Biết \(\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{3\sin x + \cos x}}{{2\sin x + 3\cos x}}dx =  - \dfrac{7}{{13}}\ln 2 + b\ln 3 + c\pi \,\,\left( {b,c \in \mathbb{Q}} \right).} \) Tính \(\dfrac{b}{c}.\)

Ta có \(3\sin x + \cos x = A\left( {2\sin x + 3\cos x} \right) + B\left( {2\cos x - 3\sin x} \right)\)

    \(\begin{array}{l} \Leftrightarrow 3\sin x + \cos x = \left( {2A - 3B} \right)\sin x + \left( {3A + 2B} \right)\cos x\\ \Leftrightarrow \left\{ \begin{array}{l}2A - 3B = 3\\3A + 2B = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = \dfrac{9}{{13}}\\B =  - \dfrac{7}{{13}}\end{array} \right.\end{array}\)

Nên \(3\sin x + \cos x = \dfrac{9}{{13}}\left( {2\sin x + 3\cos x} \right) - \dfrac{7}{{13}}\left( {2\cos x - 3\sin x} \right)\)

Từ đó ta có

\(\begin{array}{l}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{3\sin x + \cos x}}{{2\sin x + 3\cos x}}dx = } \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\dfrac{9}{{13}}\left( {2\sin x + 3\cos x} \right) - \dfrac{7}{{13}}\left( {2\cos x - 3\sin x} \right)}}{{2\sin x + 3\cos x}}dx = } \\ = \dfrac{9}{{13}}\int\limits_0^{\dfrac{\pi }{2}} {dx}  - \dfrac{7}{{13}}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{2\cos x - 3\sin x}}{{2\sin x + 3\cos x}}dx} \\ = \dfrac{{9\pi }}{{26}} - \dfrac{7}{{13}}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{1}{{2\sin x + 3\cos x}}d\left( {2\sin x + 3\cos x} \right)} \\ = \dfrac{{9\pi }}{{26}} - \dfrac{7}{{13}}\left. {\ln \left| {2\sin x + 3\cos x} \right|} \right|_0^{\dfrac{\pi }{2}}\\ = \dfrac{{9\pi }}{{26}} - \dfrac{7}{{13}}\ln 2 + \dfrac{7}{{13}}\ln 3\end{array}\)

Suy ra \(b = \dfrac{7}{{13}};c = \dfrac{9}{{26}} \Rightarrow \dfrac{b}{c} = \dfrac{{14}}{9}\).