A2 June 2025 Paper 2 Q12

AQACurrent spec5 marksHyperbolic Functions Integration

12 Find the value of

\[\int_1^5 \frac{1}{\sqrt{x^2 + 6x + 5}}\,\mathrm{d}x\]

Give your answer in the form

\[\ln\left(8 + a\sqrt{3} + b\sqrt{5} + c\sqrt{15}\right)\]

where \(a\), \(b\) and \(c\) are integers. [5 marks]

Mark scheme

Scheme	Marks	AO
Completes the square to obtain \((x + p)^2 - q^2\)	M1	3.1a
Integrates to obtain \(\cosh^{-1}\left(\dfrac{x + p}{q}\right)\) or \(\ln\left(x + p + \sqrt{(x + p)^2 - q^2}\right)\)	A1F	1.1b
Substitutes the limits and deduces \(\ln(4 + \sqrt{15}) - \ln(2 + \sqrt{3})\) or \(\ln(8 + \sqrt{60}) - \ln(4 + \sqrt{12})\)	A1	2.2a
Uses laws of logs to simplify result of their integration.	M1	1.1a
Completes a reasoned argument to obtain \(\ln(8 - 4\sqrt{3} - 3\sqrt{5} + 2\sqrt{15})\)	R1	2.1
	(5 marks)

Typical solution

\[\int_1^5 \frac{1}{\sqrt{x^2 + 6x + 5}}\,\mathrm{d}x = \int_1^5 \frac{1}{\sqrt{(x + 3)^2 - 2^2}}\,\mathrm{d}x\]\[\begin{aligned} &= \left[\cosh^{-1}\left(\frac{x + 3}{2}\right)\right]_1^5 \\ &= \cosh^{-1} 4 - \cosh^{-1} 2 \\ &= \ln(4 + \sqrt{15}) - \ln(2 + \sqrt{3}) \\ &= \ln\left(\frac{4 + \sqrt{15}}{2 + \sqrt{3}}\right) \\ &= \ln\left(\left(\frac{4 + \sqrt{15}}{2 + \sqrt{3}}\right)\left(\frac{2 - \sqrt{3}}{2 - \sqrt{3}}\right)\right) \\ &= \ln\left(\frac{8 - 4\sqrt{3} + 2\sqrt{15} - \sqrt{45}}{4 - 3}\right) \\ &= \ln(8 - 4\sqrt{3} - 3\sqrt{5} + 2\sqrt{15}) \end{aligned}\]