Lời giải chi tiết:

\(\begin{array}{l}
I = \int\limits_3^4 {\frac{{dx}}{{{x^2} + x}} = a\ln 2 + b\ln 3 + c\ln 5} \\
= \int\limits_3^4 {\frac{{dx}}{{x\left( {x + 1} \right)}} = \int\limits_3^4 {\left( {\frac{A}{x} + \frac{B}{{x + 1}}} \right)dx} } \\
\Rightarrow \frac{1}{{x\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{x + 1}}\\
\Leftrightarrow 1 = A\left( {x + 1} \right) + Bx\\
\Leftrightarrow 1 = \left( {A + B} \right)x + A\\
\Leftrightarrow \left\{ \begin{array}{l}
A + B = 0\\
A = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
A = 1\\
B = - 1
\end{array} \right.\\
\Rightarrow I = \int\limits_3^4 {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)dx} \\
= \left. {\left( {\ln \left| x \right| - \ln \left| {x + 1} \right|} \right)} \right|_3^4\\
= \ln 4 - \ln 3 - \ln 5 + \ln 4\\
= 2\ln 4 - \ln 3 - \ln 5\\
= 4\ln 2 - \ln 3 - \ln 5\\
\Rightarrow \left\{ \begin{array}{l}
a = 4\\
b = - 1\\
c = - 1
\end{array} \right.\\
\Rightarrow S = a + b + c = 4 - 1 - 1 = 2.
\end{array}\)

Chọn B.