= Solution
Assume thermal equilibrium, a three-dimensional <nondegenerate semiconductor>, one parabolic conduction extremum and one parabolic valence extremum, with two spin states in each. Use the supplied <effective masses> as scalar density-of-states parameters. With conduction energy $E=E_c+\varepsilon$, state counting gives the per-volume <density of states>
$$
D_c(\varepsilon)=\frac1{2\pi^2}\left(\frac{2m_e^*}{\hbar^2}\right)^{3/2}\sqrt\varepsilon.
$$
Since $E_c-\mu\gg k_BT$, replace the <Fermi-Dirac distribution> by its <Boltzmann factor>. Using $x=\varepsilon/(k_BT)$ and the supplied integral,
$$
n=e^{-(E_c-\mu)/(k_BT)}\int_0^\infty D_c(\varepsilon)e^{-\varepsilon/(k_BT)}d\varepsilon
=N_c e^{-(E_c-\mu)/(k_BT)},
$$
$$
N_c=2\left(\frac{m_e^*k_BT}{2\pi\hbar^2}\right)^{3/2}
=2\left(\frac{2\pi m_e^*k_BT}{h^2}\right)^{3/2}.
$$
The <electron hole> occupation is $1-f(E)$ in the <valence band>. Applying the same integration below $E_v$ gives
$$
p=N_v e^{-(\mu-E_v)/(k_BT)},\qquad
N_v=2\left(\frac{2\pi m_h^*k_BT}{h^2}\right)^{3/2}.
$$
Their product cancels the <chemical potential>, giving the <semiconductor mass-action law>
$$
\boxed{np=4\left(\frac{2\pi k_B}{h^2}\right)^3
(m_e^*m_h^*)^{3/2}T^3e^{-E_g/(k_BT)}.}
$$
Thus, for masses expressed in ordinary mass units,
$$
\boxed{A=4\left(\frac{2\pi k_B}{h^2}\right)^3
=\frac{k_B^3}{2\pi^3\hbar^6},\qquad B=\frac32,\qquad C=3.}
$$
The labels $B,C$ in this last box are exponents, unrelated to the magnetic field or <Curie constant> above. <Valley degeneracy> or multiple valence bands must be included in the <effective density of states>, or absorbed into suitable <density-of-states effective masses>, not counted twice. Real silicon has anisotropic multiple valleys, so its conductivity mass is not automatically its density-of-states mass; the numerical answers below implement the paper's supplied scalar-mass model.<ECE 340 Lecture 6 by John Dallesasse>[Slide 26, A Few Cautions: carrier counting uses the density-of-states effective mass, while silicon conductivity and cyclotron masses differ]{ref}
The formula is not an identity for arbitrarily degenerate carriers. For a concrete counterexample, take $\mu-E_c=10k_BT$ while holes remain dilute. Fermi occupation gives $n/N_c\simeq24.1$, whereas the Boltzmann replacement would give $e^{10}\simeq22026$; the predicted product then differs by about three orders of magnitude. Separate electron and hole quasi-Fermi levels out of equilibrium would also prevent the cancellation used above.
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