Confusion about which electric flux is correct Okay so electric flux density $\mathbf D$ is equal to the electric field multiplied by the permittivity of free space ($\mathbf D=\epsilon_0 \mathbf E \epsilon_r$). Therefore, $\mathbf D$ integrated over a closed surface would give you the total electric flux which also happens to be equal to the charge enclosed by the surface.
However, a lot of sites and YouTube people define the electric flux as the integral of the electric field over the closed surface, which is a direct contradiction to my first paragraph.
So my question is what is happening? Which one is the electric flux? 
I also asked my professor this question, and she said the Internet lies and the true electric flux is the one in my first paragraph and the electric field integrated over the surface area  is just a way to get the total electric field over a surface area. However, that still does not explain the contradiction. Surely thousands of people cannot be just saying something for it to be wrong? There has to be some explanation as to why this happens.
 A: To add to Dale's answer, $\vec{D}=\epsilon_0\vec{E}+\vec{P}$. 
Here, $\vec{E}$ is the Electric Field, $\vec{P}$ is the Polarization field, and $\vec{D}$
is the Electric Displacement field. 
The Polarization $\vec{P}$ is the electric dipole density. 
Electric fields can induce electric dipoles in materials. This is often proportional to the strength of the Electric Field. In this case, the material is called a linear dielectric and:
$$\vec{P}=\chi_e\epsilon_0\vec{E}. $$
Where $\chi_e$ is the electric susceptibility. 
So : $\vec{D}=\epsilon_0(1+\chi_e)\vec{E}=\epsilon_0\epsilon_r\vec{E}$
So $\vec{D}$ and $\vec{E}$ are different fields, but closely related. 
A very useful relationship, $\nabla \cdot \vec{E}=\rho_f+\rho_b$, where $\rho_f$ is the free charge in the material, and $\rho_b$ is the bound charge induced by the electric field. 
Some useful relations can be found from $\nabla \cdot \vec{D}=\rho_f$ and $\nabla \cdot \vec{E}=(\rho_f+\rho_b)/\epsilon_0$ . Only Gauss's law for $\vec{E}$ gives you the flux due to total charge. Gauss law applied to the Displacement charge only gives you the flux due to the free charge which is often zero in dielectric problems. 
So we have that $\nabla \cdot \vec{P}=-\rho_b$.
From here, it can be shown via Gauss' law that the net induced charge is on the surface of the dielectric. 
A: The equation $\mathbf D=\epsilon_0 \epsilon_r \mathbf E$ is only valid in linear isotropic homogenous media. It should not be taken as a general definition of $\mathbf D$. 
The electric flux in the macroscopic version of Gauss’ law can be written as $$\Phi_D=\iint_S \mathbf D \cdot d\mathbf A$$ where $\Phi_D=Q_f$
There is a similar quantity which arises in the microscopic version of Gauss’ law:$$\Phi_E=\iint_S \mathbf E \cdot \mathbf A$$ These two quantities are similar, but not the same. In particular the macroscopic flux is equal to the free charge, while the microscopic flux is proportional to the total charge. There is no contradiction with your first paragraph because they are talking about different things. 
