A2 June 2025 Paper 2 Q15

AQACurrent spec6 marksIntegration

15 Prove that

\[\int_5^{\infty} \frac{1}{(x + 1)(2x + 3)}\,\mathrm{d}x = \ln\left(\frac{13}{12}\right)\]

Show the limiting process clearly. [6 marks]

Mark scheme

Scheme	Marks	AO
Expresses the integrand as partial fractions.	M1	3.1a
Obtains a correct expression for the integrand as partial fractions.	A1	1.1b
Integrates their expression correctly to obtain an expression involving logs.	M1	1.1a
Clearly shows the limiting process by setting the upper limit to \(N\) or equivalent (not \(x\)) and taking the limit at some stage.	M1	2.4
Deduces \(\displaystyle\lim_{N \to \infty}\left(\ln\left(\frac{N + 1}{2N + 3}\right)\right) = \ln\left(\frac{1}{2}\right)\) Or \(\displaystyle\lim_{N \to \infty}\left(\ln\left(\frac{13(N + 1)}{6(2N + 3)}\right)\right) = \ln\left(\frac{13}{12}\right)\) OE	A1	2.2a
Completes a rigorous argument to obtain the correct result. Must include \(\ln\left(\dfrac{1 + {}^{1}\!/_{N}}{2 + {}^{3}\!/_{N}}\right)\) oe	R1	2.1
	(6 marks)

Typical solution

\[\frac{1}{(x + 1)(2x + 3)} \equiv \frac{A}{x + 1} + \frac{B}{2x + 3}\]\[1 \equiv A(2x + 3) + B(x + 1)\]\[x = -1:\ 1 = A\]\[x = 0:\ 1 = 3A + B \Rightarrow B = -2\]\[\begin{aligned} &\int_5^{\infty} \frac{1}{(x + 1)(2x + 3)}\,\mathrm{d}x \\ &= \int_5^{\infty} \left(\frac{1}{x + 1} - \frac{2}{2x + 3}\right)\mathrm{d}x \\ &= \lim_{N \to \infty} \int_5^{N} \left(\frac{1}{x + 1} - \frac{2}{2x + 3}\right)\mathrm{d}x \\ &= \lim_{N \to \infty} \left[\ln(x + 1) - \ln(2x + 3)\right]_5^N \\ &= \lim_{N \to \infty} \left[\ln\left(\frac{x + 1}{2x + 3}\right)\right]_5^N \\ &= \lim_{N \to \infty} \left(\ln\left(\frac{N + 1}{2N + 3}\right) - \ln\left(\frac{6}{13}\right)\right) \\ &= \lim_{N \to \infty} \left(\ln\left(\frac{1 + {}^{1}\!/_{N}}{2 + {}^{3}\!/_{N}}\right) - \ln\left(\frac{6}{13}\right)\right) \\ &= \ln\left(\frac{1}{2}\right) - \ln\left(\frac{6}{13}\right) \\ &= \ln\left(\frac{13}{12}\right) \end{aligned}\]