Biết rằng \[\int\limits_0^6 {\frac{{{x^3}}}dx} = a + b\ln 7,\] với \[a,{\rm{ }}b \in \mathbb{Z}.\] Tính \[S = a + 2b.\]

Đáp án C

Ta có \[\begin{array}{l}\int\limits_0^6 {\frac{{{x^3}}}dx} = \int\limits_0^6 {\frac{{{x^3} + 1 - 1}}dx} = \int\limits_0^6 {\left( {{x^2} - x + 1 - \frac{1}} \right)dx} \\ = \left( {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2} + x - \ln \left| {x + 1} \right|} \right)\left| \begin{array}{l}^6\\_0\end{array} \right. = 60 - \ln 7 \Rightarrow \left\{ \begin{array}{l}a = 60\\b =  - 1\end{array} \right. \Rightarrow S = 58.\end{array}\]