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

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

Let

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

Let $3 \sin x+2 \cos x$

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

$\Rightarrow 3 \sin x+2 \cos x =\lambda(-3 \sin x+2 \cos x)+\mu(3 \cos x+2 \sin x)$

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 $\mu=12 / 13$

∴  $I=\int \frac{-\left(\frac{5}{13}\right)(-3 \sin x+2 \cos x)+\left(\frac{12}{13}\right)(3 \cos x+2 \sin x)}{3 \cos x+2 \sin x} d x$

$\Rightarrow I=\frac{12}{13} \int 1 . d x-\frac{5}{13} \int \frac{-3 \sin x+2 \cos x}{3 \cos x+2 \sin x} d x $

$\Rightarrow I=\frac{12}{13} \int 1 . d x-\frac{5}{13} \int \frac{1}{3 \cos x+2 \sin x} d(3 \cos x+2 \sin x)$

$\Rightarrow I=\frac{12}{13} x-\frac{5}{13} \log |3 \cos x+2 \sin x|+C$

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