jmc

Let us first determine the number of cubes that are less than $2010$. We have $10^3=1000$, $11^3=1331$, and $12^3=1728$, but $13^3=2197$. So there are $12$ cubes less than $2010$. As for fifth powers, $4^5=1024$, but $5^5=3125$. There are $4$ fifth powers less than $2010$, but only $3$ of these have not already been included, since we've already counted 1. Analyzing seventh powers, $3^7=2187$, so the only new seventh power less than $2010$ is $2^7$. There are no new ninth powers since they are all cubes, and $2^{11}=2048$ is greater than 2010. Therefore, there are $12+3+1=\boxed{16}$ oddly powerful integers less than $2010$.