Lời giải:\(I = \int\limits_0^1 {\frac{{dx}}{{3x + 5\sqrt {3x + 1} + 7}} = \int\limits_0^1 {\frac{{dx}}{{3x + 1 + 5\sqrt {3x + 1} + 6}}} } \)
Đặt \(\sqrt {3x + 1} = t \Rightarrow {t^2} = 3x + 1 \Rightarrow 2tdt = 3dx \Leftrightarrow dx = \frac{2}{3}tdt.\)
Đổi cận: \(\left\{ \begin{array}{l}x = 1 \Rightarrow t = 2\\x = 0 \Rightarrow t = 1\end{array} \right..\)
\(\begin{array}{l} \Rightarrow I = \int\limits_1^2 {\frac{2}{3}\frac{{tdt}}{{{t^2} + 5t + 6}} = } \frac{2}{3}\int\limits_1^2 {\frac{{tdt}}{{\left( {t + 2} \right)\left( {t + 3} \right)}}} = \frac{2}{3}\int\limits_1^2 {\left( {\frac{3}{{t + 3}} - \frac{2}{{t + 2}}} \right)dt} \\ = \frac{2}{3}\left. {\left( {3\ln \left| {t + 3} \right| - 2\ln \left| {t + 2} \right|} \right)} \right|_1^2 = \frac{2}{3}\left( {3\ln 5 - 2\ln 4 - 3\ln 4 + 2\ln 3} \right)\\ = \frac{2}{3}\left( {3\ln 5 + 2\ln 3 - 5\ln 4} \right) = \frac{2}{3}\left( { - 10\ln 2 + 2\ln 3 + 3\ln 5} \right) = - \frac{{20}}{3}\ln 2 + \frac{4}{3}\ln 3 + 2\ln 5.\\ \Rightarrow \left\{ \begin{array}{l}a = - \frac{{20}}{3}\\b = \frac{4}{3}\\c = 2\end{array} \right. \Rightarrow a + b + c = - \frac{{10}}{3}.\end{array}\)
Chọn A.
