Integration by parts integrates a product by reversing the product rule: ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx. Choose u to be the part that becomes simpler when differentiated, usually the polynomial, and dv/dx to be the part you can integrate. The exception is a logarithm: ln x is always chosen as u.
Integration by parts
- ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx
- Choose u: the part that simplifies when differentiated (usually a polynomial)
- If ln x appears, let u = ln x
- Definite form: [uv] from a to b − ∫ v (du/dx) dx from a to b
How does integration by parts work?
Split the product into a part to differentiate, u, and a part to integrate, dv/dx. Find du/dx and v, put them into the formula, and integrate the remaining term ∫ v (du/dx) dx. If you chose well, that remaining integral is simpler than the one you started with.
How do you choose u?
Choose u to be the factor that becomes simpler when differentiated. A polynomial such as x or x² gets simpler; eˣ, sin x and cos x do not. So in x eˣ or x sin x, let u = x.
Logarithms are the exception, because ln x is hard to integrate but easy to differentiate. Whenever ln x appears, let u = ln x.
Integrating ln x on its own
ln x has no obvious second factor, so write it as 1 × ln x. Let u = ln x and dv/dx = 1, so du/dx = 1/x and v = x. The remaining integral then becomes ∫ x × (1/x) dx = ∫ 1 dx, which is straightforward. The same choice of u works for x ln x and x² ln x.
Using integration by parts twice
If u is a quadratic, one application is not enough: the remaining integral still contains a product. Apply parts again to that new integral; each round lowers the power of x by one. Keep careful track of signs, especially the minus in front of the second integral.
For a product like eˣ sin x, two rounds bring the original integral back. Call it I, apply parts twice to get an equation in I, and solve it.
Definite integrals by parts
The cleanest approach is to find the complete antiderivative first and then substitute the limits once at the end. Answers often involve π, e or ln, so keep them exact.
Worked examples
Example 1. Find ∫ x eˣ dx.
- u = x and dv/dx = eˣ, so du/dx = 1 and v = eˣ.
- Apply: x eˣ − ∫ eˣ dx.
- Integrate the remaining term.
Answer: (x − 1)eˣ + c.
Marks: M1 for the choice of u and dv/dx, M1 for applying the formula, A1 for the answer.
Example 2. Find ∫ x ln x dx.
- ln x is present, so u = ln x and dv/dx = x. Then du/dx = 1/x and v = x²/2.
- Apply: (x²/2) ln x − ∫ (x²/2)(1/x) dx = (x²/2) ln x − ∫ x/2 dx.
- Integrate the remaining term: x²/4.
Answer: (x²/2) ln x − x²/4 + c.
Marks: M1 for u = ln x, M1 for the formula, A1 for the answer.
Example 3. Find ∫ x² sin x dx.
- u = x², dv/dx = sin x, so du/dx = 2x and v = −cos x.
- Apply: −x² cos x + ∫ 2x cos x dx.
- Parts again with u = 2x, dv/dx = cos x: ∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx = 2x sin x + 2 cos x.
Answer: −x² cos x + 2x sin x + 2 cos x + c.
Marks: M1 for the first application, M1 for the second, A1 for the answer.
Example 4. Find the exact value of ∫ x cos x dx from 0 to π/2.
- u = x, dv/dx = cos x: ∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x.
- Evaluate [x sin x + cos x] from 0 to π/2: (π/2 + 0) − (0 + 1).
Answer: π/2 − 1.
Marks: M1 for x sin x + cos x, A1 for π/2 − 1.
Example 5. Find the exact area under y = x sin x from x = 0 to x = π/2.
- u = x, dv/dx = sin x gives −x cos x + sin x.
- Evaluate from 0 to π/2: (0 + 1) − (0 + 0).
Answer: 1 square unit.
Marks: M1 for −x cos x + sin x, A1 for 1.
Common mistakes
- Choosing u the wrong way round. If u = eˣ and dv/dx = x, the new integral becomes harder. Let the polynomial be u.
- Not using u = ln x. ln x is awkward to integrate, so it should always be u.
- Sign errors in the second application. The minus in uv − ∫ v du/dx dx applies to the whole of the second integral.
- Differentiating dv/dx instead of integrating it. You differentiate u and integrate dv/dx.
- Substituting the limits too early. Find the full antiderivative first, then substitute once.
How integration by parts is examined
Expect integration by parts on WMA14, the Pure 4 paper in Edexcel International A-Level; in Edexcel A-Level (9MA0) it is second-year content. Questions ask you to:
- integrate a product such as x eˣ, x sin x or x ln x
- integrate ln x or a power of x times ln x
- apply integration by parts twice
- find an exact area under a product curve
The Pure 4 integration exam questions include integration by parts questions with full mark schemes. Use the integration worksheet or the A-Level integration revision engine for more.
Integration by parts reverses the product rule. Compare it with integration by substitution, the other main Pure 4 technique.
Frequently asked questions
What is the integration by parts formula?
∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx.
How do I choose u in integration by parts?
Choose the part that becomes simpler when differentiated, usually a polynomial. If ln x appears, always choose u = ln x.
When do you use integration by parts twice?
When u is a quadratic or higher power, or for products like eˣ sin x, where the integral loops back to itself after two rounds.