Đáp án: B
Giải thích các bước giải:
$\begin{array}{l}
\dfrac{{3\sin x + \cos x}}{{2\sin x + 3\cos x}}\\
= \dfrac{{a.\left( {2\sin x + 3\cos x} \right) + b.\left( {2.\sin x + 3\cos x} \right)'}}{{2.\sin x + 3\cos x}}\\
= \dfrac{{2a.\sin x + 3a.\cos x + b.\left( {2\cos x - 3\sin x} \right)}}{{2.\sin x + 3\cos x}}\\
= \dfrac{{\left( {2a - 3b} \right).sinx + \left( {3a + 2b} \right).\cos x}}{{2.\sin x + 3\cos x}}\\
\Rightarrow 3\sin x + \cos x = \left( {2a - 3b} \right)\sin x + \left( {3a + 2b} \right).\cos x\\
\Rightarrow \left\{ \begin{array}{l}
2a - 3b = 3\\
3a + 2b = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \dfrac{9}{{13}}\\
b = - \dfrac{7}{{13}}
\end{array} \right.\\
\Rightarrow \dfrac{{3\sin x + \cos x}}{{2\sin x + 3\cos x}}\\
= \dfrac{9}{{13}} - \dfrac{7}{{13}}.\dfrac{{\left( {2.\sin x + 3\cos x} \right)'}}{{2\sin x + 3\cos x}}\\
\Rightarrow I = \int\limits_0^{\pi /2} {\left( {\dfrac{9}{{13}} - \dfrac{7}{{13}}.\dfrac{{\left( {2\sin x + 3\cos x} \right)'}}{{2\sin x + 3\cos x}}} \right)dx} \\
= \left( {\dfrac{{9x}}{{13}}} \right)_0^{\pi /2} - \dfrac{7}{{13}}.\ln \left| {2\sin x + 3\cos x} \right|_o^{\pi /2}\\
= \dfrac{{9\pi }}{{26}} - \dfrac{7}{{13}}.ln2 + \dfrac{7}{{13}}.\ln 3\\
= - \dfrac{7}{{13}}\ln 2 + \dfrac{7}{{13}}\ln 3 + \dfrac{9}{{26}}\pi \\
\Rightarrow \left\{ \begin{array}{l}
b = \dfrac{7}{{13}}\\
c = \dfrac{9}{{26}}
\end{array} \right.\\
\Rightarrow \dfrac{b}{c} = \dfrac{{14}}{9}
\end{array}$
