Integrate linear partial fractions including repeated factors
Decompose first, integrate simple and repeated poles with different rules, and preserve the original excluded inputs.
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
- Decompose the linear-factor forms in Paper 3 Algebra.
- Integrate linear reciprocals with logarithms and other negative powers with the power rule.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which terms are needed for a denominator x(x + 1)²?
What you will learn
- Integrate rational functions with distinct or repeated linear factors after a valid decomposition.
- Handle an equal-degree numerator by division and check the full definite interval for poles.
The denominator power decides the primitive
Resolve a rational function into the allowed partial fractions before integrating its terms. For a ≠ 0, ∫A/(ax + b) dx = (A/a)ln|ax + b| + C on each connected interval avoiding its pole. For a repeated square, ∫B/(ax + b)² dx = −B/[a(ax + b)] + C. The square needs a reciprocal primitive rather than the same logarithm used for the first power.
For (3x + 5)/[(x + 1)(x + 2)], the decomposition is 2/(x + 1) + 1/(x + 2). Recombining gives numerator 2(x + 2) + (x + 1) = 3x + 5, an independent algebra check. Integration gives 2ln|x + 1| + ln|x + 2| + C on an interval excluding −1 and −2. The absolute values provide positive log arguments; they do not remove the original poles.
Division and endpoint checks are separate steps
An equal-degree case such as x²/(x² − 1) first becomes 1 + 1/(x² − 1), then 1 + 1/[2(x − 1)] − 1/[2(x + 1)]. Its primitive is x + (1/2)ln|(x − 1)/(x + 1)| + C, for x ≠ ±1 on separate intervals. The polynomial term x would be lost if you treated the original function as a proper fraction without dividing. Section 3.1 excludes higher-degree numerators; this example uses equal degrees.
Before evaluating F(b) − F(a), check that the original denominator never vanishes between the endpoints. For ∫ from 2 to 3 of x²/(x² − 1) dx, there is no pole inside the interval, so the result is 1 + (1/2)ln(3/2). For an interval crossing 1 or −1, ordinary endpoint subtraction is not justified. When a factor cancels algebraically, a missing original input remains excluded unless the problem explicitly defines a continuous extension.
Pause and explain
What should happen before integrating x²/(x² − 1)?
Worked example
Integrate (3x + 5)/[(x + 1)(x + 2)] and state its exclusions.
Show the worked solution
- Write A/(x + 1) + B/(x + 2), giving A + B = 3 and 2A + B = 5.
- Hence A = 2 and B = 1; check by recombining the two fractions.
- Integrate the two first-power reciprocals and retain the poles x = −1 and −2.
Answer 2ln|x + 1| + ln|x + 2| + C, on separate intervals excluding −1 and −2.
Try a topic test
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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.
Evaluate ∫ from 1 to 2 of 1/[x(x + 1)(x + 2)] dx exactly.
Give me a hint
Clear the three distinct linear factors to find coefficients 1/2, −1 and 1/2; check every pole is outside the interval.
Compare my reasoning
- Recombination verifies 1/[x(x + 1)(x + 2)] = 1/(2x) − 1/(x + 1) + 1/[2(x + 2)].
- A primitive on this positive interval is (1/2)ln x − ln(x + 1) + (1/2)ln(x + 2).
- Subtract endpoints to get (5/2)ln 2 − (3/2)ln 3 = (1/2)ln(32/27); the poles 0, −1 and −2 are outside [1, 2].
(1/2)ln(32/27).
Look for these in your work
- Verify all three fraction coefficients and the pole-free interval.
- Subtract complete endpoints in the right order and simplify the logarithm.
Integrate (x + 2)/[x(x + 1)²], giving all repeated-factor terms and exclusions.
Give me a hint
Clear the denominator and compare coefficients, or substitute the roots and one additional input.
Compare my reasoning
- The decomposition is 2/x − 2/(x + 1) − 1/(x + 1)²; recombination gives x + 2.
- Integrate term by term: 2ln|x| − 2ln|x + 1| + 1/(x + 1) + C.
- Differentiation recovers the three fractions; retain original exclusions x = 0 and −1 on separate intervals.
2ln|x| − 2ln|x + 1| + 1/(x + 1) + C, with x ≠ 0, −1.
Look for these in your work
- Include both powers of the repeated factor.
- Use the correct reciprocal sign and preserve the original domain.
A supplied positive rate is q(t) = t²/(t² − 1) litres per minute for 2 ≤ t ≤ 3 minutes. Find the modelled volume added over this interval and explain why endpoint subtraction is valid.
Give me a hint
Divide the equal-degree fraction before decomposing its remainder.
Compare my reasoning
- Write q(t) = 1 + 1/[2(t − 1)] − 1/[2(t + 1)].
- A primitive is t + (1/2)ln|(t − 1)/(t + 1)|; neither pole ±1 lies in [2, 3].
- Evaluation gives 1 + (1/2)ln[(2/4)/(1/3)] = 1 + (1/2)ln(3/2) litres.
Added volume 1 + (1/2)ln(3/2) litres.
Look for these in your work
- Retain the divided polynomial part and all fraction coefficients.
- Check poles and interpret a rate integral with matching time units.
Common mistakes
- A repeated square integrates by the power rule, not by the same logarithm as a simple pole.
- Keep the polynomial term after equal-degree division.
- A finite pair of endpoint primitives cannot justify crossing an interior pole.
What can you explain now?
Find ∫1/(x + 1)² dx.
Compare with the explanation
−1/(x + 1) + C, with x ≠ −1.
Differentiating the negative reciprocal gives the positive inverse square.
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Make it stick
Explain the different primitives for simple and repeated linear factors. Decompose, integrate, differentiate back and check the interval before evaluating endpoints.
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.