(a)     Show that \(\frac{3}{{x + 1}} + \frac{2}{{x + 3}} = \frac{{5x + 11}}{{{x^2} + 4x + 3}}\).

(b)     Hence find the value of k such that \(\int_0^2 {\frac{{5x + 11}}{{{x^2} + 4x + 3}}{\text{d}}x = \ln k} \) .

(a)     \(\frac{3}{{x + 1}} + \frac{2}{{x + 3}} = \frac{{3(x + 3) + 2(x + 1)}}{{(x + 1)(x + 3)}}\)     M1

\( = \frac{{3x + 9 + 2x + 2}}{{{x^2} + 4x + 3}}\)     A1

\( = \frac{{5x + 11}}{{{x^2} + 4x + 3}}\)     AG

(b)     \(\int_0^2 {\frac{{5x + 11}}{{{x^2} + 4x + 3}}{\text{d}}x} = \int_0^2 {\left( {\frac{3}{{x + 1}} + \frac{2}{{x + 3}}} \right){\text{d}}x} \)     M1

\( = \left[ {3\ln (x + 1) + 2\ln (x + 3)} \right]_0^2\)     A1

\( = 3\ln 3 + 2\ln 5 - 3\ln 1 - 2\ln 3\,\,\,\,\,( = 3\ln 3 + 2\ln 5 - 2\ln 3)\)     A1

\( = \ln 3 + 2\ln 5\)

\( = \ln 75\,\,\,\,\,(k = 75)\)     A1

[6 marks]

Many students did not ‘Show’ enough in a) in order to be convincing. The need for the steps of the simplification to be shown was not clear. Too many did not link a) to b) and seemed to not be aware of the Command Term ‘hence’ and its implication for marking ( no marks will be awarded to alternative methods). The simplifications of the log expressions were done poorly by many and the fact that \({3^3} = 9\) was noted by too many. There were very few elegant solutions to this question.