Deriving Maxwell-Boltzmann momentum distribution

I am studying statistical mechanics and I saw the following statement in my notes:

$$frac{drho}{rho} = frac{e^{-beta p^2/2m}}{(2pi m k_B T)^{3/2}} 4pi p^2 dp quad ldots (1)$$

where $rho = langle N rangle /V$, the particle density and $p$ is momentum, $beta = 1/k_B T$ where $T$ is the temperature, $k_B$ is Boltzmann’s constant, $m$ is mass of the particle. We can see that the above equation denotes the momentum distribution function for an ideal gas.

I want to derive this equation from the Fermi distribution for particle density:

$$rho = frac{g}{h^3} int frac{1}{e^{beta(p^2/2m – mu)}+1} dmathbf{p} quad ldots (2)$$
in the limit as $e^{beta(p^2/2m – mu)} gg 1$, where $g=2$ for a Fermi gas.

My question is, how do I get equation $(1)$ from the above relation?

My attempt: Since we know that $e^{beta(p^2/2m – mu)} gg 1$, equation $(2)$ becomes,

$$rho = frac{2}{h^3} int_{0}^{infty} int_{0}^{infty} int_{0}^{infty} e^{-beta(p^2/2m – mu)} dp_xdp_ydp_z
=frac{2e^{beta mu}}{h^3} int_{0}^{infty} int_{0}^{infty} int_{0}^{infty} e^{-beta(p^2/2m)} dp_xdp_ydp_z
=frac{e^{beta mu}}{sqrt{2}h^3} cdot (2pi m k_B T)^{3/2} $$

However, I don’t see a way to make it to equation $(1)$.

I would appreciate any advice you have for me.