Repeat integration by parts and combine it with inverse tangent
Reduce a polynomial product step by step, and simplify the rational remainder created by an inverse-tangent factor.
Section 3.5 · Integration
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
- Use the parts identity with a useful choice of factors.
- Differentiate inverse tangent and integrate positive quadratic reciprocals.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the derivative of arctan x in radians?
What you will learn
- Apply parts more than once when a polynomial factor remains.
- Integrate x arctan x by parts, then split the remaining rational function into standard terms.
Each application must retain its own subtraction sign
For ∫x²e^(−x) dx, choose u = x² and v′ = e^(−x), so u′ = 2x and v = −e^(−x). The identity gives −x²e^(−x) + 2∫x e^(−x) dx. The remaining product still needs parts: ∫x e^(−x) dx = −x e^(−x) − e^(−x) + C. Combining gives −e^(−x)(x² + 2x + 2) + C. Differentiate the factored expression to check the signs and cancellation.
For ∫x²cos x dx, the first application gives x²sin x − 2∫x sin x dx. A second application gives ∫x sin x dx = −x cos x + sin x, so the complete primitive is x²sin x + 2x cos x − 2sin x + C. Stop when the remaining polynomial has become constant and the standard integral is manageable. One final arbitrary constant suffices on a connected interval.
The remaining integral may need a different method
For ∫x arctan x dx, choose u = arctan x and v′ = x. Then u′ = 1/(1 + x²) and v = x²/2, giving (x²/2)arctan x − (1/2)∫x²/(1 + x²) dx. Rewrite the remaining fraction as 1 − 1/(1 + x²). Its primitive is x − arctan x, so the original primitive is [(x² + 1)/2]arctan x − x/2 + C. Parts has converted the product into a rational integrand; the inverse-tangent rule finishes it.
Over [0, 1], ∫x arctan x dx = π/4 − 1/2, a positive number consistent with the positive integrand. For a supplied inflow t²e^(−t) litres per minute over [0, 1], the first primitive above gives volume 2 − 5/e litres. Retain this small positive result instead of rounding intermediate values: subtracting nearby endpoint values can amplify rounding error. An initial stock belongs in a separate final-total calculation.
Pause and explain
After parts in ∫x arctan x dx, how should x²/(1 + x²) be handled?
Worked example
Find ∫x²e^(−x) dx and check the complete primitive.
Show the worked solution
- Parts with u = x² gives −x²e^(−x) + 2∫x e^(−x) dx.
- Repeat parts to obtain ∫x e^(−x) dx = −x e^(−x) − e^(−x).
- Combine into −e^(−x)(x² + 2x + 2) + C; its derivative is e^(−x)[x² + 2x + 2 − (2x + 2)] = x²e^(−x).
Answer −e^(−x)(x² + 2x + 2) + C.
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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 ∫x²cos x dx by repeating parts and check the cancellation.
Give me a hint
Differentiate the polynomial factor on each application.
Compare my reasoning
- First obtain x²sin x − 2∫x sin x dx.
- The remaining primitive is −x cos x + sin x.
- Combine to get x²sin x + 2x cos x − 2sin x + C; differentiation leaves x²cos x after the other terms cancel.
x²sin x + 2x cos x − 2sin x + C.
Look for these in your work
- Retain the outside −2 during the second application.
- Differentiate every product in the final check.
Find a primitive of x arctan x and evaluate its integral from 0 to 1 exactly.
Give me a hint
Differentiate arctan x in parts, then divide the rational remainder.
Compare my reasoning
- Parts gives (x²/2)arctan x − (1/2)∫x²/(1 + x²) dx.
- Using x²/(1 + x²) = 1 − 1/(1 + x²) gives F(x) = [(x² + 1)/2]arctan x − x/2.
- F(1) − F(0) = π/4 − 1/2, positive as expected.
Primitive [(x² + 1)/2]arctan x − x/2 + C; definite integral π/4 − 1/2.
Look for these in your work
- Combine parts with the standard inverse-tangent primitive.
- Use exact radian endpoints and check the result's sign.
A vessel starts with 10 litres and has supplied inflow q(t) = t²e^(−t) litres per minute on 0 ≤ t ≤ 1, with no outflow. Find the exact added volume and final total.
Give me a hint
Use the repeated-parts primitive and keep both endpoint values exact.
Compare my reasoning
- A primitive is F(t) = −e^(−t)(t² + 2t + 2).
- The added volume is F(1) − F(0) = −5/e − (−2) = 2 − 5/e litres, approximately 0.161 litres.
- The final total is 10 + 2 − 5/e = 12 − 5/e litres.
Added volume 2 − 5/e litres; final total 12 − 5/e litres.
Look for these in your work
- Preserve all signs across repeated parts and endpoint subtraction.
- Retain exact values until final rounding and separate increase from total.
Common mistakes
- Do not stop after the first application if the remaining product still needs parts.
- Keep outside multipliers and nested subtraction signs when combining results.
- Simplify a rational remainder rather than forcing every step to use the same method.
What can you explain now?
Rewrite x²/(1 + x²) as a constant plus a simple reciprocal.
Compare with the explanation
1 − 1/(1 + x²).
Putting the two terms over their common denominator recovers x²; this makes the remainder integrable.
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Make it stick
Repeat parts until a product becomes manageable. Explain how the x arctan x example changes method for its rational remainder and verify the final derivative.
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.