Hướng dẫn giải

Ta có \(\int\limits_2^3 {\frac{{2x + 3}}{{{x^2} + x}}dx} = \int\limits_2^3 {\frac{{2x + 1 + 2}}{{{x^2} + x}}dx} = \int\limits_2^3 {\left( {\frac{{2x + 1}}{{{x^2} + x}} + \frac{2}{{{x^2} + x}}} \right)dx} \)

\( = \int\limits_2^3 {\left( {\frac{{2x + 1}}{{{x^2} + x}} + \frac{2}{x} - \frac{2}{{x + 1}}} \right)dx = \left( {\ln \left| {{x^2} + x} \right| + 2\ln \left| x \right| - 2\ln \left| {x + 1} \right|} \right)\left| {_{\scriptstyle\atop\scriptstyle2}^{\scriptstyle3\atop\scriptstyle}} \right. = - 5\ln 2 + 4\ln 3} \)

\( \Rightarrow \left\{ \begin{array}{l}a = - 5\\b = 4\end{array} \right. \to {a^2} - ab - b = 41.\)

Chọn D.