Suppose that the smallest of the five odd integers is $x$. Since consecutive odd integers differ by 2, the other four odd integers are $x+2, x+4, x+6$, and $x+8$. Therefore, $x + (x+2) + (x+4) + (x+6) + (x+8) = 125$. From this, we obtain $5x + 20 = 125$ and so $5x = 105$, which gives $x = 21$. Thus, the smallest of the five integers is 21.