My deepest apologies, there is a typo in the final expression! Indeed, here is the complete text from my book (p.231):
Definition 5.5. The Bayes factor is the ratio of the posterior probabilities of the null and the alternative hypotheses over the
ratio of the prior probabilities of the null and the alternative
hypotheses, i.e., $$ B^\pi_{01}(x) = {P(\theta \in \Theta_ 0\mid x)
\over P(\theta \in \Theta_1\mid x)} \bigg/ {\pi(\theta \in \Theta_ 0)
\over \pi(\theta \in \Theta_ 1)}. $$ This ratio evaluates the
modification of the odds of $\Theta_0$ against $\Theta_1$ due to the
\obs\ and can naturally be compared to $1$, although an exact
comparison scale can only be based upon a loss function. In the
particular case where $\Theta_0=\{\theta_0\}$ and
$\Theta_1=\{\theta_1\}$, the Bayes factor simplifies to the usual
likelihood ratio $$ B^\pi_{01} (x) = {f(x|\theta_0)\over
f(x|\theta_1)}. $$ In general, the Bayes factor depends on prior
information, but is still proposed as an ``objective'' Bayesian
answer, since it partly eliminates the influence of the prior modeling
and emphasizes the role of the observations. Actually, it can be
perceived as a Bayesian likelihood ratio since, if $\pi_0$ is the
prior distribution under $H_0$ and $\pi_1$ the prior distribution
under $H_1$, $B^\pi_{01}(x)$ can be written as \begin{equation}
B^\pi_{01} (x) = {\int_{\Theta_0} f(x|\theta_0)\pi_0(\theta)
\,\text{d}\theta \over \int_{\Theta_1} f(x|\theta_1)\pi_1(\theta)
\,\text{d}\theta}
=\frac{m_0(x)}{m_1(x)}\,, \end{equation} thus replacing the likelihoods with the marginals under both hypotheses.
The prior is thus defined as a mixture:
$$\pi(\theta)=\pi(\theta \in \Theta_ 0)\times\pi_0(\theta)\times\mathbb{I}_{\Theta_0}(\theta)+\pi(\theta \in \Theta_ 1)\times\pi_1(\theta)\times\mathbb{I}_{\Theta_1}(\theta)$$where
$$\pi(\theta \in \Theta_ 0)=\rho_0\qquad\text{and}\qquad\pi(\theta \in \Theta_ 1)=1-\rho_0\stackrel{\text{def}}{=}\rho_1$$
are the prior weights of both hypotheses and
$$\int_{\Theta_0} \pi_0(\theta_0)\text{d}\theta_0=\int_{\Theta_1} \pi_1(\theta_1)\text{d}\theta_1=1$$Therefore
\begin{align*}P(\theta \in \Theta_ 0|x)=\int_{\Theta_0} \pi(\theta_0|x)\text{d}\theta_0&=\int_{\Theta_0} \pi(\theta_0|x)\text{d}\theta_0\\
&= \frac{\int_{\Theta_0} \pi(\theta_0)f(x|\theta_0)\text{d}\theta_0}{\int_{\Theta_0} \pi(\theta_0)f(x|\theta_0)\text{d}\theta_0+\int_{\Theta_1} \pi(\theta_1)f(x|\theta_1)\text{d}\theta_1}\\
&=\frac{\int_{\Theta_0} \rho_0\pi_0(\theta_0)f(x|\theta_0)\text{d}\theta_0}{\int_{\Theta_0} \rho_0\pi_0(\theta_0)f(x|\theta_0)\text{d}\theta_0+\int_{\Theta_1} \rho_1\pi_1(\theta_1)f(x|\theta_1)\text{d}\theta_1}
\end{align*}
and
$$P(\theta \in \Theta_1\mid x)=\frac{\int_{\Theta_1} \rho_1\pi_1(\theta_1)f(x|\theta_1)\text{d}\theta_1}{\int_{\Theta_0} \rho_0\pi_0(\theta_0)f(x|\theta_0)\text{d}\theta_0+\int_{\Theta_1} \rho_1\pi_1(\theta_1)f(x|\theta_1)\text{d}\theta_1}$$Hence,
$$\frac{P(\theta \in \Theta_ 0|x)}{P(\theta \in \Theta_1\mid x)}=\frac{\rho_0\int_{\Theta_0} \pi_0(\theta_0)f(x|\theta_0)\text{d}\theta_0}{\rho_1\int_{\Theta_1} \pi_1(\theta_1)f(x|\theta_1)\text{d}\theta_1}=\frac{\rho_0}{1-\rho_0}B^\pi_{01}(x)$$
This hopefully explains where the final expression comes from.
Alas, there is a mistake in the last ratio of integrals in the quoted text (presumably due to a cut & paste)
$$B_{01}^{\pi}=\frac{\int_{\Theta_{0}}f(x|\theta_{0})\pi_{0}(\theta)d\theta}{\int_{\Theta_{1}}f(x|\theta_{1})\pi_{1}(\theta)d\theta}$$
which should be (when identifying the integrands differently in the two integrals to signify that the integrals are over two different parameter spaces)
$$B_{01}^{\pi}=\frac{\int_{\Theta_{0}}f(x|\theta_{0})\pi_{0}(\theta_0)d\theta_0}{\int_{\Theta_{1}}f(x|\theta_{1})\pi_{1}(\theta_1)d\theta_1}$$or (with another choice of representation of the integrands)
$$B_{01}^{\pi}=\frac{\int_{\Theta_{0}}f(x|\theta)\pi_{0}(\theta)d\theta}{\int_{\Theta_{1}}f(x|\theta)\pi_{1}(\theta)d\theta}$$
The only case when the notations $\theta_0$ and $\theta_1$ are important is when both null and alternative hypotheses are point hypotheses, i.e., when $\Theta_0=\{\theta_0\}$ and $\Theta_1=\{\theta_1\}$, since both symbols then take specific values, like $\theta_0=3$ and $\theta_1=-2$. In this specific case, the posterior is concentrated on $\{\theta_0,\theta_1\}$ and the Bayes factor writes $$ B^\pi_{01} (x) = {f(x|\theta_0)\over f(x|\theta_1)}$$ since
$$P(\theta=\theta_0|x)=\frac{\overbrace{\pi(\theta_0)}^{\rho_0}f(x|\theta_0)}{\pi(\theta_0)f(x|\theta_0)+\underbrace{\pi(\theta_1)}_{\rho_1}f(x|\theta_1)}$$
Thank you for pointing out this error, to be added to the list of typos.
