jmc

First, we can write $$\frac{3x^2+9x+17}{3x^2+9x+7}=\frac{(3x^2+9x+7)+10}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}.$$Thus, we want to minimize $3x^2+9x+7.$

Completing the square, we get $$3x^2+9x+7=3{\left(x+\frac{3}{2}\right)}^2+\frac{1}{4},$$so the minimum value of $3x^2+9x+7$ is $\frac{1}{4}.$

Therefore, the maximum integer value of $$1+\frac{10}{3x^2+9x+7}$$is $1+\frac{10}{1/4}=\boxed{41}.$