Given that $\int_0^5 {\frac{2}{{2x + 5}}} {\rm{d}}x = \ln k$ , find the value of k .

correct integration, $2 \times \frac{1}{2}\ln (2x + 5)$     A1A1

Note: Award A1 for $2 \times \frac{1}{2}( = 1)$ and A1 for $\ln (2x + 5)$ .

evidence of substituting limits into integrated function and subtracting     (M1)

e.g. $\ln (2 \times 5 + 5) - \ln (2 \times 0 + 5)$

correct substitution     A1

e.g. $\ln 15 - \ln 5$

correct working     (A1)

e.g. $\ln \frac{{15}}{5},\ln 3$

$k = 3$     A1     N3 

[6 marks]

Knowing that the answer to this integration led to a natural logarithm function helped many candidates make progress on this more challenging question, although some candidates simply substituted the limits straight away without integrating. Although some candidates incorrectly simplified $\ln 15 - \ln 5$ as $\ln 10$ or $\frac{{\ln 15}}{{\ln 5}} = \ln 3$ , a pleasing number applied the logarithm property correctly. Some candidates had difficulty with missing brackets which typically led to $\ln 0$ in their answer.