A2 June 2025 Paper 1 Q7
7 In this question you must show detailed reasoning.
By first expressing \(\dfrac{1}{x^2 - 4}\) in partial fractions, show that \(\displaystyle\int_3^{\infty} \frac{1}{x^2 - 4}\,\mathrm{d}x = \frac{1}{m}\ln n\), where \(m\) and \(n\) are integers to be determined. [8]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{1}{x^2 - 4} = \dfrac{1}{(x - 2)(x + 2)}\) | B1 | 1.1 |
| \(\dfrac{1}{(x - 2)(x + 2)} = \dfrac{A}{x - 2} + \dfrac{B}{x + 2}\) | M1 | 1.1 |
| \(1 = A(x + 2) + B(x - 2)\) | M1 | 1.1 |
| \(x = 2 \Rightarrow A = \tfrac{1}{4},\; x = -2 \Rightarrow B = -\tfrac{1}{4}\) | A1 | 1.1 |
| \(\displaystyle\int_3^{\infty} \frac{1}{x^2 - 4}\,\mathrm{d}x = \frac{1}{4}\int_3^{\infty} \left(\frac{1}{x - 2} - \frac{1}{x + 2}\right)\mathrm{d}x\) \(= \dfrac{1}{4}\displaystyle\lim_{k \to \infty}\big[\ln(x - 2) - \ln(x + 2)\big]_3^k\) | M1* | 2.1 |
| \(= \dfrac{1}{4}\displaystyle\lim_{k \to \infty}\left[\ln\left(\frac{x - 2}{x + 2}\right)\right]_3^k\) | A1 | 1.1 |
| As \(k \to \infty\) \(\dfrac{k - 2}{k + 2} \to 1\) or \(\ln\left(\dfrac{k - 2}{k + 2}\right) \to 0\) | A1 | 2.5 |
| so \(\dfrac{1}{4}\displaystyle\lim_{k \to \infty}\left[\ln\left(\frac{x - 2}{x + 2}\right)\right]_3^k = -\frac{1}{4}\ln\frac{1}{5} = \frac{1}{4}\ln 5\) | B1dep | 2.1 |
| [8] |
Notes
B1: rewriting \(\frac{1}{x^2 - 4}\) as \(\frac{1}{(x - 2)(x + 2)}\) soi
M1: rewriting as partial fractions
M1: evaluating their \(A\) and \(B\) using substitution, or equating coeffs
A1: \(A = \frac{1}{4}\) and \(B = -\frac{1}{4}\) or \(\frac{1}{4(x - 2)} - \frac{1}{4(x + 2)}\)
M1*: integrate their \(\left(\frac{1}{x - 2} - \frac{1}{x + 2}\right)\) correctly (condone missing \(\frac{1}{4}\) or incorrect multiples). \(\frac{1}{4}\) could be incorporated into logarithms.
A1: \(\frac{1}{4}\left[\ln\left(\frac{x - 2}{x + 2}\right)\right]\). Fraction could be unsimplified.
A1: clear limit argument used to evaluate limit as \(k \to \infty\) or \(\lim_{k \to \infty}\left(\ln\frac{k - 2}{k + 2}\right) = 0\). Must work with a single term.
\(\to\) or \(=\) must be used correctly, e.g. do not condone \(\lim_{k \to \infty}\left[\ln\frac{k - 2}{k + 2}\right] \to 0\) or “\(k \to \infty, \frac{k - 2}{k + 2} = 1\)”
B1dep: Or \(m = 4, n = 5\).