QED lagrangian: gauge fixing term

Question

I have a question about the structure of the QED lagrangian, in particular the free photon lagrangian which is contained in it. My premise is: I only know how to exploit canonical quantization in order to quantize a theory; I don't know how to use the path integral formulation.

The QED lagrangian is: $$ \mathcal{L}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}+\bar{\psi}(i\gamma^{\mu}D_{\mu}-m)\psi, $$ so I assume that the free photon theory exploited here is $$ \mathcal{L_{free}}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}. $$ However, I also learnt that $\mathcal{L_{free}}$ jointed with Lorenz's gauge cannot give us a covariant quantization for the electromagnetic field (by means of the canonical quantization, at least). In fact, we introduce the following lagrangian: $$ \mathcal{L_{feyn}}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}-\frac{1}{2 \xi}(\partial_{\mu}A^{\mu})^2 $$ with Feynman gauge choice $\xi=1$. This, jointed with Gupta-Bleuer constraint, gives us the physical states of the electromagnetism.

So: why do we adopt $\mathcal{L_{free}}$ instead of $\mathcal{L_{feyn}}$? I know that the latter is not gauge-invariant, but the covariant quantization of the theory is achieved through that, so this point is not clear to me.