Đáp án đúng: B
Giải chi tiết:Ta có:
\(\begin{array}{l}I = \int\limits_0^3 {\dfrac{{\left| {{x^2} - x} \right|}}{{x + 3}}dx} = \int\limits_0^1 {\dfrac{{\left| {{x^2} - x} \right|}}{{x + 3}}dx} + \int\limits_1^3 {\dfrac{{\left| {{x^2} - x} \right|}}{{x + 3}}dx} \\\,\,\,\, = - \int\limits_0^1 {\dfrac{{{x^2} - x}}{{x + 3}}dx} + \int\limits_1^3 {\dfrac{{{x^2} - x}}{{x + 3}}dx} \\\,\,\,\, = - \int\limits_0^1 {\left( {x - 4 + \dfrac{{12}}{{x + 3}}} \right)dx} + \int\limits_1^3 {\left( {x - 4 + \dfrac{{12}}{{x + 3}}} \right)dx} \\\,\,\,\, = - \left. {\left( {\dfrac{{{x^2}}}{2} - 4x + 12\ln \left| {x + 3} \right|} \right)} \right|_0^1 + \left. {\left( {\dfrac{{{x^2}}}{2} - 4x + 12\ln \left| {x + 3} \right|} \right)} \right|_1^3\\\,\,\,\, = - \left[ {\dfrac{7}{2} + 12\ln 4 - 12\ln 3} \right] + \left( { - \dfrac{{15}}{2} + 12\ln 6 + \dfrac{7}{2} - 12\ln 4} \right)\\\,\,\,\, = - \dfrac{{15}}{2} - 24\ln 4 + 12\ln 3 + 12\ln 6\\\,\,\,\, = - \dfrac{{15}}{2} - 12\ln 16 + 12\ln 3 + 12\ln 6\\\,\,\,\, = - \dfrac{{15}}{2} + 12\ln \dfrac{{27}}{{16}}\end{array}\)
Chọn B.