A2 June 2019 Paper 1 Q2
2. Show that
\[\int_0^{\infty} \frac{8x - 12}{\left(2x^2 + 3\right)(x + 1)}\,\mathrm{d}x = \ln k\]where \(k\) is a rational number to be found. (7)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{8x - 12}{\left(2x^2 + 3\right)(x + 1)} = \dfrac{Ax + B}{2x^2 + 3} + \dfrac{C}{x + 1}\) | M1 | 3.1a |
| \(8x - 12 = (Ax + B)(x + 1) + C\left(2x^2 + 3\right)\) E.g. \(x = -1 \Rightarrow C = -4,\ x = 0 \Rightarrow B = 0,\ x = 1 \Rightarrow A = 8\) Or Compares coefficients and solves \((A + 2C = 0 \quad A + B = 8 \quad B + 3C = -12)\) \(\Rightarrow A = \ldots,\ B = \ldots,\ C = \ldots\) | dM1 | 1.1b |
| \(A = 8 \quad B = 0 \quad C = -4\) | A1 | 1.1b |
| \(\displaystyle\int \left(\frac{8x}{2x^2 + 3} - \frac{4}{x + 1}\right)\mathrm{d}x = 2\ln\left(2x^2 + 3\right) - 4\ln(x + 1)\) | A1ft | 1.1b |
| \(2\ln\left(2x^2 + 3\right) - 4\ln(x + 1) = \ln\left(\dfrac{\left(2x^2 + 3\right)^2}{(x + 1)^4}\right)\) or \(2\ln\left(2x^2 + 3\right) - 4\ln(x + 1) = 2\ln\left(\dfrac{\left(2x^2 + 3\right)}{(x + 1)^2}\right)\) | M1 | 2.1 |
| \(\displaystyle\lim_{x \to \infty}\left\{\ln\frac{\left(2x^2 + 3\right)^2}{(x + 1)^4}\right\} = \ln 4\) or \(\displaystyle\lim_{x \to \infty}\left\{2\ln\frac{\left(2x^2 + 3\right)}{(x + 1)^2}\right\} = 2\ln 2\) | B1 | 2.2a |
| \(\displaystyle\Rightarrow \int_0^{\infty} \frac{8x - 12}{\left(2x^2 + 3\right)(x + 1)}\,\mathrm{d}x = \ln\frac{4}{9}\) cao | A1 | 1.1b |
| (7) | ||
| (7 marks) |
Notes
M1: Selects the correct form for partial fractions.
dM1: Full method for finding values for all three constants. Dependent on having the correct form for the partial fractions. Allow slips as long as the intention is clear.
A1: Correct constants or partial fractions.
A1ft: Integrates \(\displaystyle\int \frac{px}{2x^2 + 3} - \frac{q}{x + 1}\,\mathrm{d}x = \frac{p}{4}\ln\left(2x^2 + 3\right) - q\ln(x + 1)\) and no extra terms
M1: Combines two algebraic log terms correctly.
B1: Correct upper limit for \(x \to \infty\) by recognising the dominant terms. (Simply replacing \(x\) with \(\infty\) scores B0). This can be implied.
A1: Deduces the correct value for the improper integral in the correct form, cao A0 for \(2\ln\dfrac{2}{3}\)
Correct answer with no working seen is no marks.
Note: Incorrect partial fraction form, \(\dfrac{A}{2x^2 + 3} + \dfrac{B}{x + 1}\) or \(\dfrac{Ax}{2x^2 + 3} + \dfrac{B}{x + 1}\) the maximum it can score is M0M0A0A0M1B1A0