Đáp án đúng: A
Giải chi tiết:Ta có:
\(\begin{array}{l}\int\limits_0^1 {\dfrac{{dx}}{{{x^2} + 7x + 12}}} = \int\limits_0^1 {\dfrac{{dx}}{{\left( {x + 3} \right)\left( {x + 4} \right)}}} = \left. {\dfrac{1}{{\left( { - 3} \right) - \left( { - 4} \right)}}\ln \left| {\dfrac{{x + 3}}{{x + 4}}} \right|} \right|_0^1 = \ln \dfrac{4}{5} - \ln \dfrac{3}{4} = \ln \dfrac{{16}}{{15}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \ln 16 - \ln 15 = 2\ln 4 - \ln 5 - \ln 3\\ \Rightarrow a = - 1;\,\,b = 2;\,\,c = - 1 \Rightarrow a + 3b + 5c = - 1 + 3.2 + 5.\left( { - 1} \right) = 0\end{array}\)
Chọn A.