A2 June 2024 Paper 1 Q16
16 In this question you must show detailed reasoning.
Show that \(\displaystyle\int_0^1 \frac{1}{\sqrt{x^2 + x + 1}}\,\mathrm{d}x = \ln\left(\dfrac{a + b\sqrt{3}}{c}\right)\), where \(a\), \(b\) and \(c\) are integers to be determined. [6]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\displaystyle\int_0^1 \frac{1}{\sqrt{x^2 + x + 1}}\,\mathrm{d}x = \int_0^1 \frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}}}\,\mathrm{d}x\) | B1 | 3.1a |
| let \(u = x + \frac{1}{2} \Rightarrow \mathrm{d}u = \mathrm{d}x\), giving \(\displaystyle\int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{\sqrt{u^2 + \frac{3}{4}}}\,\mathrm{d}u\) | M1* | 2.1 |
| \(= \left[\operatorname{arsinh}\dfrac{u}{\frac{1}{2}\sqrt{3}}\right]_{\frac{1}{2}}^{\frac{3}{2}}\) or \(\left[\ln\left(u + \sqrt{u^2 + \frac{3}{4}}\right)\right]_{\frac{1}{2}}^{\frac{3}{2}}\) \(\left[= \operatorname{arsinh}\sqrt{3} - \operatorname{arsinh}\frac{1}{\sqrt{3}}\right]\) | A1 | 2.1 |
| \(= \ln\left(\sqrt{3} + 2\right) - \ln\dfrac{3}{\sqrt{3}}\) or \(\ln\left(\frac{3}{2} + \sqrt{3}\right) - \ln\frac{3}{2}\) | A1 | 2.1 |
| \(= \ln\left(\dfrac{3 + 2\sqrt{3}}{3}\right)\) | M1dep A1 | 2.1 2.1 |
| [6] |
Notes
B1: Completing the square
M1*: Any correct substitution for their integral, e.g. \(x + \frac{1}{2} = \frac{\sqrt{3}}{2}\sinh u\)
A1: e.g. \([u]_{\operatorname{arsinh}\frac{1}{\sqrt{3}}}^{\operatorname{arsinh}\sqrt{3}}\), ignore limits. Correct substitution must have been made for this mark.
A1: A correct expression in logarithmic form (need not be simplified)
M1dep: Combining logs.
A1: www. Can be implied by correct final answer provided two log terms shown.
Alternative method
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^1 \frac{1}{\sqrt{x^2 + x + 1}}\,\mathrm{d}x = \int_0^1 \frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}}}\,\mathrm{d}x\) | B1 |
| \(\left[\operatorname{arsinh}\dfrac{x + \frac{1}{2}}{\frac{1}{2}\sqrt{3}}\right]_0^1\) or \(\left[\ln\left(x + \frac{1}{2} + \sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}}\right)\right]_0^1\) | M1* A1 |
| \(= \ln\left(\sqrt{3} + 2\right) - \ln\dfrac{3}{\sqrt{3}}\) or \(\ln\left(\frac{3}{2} + \sqrt{3}\right) - \ln\frac{3}{2}\) | A1 |
| \(= \ln\left(\dfrac{3 + 2\sqrt{3}}{3}\right)\) | M1dep A1 |
| [6] |
B1: Completing the square
M1*: By inspection for their integral, no slips
A1: Correct, ignore limits
A1: A correct expression in logarithmic form (need not be simplified)
M1dep: Combining logs.
A1: www. Can be implied by correct final answer provided two log terms shown.