Cho \(\int_1^3 {\frac{{{\rm{d}}x}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}} = a\ln 2 + b\ln 5 + c\ln 7\,\,\,\left( {a,b,c \in

Cho \(\int_1^3 {\frac{{{\rm{d}}x}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}} = a\ln 2 + b\ln 5 + c\ln 7\,\,\,\left( {a,b,c \in \mathbb{Q}} \right)\). Tính \(S = a + 4b - c\)
A. \(1.\)
B. \(\frac{4}{3}.\)
C. \(\frac{7}{3}.\)
D. \(2.\)

Ta có \(\frac{1}{{\left( {x + 1} \right)\left( {x + 4} \right)}} = \frac{1}{3}\left[ {\frac{1}{{x + 1}} - \frac{1}{{x + 4}}} \right]\).
Do đó:
\(\begin{array}{l}\int_1^3 {\frac{{{\rm{d}}x}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}} = \frac{1}{3}\ln \left| {\frac{{x + 1}}{{x + 4}}} \right|\left| \begin{array}{l}3\\1\end{array} \right. = \frac{1}{3}\left( {\ln \frac{4}{7} - \ln \frac{2}{5}} \right)\\ = \frac{1}{3}\left( {\ln 2 + \ln 5 - \ln 7} \right) = \frac{1}{3}\ln 2 + \frac{1}{3}\ln 5 - \frac{1}{3}\ln 7 \Rightarrow \left\{ \begin{array}{l}a = \frac{1}{3}\\b = \frac{1}{3}\\c = - \frac{1}{3}\end{array} \right.\end{array}\)
Vậy: \(S = \frac{1}{3} + 4.\frac{1}{3} - \frac{1}{3} = \frac{4}{3}\)