A2 June 2024 Paper 1 Q11

AQACurrent spec5 marksDifferentiation & Maclaurin Integration

(a) Find \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^2\tan^{-1}x\right)\) [1 mark]

(b) Hence find \(\displaystyle\int 2x\tan^{-1}x\,\mathrm{d}x\) [4 marks]

Scheme	Marks	AO
Obtains \(2x\tan^{-1}x + \dfrac{x^2}{1 + x^2}\)	B1	1.1b
	(1)

Mark scheme (b)

Scheme	Marks	AO
Uses result of part (a) to obtain an equation involving the required integral.	M1	2.2a
Deduces that \(\dfrac{x^2}{1 + x^2} = 1 - \dfrac{1}{1 + x^2}\) or uses the substitution \(x = \tan u\) to obtain \(\sec^2 u - 1\)	M1	2.2a
Obtains \(x - \tan^{-1}x\)	A1	1.1b
Completes a reasoned argument starting with the result from part (a) to obtain \(x^2\tan^{-1}x - x + \tan^{-1}x + c\)	R1	2.1
	(4)
	(5 marks)

Typical solution

\[\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\tan^{-1}x\right) = 2x\tan^{-1}x + \frac{x^2}{1 + x^2}\]\[\therefore \int 2x\tan^{-1}x\,\mathrm{d}x + \int \frac{x^2}{1 + x^2}\,\mathrm{d}x = x^2\tan^{-1}x\]\[\begin{aligned}\int \frac{x^2}{1 + x^2}\,\mathrm{d}x &= \int \frac{1 + x^2 - 1}{1 + x^2}\,\mathrm{d}x \\ &= \int 1\,\mathrm{d}x - \int \frac{1}{1 + x^2}\,\mathrm{d}x \\ &= x - \tan^{-1}x + c\end{aligned}\]\[\begin{aligned}\int 2x\tan^{-1}x\,\mathrm{d}x &= x^2\tan^{-1}x - \int \frac{x^2}{1 + x^2}\,\mathrm{d}x \\ &= x^2\tan^{-1}x - x + \tan^{-1}x + c\end{aligned}\]