Split quadratic partial fractions into logarithmic and arctangent parts
Use a linear numerator over an irreducible quadratic, separating its derivative-shaped term from its constant term.
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 a linear numerator over an irreducible quadratic in partial fractions.
- Integrate linear reciprocals and positive quadratic reciprocals.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What numerator belongs above x² + 1 in the decomposition of a general fraction with denominator x(x² + 1)?
What you will learn
- Integrate the allowed linear-times-irreducible-quadratic partial-fraction form.
- Separate a quadratic's derivative term for a logarithm from its positive-reciprocal term for inverse tangent.
One quadratic fraction can require two primitives
For (2x + 3)/(x² + 4), split the integrand into 2x/(x² + 4) + 3/(x² + 4). The first numerator is the derivative of its denominator, giving ln(x² + 4). The constant part gives (3/2)arctan(x/2). Thus the full primitive is ln(x² + 4) + (3/2)arctan(x/2) + C. A single logarithm cannot supply the constant numerator 3.
In the allowed denominator form (ax + b)(cx² + d), the quadratic fraction needs a general linear numerator, not just a constant. For (3x + 1)/[x(x² + 1)], write A/x + (Bx + D)/(x² + 1). Clearing denominators gives (A + B)x² + Dx + A = 3x + 1, so A = 1, B = −1 and D = 3. Recombine before integrating to check all three coefficients.
Preserve exclusions while combining exact endpoint values
The preceding decomposition integrates to F(x) = ln|x| − (1/2)ln(x² + 1) + 3arctan x + C. Its original denominator excludes x = 0, while x² + 1 is positive for every real x. Work on either side of the pole separately. The factor 1/2 in the quadratic logarithm comes from the derivative 2x; the arctan coefficient 3 comes from the remaining constant numerator.
On [1, √3] there is no pole. Using arctan 1 = π/4 and arctan √3 = π/3 gives ∫(3x + 1)/[x(x² + 1)] dx over that interval as (1/2)ln(3/2) + π/4. The positive integrand has a positive result. Exact inverse-tangent angles and logarithmic ratios should be retained until a requested numerical rounding; a calculator's degree setting must not change the radian primitive.
Pause and explain
Why does ∫(2x + 3)/(x² + 4) dx need both a logarithm and arctan?
Worked example
Integrate (3x + 1)/[x(x² + 1)] and state its domain.
Show the worked solution
- Decompose into 1/x + (−x + 3)/(x² + 1).
- The −x term supplies −(1/2)ln(x² + 1), and the constant 3 supplies 3arctan x.
- Add ln|x| and C; differentiate and retain the original exclusion x = 0.
Answer ln|x| − (1/2)ln(x² + 1) + 3arctan x + C, separately on x < 0 or x > 0.
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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.
Integrate (2x + 3)/(x² + 4) and verify both numerator terms.
Give me a hint
Split the fraction into its 2x and constant-numerator parts.
Compare my reasoning
- The 2x term gives ln(x² + 4).
- The constant 3 gives (3/2)arctan(x/2).
- Differentiate the sum to recover 2x/(x² + 4) + 3/(x² + 4).
ln(x² + 4) + (3/2)arctan(x/2) + C.
Look for these in your work
- Keep the quadratic-log chain factor.
- Check the scaled arctan part independently.
Evaluate ∫ from 1 to √3 of (3x + 1)/[x(x² + 1)] dx exactly.
Give me a hint
Use the verified primitive and the exact angles π/4 and π/3.
Compare my reasoning
- The interval is positive and does not cross the pole x = 0.
- Evaluate F = ln x − (1/2)ln(x² + 1) + 3arctan x at both endpoints.
- The logarithmic difference is (1/2)ln(3/2), and the angular difference is 3(π/3 − π/4) = π/4.
(1/2)ln(3/2) + π/4.
Look for these in your work
- Use a valid decomposition and pole-free interval.
- Combine exact logarithmic and radian endpoint values.
A supplied flow is q(t) = (t² + 1)/[t(t² + 4)] litres per minute for 1 ≤ t ≤ 2. Find the modelled added volume exactly.
Give me a hint
Try A/t + Bt/(t² + 4), then compare the constant and quadratic coefficients.
Compare my reasoning
- Clearing denominators gives 4A = 1 and A + B = 1, so A = 1/4 and B = 3/4.
- A primitive on this positive interval is (1/4)ln t + (3/8)ln(t² + 4); the pole t = 0 is outside it.
- Subtract endpoints to obtain (1/4)ln 2 + (3/8)ln(8/5) litres.
Added volume (1/4)ln 2 + (3/8)ln(8/5) litres.
Look for these in your work
- Recombine the fractions to verify both coefficients.
- Use the quadratic-log factor 1/2 and match rate and time units.
Common mistakes
- Use a general linear numerator above the irreducible quadratic.
- Split the numerator before deciding which primitive applies.
- Positive quadratics have no real poles, but a separate linear factor may still have one.
What can you explain now?
Find ∫x/(x² + 4) dx.
Compare with the explanation
(1/2)ln(x² + 4) + C.
The denominator derivative is 2x, so the given numerator is one half of it.
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Make it stick
Decompose a linear-times-quadratic rational function, separate the quadratic numerator and identify which term needs a logarithm or inverse tangent.
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