Integrate positive quadratic reciprocals with inverse tangent
Recognise the inverse-tangent derivative, scale the denominator correctly and check definite accumulation against a positive integrand.
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
- Differentiate arctan of an inner function.
- Reverse a chain multiplier and subtract definite-integral endpoints.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is d(arctan x)/dx with radian outputs?
What you will learn
- Integrate 1/(x² + a²) for a > 0 and scaled positive quadratic reciprocals.
- Check the primitive by differentiation and evaluate exact definite integrals in radians.
Recognise the derivative before choosing a method
Since d(arctan x)/dx = 1/(1 + x²), its reverse gives ∫1/(1 + x²) dx = arctan x + C. For a > 0, differentiation of arctan(x/a) gives (1/a)/[1 + x²/a²] = a/(x² + a²). Therefore ∫1/(x² + a²) dx = (1/a)arctan(x/a) + C. The outside factor and inside scale work together; just replacing 1 by a² in the denominator is insufficient.
For A > 0 and B > 0, ∫1/(A + Bx²) dx = arctan(x√(B/A))/√(AB) + C. Check this by differentiation rather than memorising it without its conditions. Thus ∫1/(2 + 3x²) dx = arctan(x√(3/2))/√6 + C. Every real x is permitted because the denominator stays positive. A difference of squares, such as x² − a², needs a different method and has poles; do not apply the positive-quadratic rule to it.
The integral is accumulation, not a reciprocal of an angle
The inverse notation tan⁻¹ names arctan, not 1/tan. The standard primitive outputs radian values. On 0 ≤ x ≤ 1, 1/(1 + x²) is positive, so its signed integral is also the geometric area beneath it. The exact result is arctan 1 − arctan 0 = π/4. A drawing helps you see its sign and size; exact endpoint values supply the exact answer.
A supplied angular rate θ′(t) = 2/(t² + 4) rad/s integrates to θ(t) = arctan(t/2) + C radians when t is measured numerically in seconds. If θ(0) = π/6, then C = π/6 and θ(2) = 5π/12. The integral of the rate over [0, 2] is the increase π/4, whereas the final angle also includes the initial angle. Use the supplied model and units explicitly.
Pause and explain
Which primitive differentiates to 1/(x² + 4)?
Worked example
Find ∫1/(2 + 3x²) dx and verify the coefficient.
Show the worked solution
- Write the denominator as 2[1 + (x√(3/2))²].
- The inner derivative is √(3/2), so reverse it as well as the outside factor 1/2.
- Differentiating arctan(x√(3/2))/√6 gives exactly 1/(2 + 3x²).
Answer arctan(x√(3/2))/√6 + C, for all real x.
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 ∫5/(x² + 4) dx and check it by differentiation.
Give me a hint
Identify a = 2, then retain the numerator multiplier 5.
Compare my reasoning
- The base reciprocal primitive is (1/2)arctan(x/2).
- Multiply by 5 and add C to get (5/2)arctan(x/2) + C.
- Its derivative is (5/4)/[1 + x²/4] = 5/(x² + 4).
(5/2)arctan(x/2) + C.
Look for these in your work
- Scale both the argument and the outside coefficient.
- Verify the exact integrand and retain C.
Evaluate ∫ from 0 to 2 of 6/(x² + 4) dx exactly, and explain the sign.
Give me a hint
A primitive is 3arctan(x/2); use radian endpoint values.
Compare my reasoning
- Reverse differentiation gives F(x) = 3arctan(x/2).
- F(2) − F(0) = 3(π/4 − 0) = 3π/4.
- The integrand is positive throughout, so a positive signed result is consistent.
3π/4.
Look for these in your work
- Use exact inverse-tangent endpoint values in radians.
- Check the sign against the original integrand.
A supplied angular rate is θ′(t) = 2/(t² + 4) rad/s for t ≥ 0, with t measured numerically in seconds. If θ(0) = π/6 radians, find θ(2) and the angular increase from 0 to 2.
Give me a hint
Integrate the rate and determine the constant from the initial angle.
Compare my reasoning
- A primitive of the rate is arctan(t/2).
- The initial condition gives θ(t) = arctan(t/2) + π/6.
- At 2 s the angle is π/4 + π/6 = 5π/12 radians; its increase is π/4 radians.
Final angle 5π/12 radians; increase π/4 radians.
Look for these in your work
- Match the primitive to the supplied angular-rate units.
- Distinguish initial angle, increase and final angle.
Common mistakes
- Keep both the outside coefficient and the scale inside arctan.
- Use the rule only for the stated positive quadratic form.
- An integrated rate gives an increase; include an initial value separately.
What can you explain now?
Find ∫1/(x² + 9) dx.
Compare with the explanation
(1/3)arctan(x/3) + C.
The derivative is (1/9)/[1 + x²/9] = 1/(x² + 9).
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Make it stick
Derive the scaled positive-quadratic primitive from the inverse-tangent derivative, then check it and evaluate an exact definite integral.
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