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Date	May 2018	Marks available	5	Reference code	18M.2.AHL.TZ1.H_5
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find and Hence	Question number	H_5	Adapted from	N/A

Question

Given that $2{x^3} - 3x + 1$ can be expressed in the form $Ax\left( {{x^2} + 1} \right) + Bx + C$ , find the values of the constants $A$ , $B$ and $C$ .

[2]

Hence find $\int {\frac{{2{x^3} - 3x + 1}}{{{x^2} + 1}}} {\text{d}}x$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$2{x^3} - 3x + 1 = Ax\left( {{x^2} + 1} \right) + Bx + C$

$A = 2,\,\,C = 1,$ A1

$A + B = - 3 \Rightarrow B = - 5$ A1

[2 marks]

$\int {\frac{{2{x^3} - 3x + 1}}{{{x^2} + 1}}} {\text{d}}x = \int {\left( {2x - \frac{{5x}}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}} \right)} {\text{d}}x$ M1M1

Note: Award M1 for dividing by $\left( {{x^2} + 1} \right)$ to get $2x$ , M1 for separating the $5x$ and 1.

$= {x^2} - \frac{5}{2}{\text{ln}}\left( {{x^2} + 1} \right) + {\text{arctan}}\,x\left( { + c} \right)$ (M1)A1A1

Note: Award (M1)A1 for integrating ${\frac{{5x}}{{{x^2} + 1}}}$ , A1 for the other two terms.

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution

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