October 2020 Paper 2 Q6

EdexcelCurrent spec7 marksAlgebraic Fractions Integration

(a) Given that\[\frac{x^2 + 8x - 3}{x + 2} \equiv Ax + B + \frac{C}{x + 2} \qquad x \in \mathbb{R}\ \ x \ne -2\]find the values of the constants \(A\), \(B\) and \(C\) (3)

(b) Hence, using algebraic integration, find the exact value of\[\int_0^6 \frac{x^2 + 8x - 3}{x + 2}\,\mathrm{d}x\]giving your answer in the form \(a + b\ln 2\) where \(a\) and \(b\) are integers to be found. (4)

Mark scheme (a)

Scheme	Marks	AO
\(x^2 + 8x - 3 = (Ax + B)(x + 2) + C\) or \(Ax(x + 2) + B(x + 2) + C\) \(\Rightarrow A = \ldots, B = \ldots, C = \ldots\) or \[\begin{array}{rr} & x + 6 \\ x + 2\,\big) & \overline{x^2 + 8x - 3} \\ & \underline{x^2 + 2x} \\ & 6x - 3 \\ & \underline{6x + 12} \\ & -15 \end{array}\]	M1	1.1b
Two of \(A = 1, B = 6, C = -15\)	A1	1.1b
All three of \(A = 1, B = 6, C = -15\)	A1	1.1b
	(3)

Notes

M1: Multiplies by \((x + 2)\) and attempts to find values for \(A\), \(B\) and \(C\) e.g. by comparing coefficients or substituting values for \(x\). If the method is unclear, at least 2 terms must be correct on rhs.
Or attempts to divide \(x^2 + 8x - 3\) by \(x + 2\) and obtains a linear quotient and a constant remainder.
This mark may be implied by 2 correct values for \(A\), \(B\) or \(C\)

A1: Two of \(A = 1, B = 6, C = -15\). But note that just performing the division correctly is insufficient and they must clearly identify their \(A\), \(B\), \(C\) to score any accuracy marks.

A1: All three of \(A = 1, B = 6, C = -15\)
This is implied by stating \(\dfrac{x^2 + 8x - 3}{x + 2} = x + 6 - \dfrac{15}{x + 2}\) or within the integral in (b)

Mark scheme (b)

Scheme	Marks	AO
\(\displaystyle\int \dfrac{x^2 + 8x - 3}{x + 2}\,\mathrm{d}x = \int x + 6 - \dfrac{15}{x + 2}\,\mathrm{d}x = \ldots - 15\ln(x + 2)\)	M1	1.1b
\(= \dfrac{1}{2}x^2 + 6x - 15\ln(x + 2) \quad (+c)\)	A1ft	1.1b
\(\displaystyle\int_0^6 \dfrac{x^2 + 8x - 3}{x + 2}\,\mathrm{d}x = \left[\dfrac{1}{2}x^2 + 6x - 15\ln(x + 2)\right]_0^6\) \(= (18 + 36 - 15\ln 8) - (0 + 0 - 15\ln 2)\) \(= 18 + 36 + (15 - 45)\ln 2\) or e.g. \(18 + 36 + 15\ln\left(\dfrac{2}{8}\right)\) (corrected from the printed mark scheme: \(18 + 36 - (15 - 45)\ln 2\))	M1	2.1
\(= 54 - 30\ln 2\)	A1	1.1b
	(4)
	(7 marks)

Notes

M1: Integrates an expression of the form \(\dfrac{C}{x + 2}\) to obtain \(k\ln(x + 2)\).
Condone the omission of brackets around the “\(x + 2\)”

A1ft: Correct integration ft on their \(Ax + B + \dfrac{C}{x + 2},\ (A, B, C \ne 0)\) The brackets should be present around the “\(x + 2\)” unless they are implied by subsequent work.

M1: Substitutes both limits 0 and 6 into an expression that contains an \(x\) or \(x^2\) term or both and a ln term and subtracts either way round WITH fully correct log work to combine two log terms (but allow sign errors when removing brackets) leading to an answer of the form \(a + b\ln c\) (\(a\), \(b\) and \(c\) not necessarily integers)
e.g. if they expand to get \(-15\ln 8 - 15\ln 2\) followed by \(-15\ln 16\) and reach \(a + b\ln c\) then allow the M mark

A1: \(54 - 30\ln 2\) (Apply isw once a correct answer is seen)