Integrate linear reciprocals with logarithms on valid intervals
Use an absolute value to describe either side of a reciprocal's pole, keep the inner divisor and never integrate straight across a singularity.
Section 2.5 · Integration
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Shared sections 2.5 and 3.5 use these standard exponential, reciprocal and trigonometric integrals and identity methods. Paper 1 integration and the shared derivatives and identities are assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. The general method of integration by substitution is not required for Paper 2; Paper 3 uses given substitutions in its separate lesson. Paper 3 also adds positive-quadratic reciprocals, partial fractions, logarithmic derivative patterns and integration by parts. Use radians for trigonometric calculus. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Differentiate logarithms and identify positive log arguments.
- Find reciprocal exclusions and evaluate a definite integral.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why can the ordinary power integration formula not integrate x^(−1)?
What you will learn
- Find an antiderivative of 1/(ax + b) for a ≠ 0 on a valid connected interval.
- Evaluate definite logarithmic integrals and recognise when a pole prevents ordinary endpoint evaluation.
A reciprocal needs a logarithmic antiderivative
The power rule ∫xⁿ dx = x^(n + 1)/(n + 1) fails at n = −1. Instead, ∫1/x dx = ln|x| + C for x ≠ 0, applied on either connected interval x < 0 or x > 0. On the negative interval, ln|x| = ln(−x) has derivative (−1)/(−x) = 1/x. The absolute value supplies a positive log argument without making the original reciprocal defined at zero.
For a ≠ 0, ∫1/(ax + b) dx = ln|ax + b|/a + C where ax + b ≠ 0. Thus ∫6/(3x − 2) dx = 2 ln|3x − 2| + C. When the problem already restricts 3x − 2 > 0, ln(3x − 2) is an equivalent form on that interval. The constants on separate sides of the pole can differ; one initial value on one side does not determine the other side.
Check the entire integration interval before subtracting endpoints
For ∫ from −2 to −1 of 3/(2x − 1) dx, the denominator is negative but never zero. The antiderivative is (3/2)ln|2x − 1|, so the signed result is (3/2)[ln 3 − ln 5] = (3/2)ln(3/5), which is negative as expected from a negative integrand. The absolute value does not turn a signed integral into positive geometric area.
For ∫ from 0 to 1 of 1/(2x − 1) dx, the pole x = 1/2 lies inside the interval. Ordinary endpoint subtraction would misleadingly give zero because both endpoint logarithms are ln 1. Each one-sided improper integral diverges, so this ordinary improper integral does not exist. Cancellation of two infinities is invalid. A valid primitive on separated intervals cannot justify crossing their excluded input. If a = 0 and b ≠ 0, integrate the constant 1/b directly; if b = 0 too, the integrand is undefined.
Pause and explain
Does endpoint subtraction prove ∫ from 0 to 1 of 1/(2x − 1) dx equals zero?
Worked example
Find ∫6/(3x − 2) dx and state its interval restrictions.
Show the worked solution
- The original reciprocal excludes x = 2/3.
- Reverse the log chain factor: 6/3 = 2, giving 2 ln|3x − 2| + C.
- Differentiation gives 2·3/(3x − 2), confirming the integrand on either separate domain interval.
Answer 2 ln|3x − 2| + C, on x < 2/3 or x > 2/3 separately.
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 ∫4/(2x + 1) dx and state the excluded input.
Give me a hint
Use a logarithm, divide by the inner coefficient and check the reciprocal domain.
Compare my reasoning
- The pole is at x = −1/2.
- The antiderivative is 2 ln|2x + 1| + C on either separate interval.
- Differentiation gives 2·2/(2x + 1) = 4/(2x + 1).
2 ln|2x + 1| + C, with x ≠ −1/2 and separate interval constants.
Look for these in your work
- Reverse the inner log derivative by dividing by 2.
- Preserve the reciprocal exclusion after integration.
Evaluate ∫ from −2 to −1 of 3/(2x − 1) dx exactly and explain its sign.
Give me a hint
The interval stays entirely on one side of the pole; use the absolute values of both negative denominators.
Compare my reasoning
- A primitive is (3/2)ln|2x − 1|, and no pole occurs on [−2, −1].
- Endpoint subtraction gives (3/2)(ln 3 − ln 5) = (3/2)ln(3/5).
- The result is negative, agreeing with the negative integrand throughout the interval.
(3/2)ln(3/5), a negative signed integral.
Look for these in your work
- Verify continuity on the full interval before endpoint evaluation.
- Retain signed accumulation despite the logarithmic absolute value.
A vessel starts with 5 litres and has no outflow. Its supplied inflow is 6/(t + 2) litres per minute on 0 ≤ t ≤ 4. Find the increase and final total exactly.
Give me a hint
The denominator stays positive on the time interval; integrate the rate and then add the initial stock.
Compare my reasoning
- A primitive of the rate is 6 ln(t + 2), valid throughout the interval.
- The increase is 6(ln 6 − ln 2) = 6 ln 3 litres.
- The final total is 5 + 6 ln 3 litres; the reciprocal pole at t = −2 is outside the supplied model domain.
Increase 6 ln 3 litres; final total 5 + 6 ln 3 litres.
Look for these in your work
- Use a valid connected interval and a correct logarithmic difference.
- Report the increase and total as separate volumes.
Common mistakes
- Do not apply the power formula with new exponent zero.
- An absolute value in the log keeps its argument positive; it does not remove the original pole.
- Check for a singularity inside a definite integral before evaluating endpoints.
What can you explain now?
Find ∫1/(3 − 2x) dx on a valid interval.
Compare with the explanation
−(1/2)ln|3 − 2x| + C, with x ≠ 3/2.
The logarithmic chain factor −2 is reversed by the divisor −2.
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Make it stick
Derive the reciprocal-linear log formula by differentiating it. Explain the absolute value and why a pole between the endpoints invalidates ordinary endpoint subtraction.
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.