Find  \(\int {\frac{1}{{2x + 3}}} {\rm{d}}x\) .

Given that \(\int_0^3 {\frac{1}{{2x + 3}}} {\rm{d}}x = \ln \sqrt P \) , find the value of P.

\(\int {\frac{1}{{2x + 3}}} {\rm{d}}x = \frac{1}{2}\ln (2x + 3) + C\)  (accept \(\frac{1}{2}\ln |(2x + 3)| + C\) )    A1A1     N2

[2 marks]

\(\int_0^3 {\frac{1}{{2x + 3}}} {\rm{d}}x = \left[ {\frac{1}{2}\ln (2x + 3)} \right]_0^3\)

evidence of substitution of limits     (M1)

e.g.\(\frac{1}{2}\ln 9 - \frac{1}{2}\ln 3\)

evidence of correctly using \(\ln a - \ln b = \ln \frac{a}{b}\) (seen anywhere)     (A1)

e.g. \(\frac{1}{2}\ln 3\)

evidence of correctly using \(a\ln b = \ln {b^a}\) (seen anywhere)     (A1)

e.g. \(\ln \sqrt {\frac{9}{3}} \)

\(P = 3\) (accept \(\ln \sqrt 3 \) )     A1     N2

[4 marks]

Many candidates were unable to correctly integrate but did recognize that the integral involved the natural log function; they most often missed the factor \(\frac{1}{2}\) or replaced it with 2.

Part (b) proved difficult as many were unable to use the basic rules of logarithms.