22M.1.AHL.TZ1.14

$\frac{1}{u^2}+\frac{2}{u}+1$             A1

 

[1 mark]

$\int {\left(\frac{1}{\left(x+2\right)}+1\right)}^{2}dx$

$=\int \left(\frac{1}{{\left(x+2\right)}^2}+\frac{2}{x+2}+1\right)dx$   OR   $\int \left(\frac{1}{u^2}+\frac{2}{u}+1\right)du$             (M1)

$=-\frac{1}{\left(x+2\right)}+2 \ln \left|x+2\right|+x\left(+c\right)$             A1A1

Note: Award A1 for first expression, A1 for second two expressions.
Award A1A0 for a final answer of $=-\frac{1}{u}+2 \ln \left(u\right)+u\left(+c\right)$.

[3 marks]

volume $=\mathrm{\pi }{\left[-\frac{1}{\left(x+2\right)}+2 \ln \left(x+2\right)+x\right]}_0^2$             M1

$=\mathrm{\pi }\left(-\frac{1}{4}+2 \ln \left(4\right)+2+\frac{1}{2}-2 \ln 2\right)$             A1

$=\mathrm{\pi }\left(\frac{9}{4}+2 \ln \left(4\right)-2 \ln 2\right)$

use of log laws seen, for example             M1

$\mathrm{\pi }\left(\frac{9}{4}+4 \ln \left(2\right)-2 \ln 2\right)$   OR   $\mathrm{\pi }\left(\frac{9}{4}+2 \ln \left(\frac{4}{2}\right)\right)$

$=\frac{\mathrm{\pi }}{4}\left(9+8 \ln \left(2\right)\right)$   OR   $a=9, b=8$ and $c=2$             A1

Note: Other correct integer solutions are possible and should be accepted for example $a=9, b=c=4$.

 

[4 marks]

Some candidates could answer part (a) (i). The link between parts (i) and (ii) was, however, lost to the majority. Those who did see the link were often able to give a reasonable answer to (ii). But some candidates lacked the skills to integrate without the use of technology, so an indefinite integral presented many problems. Even those who successfully navigated part (a) went on to fail to see the link to part (b). Either the integration was attempted as something totally new and unconnected or it was simply found with the GDC which did not lead to a final answer in the required form.

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