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Date	May 2019	Marks available	6	Reference code	19M.2.AHL.TZ1.H_5
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	H_5	Adapted from	N/A

Question

The function $f$ is defined by $f\left( x \right) = {\text{sec}}\,x + 2$ , $0 \leqslant x < \frac{\pi }{2}$ .

Use integration by parts to find $\int {{{\left( {{\text{ln}}\,x} \right)}^2}} {\text{d}}x$ .

Markscheme

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

METHOD 1

write as $\int {1 \times {{\left( {{\text{ln}}\,x} \right)}^2}} {\text{d}}x$ (M1)

$= x{\left( {{\text{ln}}\,x} \right)^2} - \int {x \times \frac{{2\left( {{\text{ln}}\,x} \right)}}{x}} {\text{d}}x\left( { = x{{\left( {{\text{ln}}\,x} \right)}^2} - \int {2\,{\text{ln}}\,x} } \right)$ M1A1

$= x{\left( {{\text{ln}}\,x} \right)^2} - 2x\,{\text{ln}}\,x + \int 2 \,{\text{d}}x$ (M1)(A1)

$= x{\left( {{\text{ln}}\,x} \right)^2} - 2x\,{\text{ln}}\,x + 2x + c$ A1

METHOD 2

let $u = {\text{ln}}\,x$ M1

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}$

$\int {{u^2}{{\text{e}}^u}} {\text{d}}u$ A1

$= {u^2}{{\text{e}}^u} - \int {2u{{\text{e}}^u}} {\text{d}}u$ M1

$= {u^2}{{\text{e}}^u} - 2u{{\text{e}}^u} + \int {2{{\text{e}}^u}} {\text{d}}u$ A1

$= {u^2}{{\text{e}}^u} - 2u{{\text{e}}^u} + 2{{\text{e}}^u} + c$

$= x{\left( {{\text{ln}}\,x} \right)^2} - 2x\,{\text{ln}}\,x + 2x + c$ M1A1

METHOD 3

Setting up $u = {\text{ln}}\,x$ and $\frac{{{\text{d}}v}}{{{\text{d}}x}} = {\text{ln}}\,x$ M1

${\text{ln}}\,x\left( {x\,{\text{ln}}\,x - x} \right) - \int {\left( {{\text{ln}}\,x - 1} \right)} {\text{d}}x$ M1A1

$= x{\left( {{\text{ln}}\,x} \right)^2} - x\,{\text{ln}}\,x - \left( {x\,{\text{ln}}\,x - x} \right) + x + c$ M1A1

$= x{\left( {{\text{ln}}\,x} \right)^2} - 2x\,{\text{ln}}\,x + 2x + c$ A1

[6 marks]

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Syllabus sections

Topic 5 —Calculus » AHL 5.16—Integration by substitution, parts and repeated parts

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19M.1.AHL.TZ2.H_4:
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17M.1.AHL.TZ2.H_6a:
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17N.2.AHL.TZ0.H_8:
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18N.1.AHL.TZ0.H_10d:
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