Integration by parts, explained with examples

By Ashraf Bhura, Co-Founder, STEM-Tutor · Updated October 2026

Integration by parts lets you integrate a product of two different kinds of function, such as x e^x or x sin x. It is in AP Calculus BC and in the second year of A level Maths.

The formula

∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx, often written more briefly as ∫ u dv = uv − ∫ v du.

You split the integrand into two parts. One part, u, you will differentiate. The other part, dv, you will integrate to get v. The formula swaps your original integral for a new one, ∫ v du, which should be easier. If it is not easier, you probably chose u and dv the wrong way round.

Where the formula comes from

Integration by parts is the product rule in reverse. The product rule says d(uv)/dx = u(dv/dx) + v(du/dx). Integrate both sides with respect to x: uv = ∫ u(dv/dx) dx + ∫ v(du/dx) dx. Rearranging gives the formula. Knowing this helps you remember it, and it is a short proof that A level questions occasionally ask for.

Choosing u: the LIATE rule

Choose u to be whichever part comes first in this list, because it gets simpler when you differentiate it:

1. Logarithms, such as ln x
2. Inverse trig functions, such as arctan x
3. Algebraic functions, such as x or x²
4. Trig functions, such as sin x
5. Exponentials, such as e^x

LIATE is a guide, not a law, but it works for nearly every exam question. The main idea is that polynomials shrink when differentiated, logs turn into simple fractions, while exponentials and sine or cosine stay the same size whichever way you go.

Worked examples

Example 1: x times an exponential

Find ∫ x e^x dx.

• u = x (algebraic), dv = e^x dx. Then du = dx and v = e^x.
• ∫ x e^x dx = x e^x − ∫ e^x dx = x e^x − e^x + C.
• Check by differentiating: e^x + x e^x − e^x = x e^x. Correct.

Example 2: x times a trig function

Find ∫ x cos x dx.

• u = x, dv = cos x dx, so du = dx and v = sin x.
• = x sin x − ∫ sin x dx = x sin x + cos x + C.

Example 3: ln x on its own

Find ∫ ln x dx.

• Write it as ∫ ln x × 1 dx. Take u = ln x and dv = 1 dx, so du = (1/x) dx and v = x.
• = x ln x − ∫ x × (1/x) dx = x ln x − x + C.

Example 4: a power times ln x

Find ∫ x² ln x dx.

• Logs come first in LIATE, so u = ln x and dv = x² dx. Then du = (1/x) dx and v = x³/3.
• = (x³/3) ln x − ∫ (x³/3)(1/x) dx = (x³/3) ln x − ∫ x²/3 dx.
• = (x³/3) ln x − x³/9 + C.

Example 5: using parts twice

Find ∫ x² e^x dx.

• First: u = x², dv = e^x dx gives x²e^x − ∫ 2x e^x dx.
• Second: ∫ 2x e^x dx = 2(x e^x − e^x), from Example 1.
• Answer: e^x(x² − 2x + 2) + C.

Example 6: the loop trick

Find I = ∫ e^x sin x dx.

• Parts with u = sin x: I = e^x sin x − ∫ e^x cos x dx.
• Parts again with u = cos x: ∫ e^x cos x dx = e^x cos x + ∫ e^x sin x dx = e^x cos x + I.
• So I = e^x sin x − e^x cos x − I, which gives 2I = e^x(sin x − cos x).
• I = ½e^x(sin x − cos x) + C. The original integral came back, so you solve for it like an equation.

Example 7: a definite integral

Evaluate ∫₀¹ x e^2x dx.

• u = x, dv = e^2x dx, so v = ½e^2x.
• = [½x e^2x]₀¹ − ∫₀¹ ½e^2x dx = ½e² − [¼e^2x]₀¹.
• = ½e² − ¼e² + ¼ = ¼(e² + 1) ≈ 2.10.

The tabular method

When u is a polynomial that eventually differentiates to zero, a table saves time. Write u and its derivatives down one column, and dv and its repeated integrals down another. Multiply diagonally, with signs alternating +, −, +. For ∫ x² e^x dx, the u column is x², 2x, 2, 0 and the dv column is e^x, e^x, e^x, e^x. The diagonals give x²e^x − 2x e^x + 2e^x, the same answer as Example 5. In an exam, show enough working that the method is clear.

Parts or substitution?

Use substitution when one part of the integrand is the derivative of another part, such as ∫ 2x e^x² dx, where 2x is the derivative of x². Use parts when the two factors are unrelated kinds of function, such as x and e^x. If you are unsure, try substitution first, because it is usually quicker when it works.

Common mistakes

• Choosing u = e^x in ∫ x e^x dx, which makes the new integral harder.
• Sign errors in the minus part of the formula, especially when using parts twice.
• Forgetting the + C, or forgetting to apply limits to the uv part in a definite integral.
• Not spotting that ln x on its own needs the trick of writing it as ln x × 1.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Is integration by parts on AP Calculus AB?

No. It is a BC-only topic, in Unit 6. AB students do not need it.

Is integration by parts in A level Maths?

Yes. It is in the second year of A level Maths, and the formula is in the formula booklet for most boards.

How do I know if I chose u correctly?

The new integral, ∫ v du, should be simpler than the one you started with. If it is more complicated, swap your choice.

Key takeaways

• ∫ u dv = uv − ∫ v du.
• It is the product rule in reverse.
• Choose u with LIATE: logs, inverse trig, algebraic, trig, exponential.
• Some integrals need parts twice, or the loop trick.
• Check answers by differentiating.

Related guides

• The chain rule
• AP Calculus BC units
• A level Maths integration