Use integration by parts to simplify a product
Reverse the product rule, choose useful differentiated and integrated factors, and verify the resulting primitive.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.5. Paper 1 integration, shared standard integrals and identities, Paper 3 partial fractions and derivatives are assumed. Rational decompositions follow the three denominator forms in section 3.1; higher-degree numerators are excluded, while equal-degree cases may need division first. Substitution questions use a given substitution. Trigonometric and inverse-tangent outputs use radians. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Differentiate products and integrate standard exponential and trigonometric functions.
- Keep signs, inner scale factors, initial values and definite endpoints.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which identity is the product rule?
What you will learn
- Apply ∫uv′ dx = uv − ∫u′v dx to polynomial–exponential and polynomial–trigonometric products.
- Integrate ln x by treating it as a product with 1, and evaluate a definite rate integral.
The product rule supplies the identity
Start from (uv)′ = u′v + uv′. Rearranging and integrating gives ∫uv′ dx = uv − ∫u′v dx. You choose a factor u to differentiate and a factor v′ to integrate; v must be an actual primitive of v′. The remaining integral should become simpler or otherwise manageable. Keep the subtraction sign outside the entire remaining integral.
For ∫x sin(2x) dx, choose u = x and v′ = sin(2x). Then u′ = 1 and v = −cos(2x)/2. Substitution into the identity gives −x cos(2x)/2 + (1/2)∫cos(2x) dx = −x cos(2x)/2 + sin(2x)/4 + C. The first factor's derivative becomes simpler, and both trigonometric primitives retain their inner scale 2.
A logarithm can be a product even when only one factor is written
To integrate ln x for x > 0, write it as (ln x)·1, choose u = ln x and v′ = 1, then u′ = 1/x and v = x. The identity gives x ln x − ∫1 dx = x ln x − x + C. Differentiating the result gives ln x + 1 − 1 = ln x. Integrating ln x as 1/x would confuse its derivative with its primitive.
For a supplied rate q(t) = t eᵗ litres per minute on [0, 1], choose u = t and v′ = eᵗ. A primitive is (t − 1)eᵗ. Endpoint subtraction gives 0 − (−1) = 1 litre added. A primitive's negative value at one endpoint does not make the definite integral negative: its change is positive, consistent with the non-negative rate. Add a stated initial stock only when finding the final total.
Pause and explain
For ∫x sin(2x) dx with u = x, what is v?
Worked example
Find ∫x sin(2x) dx by parts and verify the signs.
Show the worked solution
- Choose u = x, u′ = 1 and v′ = sin(2x), giving v = −cos(2x)/2.
- Use uv − ∫u′v dx to obtain −x cos(2x)/2 + (1/2)∫cos(2x) dx.
- The remaining integral gives sin(2x)/4; differentiate the complete sum to see the cosine terms cancel.
Answer −x cos(2x)/2 + sin(2x)/4 + C.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find ∫x e^(2x) dx using parts and verify it.
Give me a hint
Use u = x and v′ = e^(2x), with v = e^(2x)/2.
Compare my reasoning
- The parts formula gives x e^(2x)/2 − (1/2)∫e^(2x) dx.
- The remaining primitive is e^(2x)/4, so the result is x e^(2x)/2 − e^(2x)/4 + C.
- Differentiation gives e^(2x)/2 + x e^(2x) − e^(2x)/2, leaving the required product.
(x/2 − 1/4)e^(2x) + C.
Look for these in your work
- Reverse the exponential scale in both integrations.
- Check the product-rule cancellation.
Find ∫ln x dx for x > 0, then evaluate it from 1 to e exactly.
Give me a hint
Use ln x times 1, and keep the lower endpoint even though ln 1 = 0.
Compare my reasoning
- Parts gives F(x) = x ln x − x.
- F(e) = e − e = 0 and F(1) = −1.
- The definite integral is 0 − (−1) = 1, consistent with ln x ≥ 0 on this interval.
Primitive x ln x − x + C; definite integral 1.
Look for these in your work
- Choose the unprinted factor 1 for v′.
- Evaluate both complete endpoint expressions.
A vessel holds 2 litres initially and has supplied inflow q(t) = t eᵗ litres per minute on 0 ≤ t ≤ 1, with no outflow. Find the added volume and final total.
Give me a hint
Parts gives a primitive (t − 1)eᵗ; subtract both endpoint values.
Compare my reasoning
- Choose u = t and v′ = eᵗ, obtaining F(t) = (t − 1)eᵗ.
- The increase is F(1) − F(0) = 0 − (−1) = 1 litre.
- Add the initial 2 litres to obtain 3 litres total.
Added volume 1 litre; final total 3 litres.
Look for these in your work
- Use a verified product primitive and the right endpoint order.
- Distinguish the rate's accumulation from the final stock.
Common mistakes
- Integrate v′ to obtain v; do not insert the original v′ into the uv term.
- Keep the subtraction sign and each trigonometric inner divisor.
- A primitive and an integrand can have different signs; a definite integral is an endpoint difference.
What can you explain now?
Find ∫ln x dx for x > 0.
Compare with the explanation
x ln x − x + C.
Parts with u = ln x and v′ = 1 leaves ∫1 dx; differentiation checks the cancellation.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Derive the parts identity from the product rule. Choose factors that simplify the remaining integral, then check the complete primitive by differentiation.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.