Cho tích phân sau \(I = \int\limits_3^5 {\frac{1}{{2x - 1}}dx} = a\ln 3 + b\ln 5\,\,\,\left( {a,b \in \mathbb{Q}} \right)\). Tính \(S = a + b.\)

* Hướng dẫn giải

\(\begin{array}{l}I = \int\limits_3^5 {\frac{1}{{2x - 1}}dx}  = \left. {\frac{1}{2}\ln \left| {2x - 1} \right|} \right|_3^5\\\,\,\,\, = \frac{1}{2}\left( {\ln \left| {2.5 - 1} \right| - \ln \left| {2.3 - 1} \right|} \right)\\\,\,\,\, = \frac{1}{2}.\left( {\ln 9 - \ln 5} \right) = \ln 3 - \frac{1}{2}\ln 5\\ \Rightarrow a = 1;\,\,b =  - \frac{1}{2}.\end{array}\)

Vậy \(S = a + b = 1 - \frac{1}{2} = \frac{1}{2}.\)

Chọn D.