where this leads

Integration by parts returns in BC Topic 6.13 (improper integrals such as ∫ from 0 to ∞ of x e^(−x) dx) and in the applications of integration in Unit 8, wherever an accumulation, area or volume problem has a product integrand. Topic 6.14 then asks you to select the right technique for an unfamiliar integral. Use the Calculus BC roadmap to see the order.

Try the practice questions now, then use the revision notes and the checklist to consolidate.