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.