$\begin{align*}

&\frac{dP}{dr}=\frac{-\Big(P+\rho (P)c^2 \Big)} {2} \frac {d\nu} {dr} \Rightarrow

\\ \\

& d\nu = \frac {-2dP} {P+\rho (P) c^2}

\\ \\

&\nu (r) = - \int_{0}^{P(r)} \frac {2dP}{P+\rho (P) c^2} + \nu (r_b)

\\ \\

&\large e^{\nu(r)} = e^{\nu(r_b) - \int_{0}^{P(r)} \frac {2dP}{P+\rho (P) c^2} }

\\ \\

&\large e^ {\nu (r)}= e^{\nu(r_b)} exp \Big[-\int_{0}^{P(r)} \frac {2dP}{P+\rho (P) c^2} \Big]

\end{align*}$

This do it for you?