A2 June 2022 Paper 1 Q7
7 In this question you must show detailed reasoning.
Show that \(\displaystyle\int_2^3 \frac{x + 1}{(x - 1)\left(x^2 + 1\right)}\,\mathrm{d}x = \tfrac{1}{2}\ln 2\). [9]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{x + 1}{(x - 1)\left(x^2 + 1\right)} = \dfrac{A}{x - 1} + \dfrac{Bx + C}{x^2 + 1}\) \(\Rightarrow x + 1 = A\left(x^2 + 1\right) + (Bx + C)(x - 1)\) | M1 | 2.1 |
| \(x = 1 \Rightarrow 2 = 2A \Rightarrow A = 1\) | A1 | 1.1 |
| coefficient of \(x^2\): \(0 = A + B \Rightarrow B = -1\) | A1 | 2.1 |
| constants: \(1 = A - C \Rightarrow C = 0\) | A1 | 2.1 |
| \(\displaystyle\int_2^3 \frac{x + 1}{(x - 1)\left(x^2 + 1\right)}\,\mathrm{d}x = \int_2^3\left(\frac{1}{x - 1} - \frac{x}{x^2 + 1}\right)\mathrm{d}x\) \(= \left[\ln(x - 1) - \dfrac{1}{2}\ln\left(x^2 + 1\right)\right]_2^3\) | B1ft M1 A1ft | 2.1 2.1 1.1 |
| \(= \ln 2 - \dfrac{1}{2}\ln 10 + \dfrac{1}{2}\ln 5\ [-\ln 1]\) \(= \ln 2 - \frac{1}{2}\ln\frac{10}{5}\) | M1 | 1.1 |
| \(= \ln 2 - \dfrac{1}{2}\ln 2 = \dfrac{1}{2}\ln 2\) | A1 | 2.2a |
| [9] |
Notes
M1: correct partial fractions
B1ft: \(\ln(x - 1)\)
M1: \(k\ln\left(x^2 + 1\right)\) or \(u = x^2 + 1 \Rightarrow \displaystyle\int\frac{1}{2u}\,\mathrm{d}u\)
A1ft: \(k = -\frac{1}{2}\) or \(-\frac{1}{2}\ln u\)
M1: Combining two of their logarithm terms correctly
A1: AG