Biết rằng \[\int\limits_0^1 {\frac{{{x^2} + 3x + 2}}dx} = a\ln 2 + b\ln 3,\] với \[a,{\rm{ }}b \in \mathbb{Z}.\] Tính \[S = {a^3} + {b^3}.\]

Đáp án D

Phân tích

\(\frac{{{x^2} + 3x + 2}} = \frac{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \frac{m} + \frac{n} \Rightarrow x - 1 = m\left( {x + 2} \right) + n\left( {x + 1} \right).\)

Cho

\(\left\{ \begin{array}{l}x = - 1 \Rightarrow m = - 2\\x = - 2 \Rightarrow n = 3\end{array} \right. \Rightarrow \int\limits_0^1 {\frac{{{x^2} + 3x + 2}}} dx = \int\limits_0^1 {\left( {\frac{3} - \frac{2}} \right)dx} = 3\ln \left| {x + 2} \right|\left| {_{\scriptstyle\atop\scriptstyle0}^{\scriptstyle1\atop\scriptstyle}} \right.\)

\( \Rightarrow I = \left( {3\ln 3 - 3\ln 2} \right) - 2\ln 2 = 3\ln 3 - 5\ln 2 \Rightarrow a = - 5,b = 3 \Rightarrow S = - 98\)