Well, the hardest part of this problem is determining m and M. However, we have an interval in which the function is non-increasing, [0,3]. So we know $\displaystyle f(3) \le f(x) \le f(0)$. From this we see that
$\displaystyle \frac{3}{1+3^2} \cdot (3 - 0)\le \int_0^3 \! \frac{3}{1 + x^2}\, dx \le \frac{3}{1+0^2} \cdot (3-0)$
$\displaystyle \frac{9}{10}\le \int_0^3 \! \frac{3}{1 + x^2}\, dx \le 9 $