If $\int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx=ax+b\log |3\cos x+2\sin x|+C$, then 

Solution:

Let 

    $I=\int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx$

Let $3 \sin x+2 \cos x$

      $=\lambda \frac{d}{dx}(3\cos x+2\sin x)+\mu (3\cos x+2\sin x)$

 $\begin{array}{l}\Rightarrow 3\sin x+2\cos x\\ =\lambda (-3\sin x+2\cos x)+\mu (3\cos x+2\sin x)\end{array}$

Comparing the coefficients of $\sin x$ and $\cos x$ on both sides, we get

      $-3\lambda +2\mu =3$ and $2\lambda +3\mu =2$

$\Rightarrow \lambda =-5/13$ and $\mu =12/13$

$\begin{array}{l}\therefore I\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\int -\left(\frac{5}{13}\right)(-3\sin x+2\cos x)+\left(\frac{12}{13}\right)(3\cos x+2\sin x)\\ \Rightarrow I\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{12}{13}\int 1\cdot dx-\frac{5}{13}\int \frac{-3\sin x+2\cos x}{3\cos x+2\sin x}dx\\ \Rightarrow I\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{12}{13}\int 1\cdot dx-\frac{5}{13}\int \frac{1}{3\cos x+2\sin x}d(3\cos x+2\sin x)\\ \Rightarrow I\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{12}{13}x-\frac{5}{13}\log |3\cos x+2\sin x|+C\end{array}$

Hence, $a=\frac{12}{13}, b=-\frac{5}{13}$