Biết rằng \[\int\limits_2^3 {\frac{{x\left( {x - 2} \right) + 1}}dx} = a + b\ln 2,\] với \[a,{\rm{ }}b \in \mathbb{Z}.\] Tính \[S = a + 2b.\]

Đáp án C

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

\[\begin{array}{l} = \int\limits_2^3 {\left[ {\frac{1} + \frac{2}{{{{\left( {x - 1} \right)}^2}}}} \right]dx = \left( {\ln \left| {x - 1} \right| - \frac{2}} \right)\left| \begin{array}{l}^3\\_2\end{array} \right.} \\ = \left( {\ln 2 - 1} \right) - \left( { - 2} \right) = 1 + \ln 2 \Rightarrow \left\{ \begin{array}{l}a = 1\\b = 1\end{array} \right. \Rightarrow S = 3.\end{array}\]