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

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

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

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

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

I am new to LaTex, so my codes look very childish, anyway someone please help me put these aligned