AQA A-Level Mathematics Paper 3, June 2025: Question 8

7 marks Β· Medium difficulty Β· Multi-step Problem

Express a rational algebraic fraction in partial fractions and use the result to evaluate a definite integral in the form $\ln q$.

Practise this question

Question

Question 8 contains two parts. Part (a) states: 'The expression x / (2x^2 + 3x + 1) can be written in the form A / (x + 1) + B / (2x + 1). Find the value of A and the value of B' for 3 marks. Part (b) states: 'Use your answer to part (a) to show that the integral from 0 to 4 of x / (2x^2 + 3x + 1) dx = ln q where q is a rational number to be found' for 4 marks.

Mark scheme

Show the mark scheme Mark scheme for Question 8. Part (a) awards M1 for setting x = A(2x + 1) + B(x + 1) and solving, A1 for A = 1, and A1 for B = -1. Part (b) awards M1 for obtaining an integrated form with ln terms, A1 for ln(x + 1) - (1/2)ln(2x + 1), M1 for substituting limits 0 and 4, and R1 for simplifying with logarithm rules to reach ln(5/3).

How to answer it

Partial Fractions and Definite Integration into Logarithms

πŸ“‹ What This Question Tests

This question assesses your ability to manipulate rational expressions and evaluate definite integrals using calculus and algebraic laws:

  • Decomposition into Partial Fractions: Setting up identity equations and finding unknown constants for distinct linear factors in the denominator.
  • Standard Logarithmic Integration: Using the standard result ∫ 1/(ax + b) dx = (1/a) ln|ax + b|.
  • Definite Integration: Accurately substituting limits and evaluating expressions.
  • Laws of Logarithms: Applying power laws (k ln a = ln(ak)) and quotient laws (ln a βˆ’ ln b = ln(a/b)) to express the final answer in the exact required form ln q.

Part (a) Finding Constants A and B

Expressing a rational algebraic function as partial fractions [3 Marks]

πŸ’‘ Key Knowledge

  • Notice the denominator factorisation:
    2xΒ² + 3x + 1 = (2x + 1)(x + 1)
  • Form the algebraic identity:
    x ≑ A(2x + 1) + B(x + 1)
  • The constants can be found by substituting strategic roots (values of x that make brackets zero) or by comparing coefficients.

πŸ“ Step-by-Step Calculation

  1. Identity equation:
    x = A(2x + 1) + B(x + 1)
  2. Find A by setting x = βˆ’1:
    βˆ’1 = A(2(βˆ’1) + 1) + B(0)
    βˆ’1 = A(βˆ’1) β‡’ A = 1
  3. Find B by setting x = βˆ’Β½:
    βˆ’Β½ = A(0) + B(βˆ’Β½ + 1)
    βˆ’Β½ = B(Β½) β‡’ B = βˆ’1

βœ… Correct Answer & Marks Breakdown

A = 1 and B = βˆ’1

[M1] Valid method shown to solve for either constant (substitution, equating coefficients, or inspection).
[A1] Correct value: A = 1 (or seen in numerator over x + 1).
[A1] Correct value: B = βˆ’1 (or seen in numerator over 2x + 1).

❌ Common Traps

  • Mismatched denominators: Cross-multiplying incorrectly and associating A with (x + 1) rather than (2x + 1).
  • Arithmetic / Sign Errors: Tripping up on negative fractions: βˆ’Β½ = Β½B is frequently miscalculated as B = 1 or B = βˆ’ΒΌ.

Part (b) Evaluating the Definite Integral

Integrating to logarithmic forms and simplifying using log laws [4 Marks]

πŸ’‘ Key Knowledge

  • Reverse Chain Rule:
    ∫ 1/(x + 1) dx = ln(x + 1)
    ∫ 1/(2x + 1) dx = ½ ln(2x + 1)
  • Log Law Rules:
    ½ ln(9) = ln(9½) = ln(√9) = ln(3)
    ln(1) = 0
    ln(a) βˆ’ ln(b) = ln(a/b)

πŸ“ Step-by-Step Calculation

  1. Rewrite integral using Part (a):
    βˆ«β‚€β΄ [ 1/(x + 1) βˆ’ 1/(2x + 1) ] dx
  2. Integrate:
    = [ ln(x + 1) βˆ’ Β½ ln(2x + 1) ]₀⁴
  3. Substitute upper limit (x = 4):
    ln(4 + 1) βˆ’ Β½ ln(2(4) + 1)
    = ln(5) βˆ’ Β½ ln(9) = ln(5) βˆ’ ln(3)
  4. Substitute lower limit (x = 0):
    ln(0 + 1) βˆ’ Β½ ln(0 + 1) = ln(1) βˆ’ Β½ ln(1) = 0
  5. Combine using logarithm quotient rule:
    ln(5) βˆ’ ln(3) = ln(5/3)

🧠 Exam Technique & Insight

  • Always show lower limit substitution: Even if limits yield zero, write down ln(1) βˆ’ Β½ln(1) = 0 to demonstrate a fully reasoned argument for the reasoning [R1] mark.
  • Watch the coefficient: Do not forget the 1/a coefficient when integrating linear expressions ax + b. Forgetting the Β½ before ln(2x + 1) is the single biggest mark-dropper.
  • The question requires ln q , where q is a rational number. Leaving the answer as ln 5 βˆ’ ln 3 is incomplete.

βœ… Final Value of q & Mark Breakdown

Value: ln(5/3) (hence q = 5/3 or 1.6Μ‡)

[M1] Integrating to form A ln(x + 1) or (B/2) ln(2x + 1).
[A1] Fully correct integral: ln(x + 1) βˆ’ Β½ ln(2x + 1).
[M1] Correctly substituting both limits (4 and 0) into integrated logarithmic terms.
[R1] Complete, rigorously reasoned solution (CSO) demonstrating log law simplification to reach ln(5/3).

Topics

Pure Mathematics Β· B: Algebra and functions Β· F: Exponentials and logarithms Β· H: Integration

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.