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.