I am currently looking at the exposition of the Bessel process given in Bass, Stochastic Processes, chapter 24 (page 200 onwards).
The squared Bessel process of order $\nu \geq 0$ starting as $y$ is defined as satisfying the SDE:
$$ \begin{align*} \mathrm{d} Y_t = 2\sqrt{Y_t}\mathrm{d}W_t + \nu\,\mathrm{d}t && Y_0 = y \end{align*} $$
In proposition 24.7, it is claimed that for $\nu = 2$, the quadratic variation of $\log Y$ up to $t$ is $\int_0^t\!\mathrm{d}s\,Y_s^{-2}$. Yet, from the following calculation: $$ \begin{align*} \mathrm{d} \log Y_t & = \frac{\mathrm{d} Y_t} {Y_t} - \frac{1}{2}\frac{\mathrm{d}\langle Y \rangle_t}{Y_t^2}\\ & = \frac{2}{\sqrt{Y_t}}\mathrm{d}W_t + \frac{\nu - 2}{Y_t}\mathrm{d}t\\ & = \frac{2}{\sqrt{Y_t}}\mathrm{d}W_t \end{align*} $$ I rather find $4\int_0^t\!\mathrm{d}s\,Y_s^{-1}$. Did I miss anything?
Perhaps they meant that the log of the Bessel process define as $R_t = Y_t^2$ has this particular quadratic variation.
Indeed, we know that the dynamic of $R_t$ is \begin{equation} dR_t = dW_t +\frac{\nu-1}{2}\frac{1}{R_t}dt \end{equation} Therefore, we have by Ito's lemma: \begin{equation} d\log(R_t) = \frac{1}{R_t}dW_t \end{equation} You can see the quadratic variation of the above process is$\int_0^t\!\mathrm{d}s\,R_s^{-2}$.