Recognise a logarithmic derivative inside an integral
Match a numerator to the denominator's derivative, retain its constant multiplier and use a logarithm on valid intervals.
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 composites and natural logarithms.
- Identify denominator exclusions and use radian trigonometric identities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the derivative of 3x² + 2?
What you will learn
- Recognise and integrate kf′(x)/f(x) where k is constant.
- Integrate tangent through the logarithmic pattern and preserve zeros of the original denominator.
Compare the numerator with a derivative
On a connected interval where f(x) ≠ 0, d(ln|f(x)|)/dx = f′(x)/f(x). Hence ∫kf′(x)/f(x) dx = kln|f(x)| + C for constant k. First name the denominator f, find f′ and compare it with the numerator. For 6x/(3x² + 2), f′ = 6x, so the primitive is ln(3x² + 2) + C. This quadratic is always positive, so absolute-value bars are optional in this example.
For x/(x² − 1), the numerator is one half of f′ = 2x, giving (1/2)ln|x² − 1| + C. The inputs ±1 remain excluded. A numerator such as x + 1 is not a constant multiple of 2x; split it into the matching part and remainder, or choose another suitable method. Do not treat a variable ratio as the constant k in this rule.
Tangent becomes a logarithm through its sine numerator
Since tan x = sin x/cos x where cos x ≠ 0, choose f = cos x with f′ = −sin x. Then ∫tan x dx = −ln|cos x| + C, with radians used in the derivative. A linear-composite example gives ∫tan(2x) dx = −(1/2)ln|cos(2x)| + C. The minus sign and factor 1/2 both follow from differentiation, rather than from guessing a primitive with the same function name.
For ∫ from 0 to π/4 of tan x dx, cosine stays positive, and endpoint evaluation gives −ln(√2/2) = (1/2)ln 2. An interval crossing a zero of cosine needs a singularity check and cannot be evaluated by ordinary endpoint subtraction. Similarly, ∫sin x/(2 + cos x) dx = −ln(2 + cos x) + C for every real x, because the positive denominator never vanishes and its derivative is −sin x.
Pause and explain
Why is ∫tan x dx equal to −ln|cos x| + C on a valid interval?
Worked example
Find ∫6x/(3x² + 2) dx and verify it.
Show the worked solution
- Set f(x) = 3x² + 2 and compute f′(x) = 6x.
- The integrand is exactly f′/f, so use ln(3x² + 2) + C.
- Differentiate the logarithm to recover 6x/(3x² + 2); its denominator is positive for all real inputs.
Answer ln(3x² + 2) + 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 ∫sin x/(2 + cos x) dx and state why the primitive is defined for all real x.
Give me a hint
Differentiate the denominator and compare its sign with the numerator.
Compare my reasoning
- For f = 2 + cos x, f′ = −sin x.
- The integrand is −f′/f, giving −ln(2 + cos x) + C.
- Because 1 ≤ 2 + cos x ≤ 3, the denominator and logarithm argument are always positive.
−ln(2 + cos x) + C, for all real x.
Look for these in your work
- Match the denominator derivative with the correct sign.
- Justify the domain using the denominator's range.
Evaluate ∫ from 0 to π/4 of tan x dx exactly and check the interval.
Give me a hint
Use −ln|cos x| and exact endpoint values in radians.
Compare my reasoning
- Cosine is positive throughout [0, π/4], so there is no tangent pole.
- Subtract endpoints: −ln(cos(π/4)) + ln(cos 0).
- The result is −ln(√2/2) = (1/2)ln 2, positive as the integrand is non-negative.
(1/2)ln 2.
Look for these in your work
- Check the whole interval for denominator zeros.
- Retain the sign and exact logarithmic endpoint value.
A vessel begins with 3 litres and has supplied inflow q(t) = 4t/(t² + 1) litres per minute on 0 ≤ t ≤ 2, with no outflow. Find its added volume and final total.
Give me a hint
The numerator is twice the derivative of t² + 1.
Compare my reasoning
- A primitive is 2ln(t² + 1).
- Evaluation on [0, 2] gives increase 2ln 5 litres, since ln 1 = 0.
- The final total is 3 + 2ln 5 litres; the positive denominator introduces no pole.
Added volume 2ln 5 litres; final total 3 + 2ln 5 litres.
Look for these in your work
- Find the constant derivative multiplier before integrating.
- Add the initial stock separately and keep units.
Common mistakes
- The multiplier k must be constant; compare derivatives before using the shortcut.
- Keep the minus sign when the denominator derivative is negative.
- Log absolute values do not fill zeros of the original denominator.
What can you explain now?
Find ∫x/(x² + 1) dx.
Compare with the explanation
(1/2)ln(x² + 1) + C.
The numerator is one half of the derivative 2x of the positive denominator.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Name f, compute f′ and identify the constant multiplier in a logarithmic-pattern integrand. Explain how the same rule integrates tangent and why domain checks remain necessary.
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.