Find the exact value of \(\int_0^\infty  {\frac{{{\text{d}}x}}{{(x + 2)(2x + 1)}}} \).

Let \(\frac{1}{{(x + 2)(2x + 1)}} = \frac{A}{{x + 2}} + \frac{B}{{2x + 1}} = \frac{{A(2x + 1) + B(x + 2)}}{{(x + 2)(2x + 1)}}\)     M1A1

\(x = - 2 \to A = - \frac{1}{3}\)     A1

\(x = - \frac{1}{2} \to B = \frac{2}{3}\)     A1     N3

\(I = \frac{1}{3}\int_0^h {\left[ {\frac{2}{{(2x + 1)}} - \frac{1}{{(x + 2)}}} \right]{\text{d}}x} \)     M1

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

\( = \frac{1}{3}\left[ {\mathop {\lim }\limits_{h \to \infty } \left( {\ln \left( {\frac{{2h + 1}}{{h + 2}}} \right)} \right) - \ln \frac{1}{2}} \right]\)     A1

\( = \frac{1}{3}\left( {\ln 2 - \ln \frac{1}{2}} \right)\)     A1

\( = \frac{2}{3}\ln 2\)     A1

Note: If the logarithms are not combined in the third from last line the last three A1 marks cannot be awarded.

Total [9 marks]

Not a difficult question but combination of the logarithms obtained by integration was often replaced by a spurious argument with infinities to get an answer. \(\log (\infty + 1)\) was often seen.