f(x)=(3x+2)/(x+1)的反函数的单调区间为

y=(3x+2)/(x+1)
y(x+1)=3x+2
yx+y=3x+2
(y-3)x=2-y
x=(2-y)/(y-3)
所以反函数：
y=(2-x)/(x-3)
=(3-x-1)/(x-3)
=-1-1/(x-3)
所以单调区间为(-∞,3)∪(3,+∞)

故选B