# Why don't counterterms appear in one-loop correction for 1PI effective action?

In Zee's "QFT in a Nutshell: Second Edition", section IV.3, the author calculates the 1PI effective potential for a single real scalar field. The full Lagrangian is given by equation (1):

$$\mathcal{L}=\frac 1 2 \left(\partial \phi\right)^2\underset{V_0(\phi)}{\underbrace{-\frac 1 2 \mu^2 \phi^2 +\frac{1}{4!}\lambda \phi^4}}+\underset{\Delta V(\phi)}{\underbrace{A\left(\partial \phi\right)^2+B\phi^2 +C \phi^4}} \tag{1}$$

where $$V_0(\phi)$$ is the renormalized classical potential, and $$\Delta V(\phi)$$ are the relevant counterterms.

In equation (14) the author presents the 1PI effective potential ignoring the counterterms $$\Delta V(\phi)$$:

$$V_{\textrm{eff}}(\phi_c)=V_0(\phi_c)+\frac{\hbar}{2}\int \frac{d^4k}{(2\pi)^4}\log \left[\frac{k^2-V_0''(\phi_c)}{k^2}\right]\tag{14}$$

Immediately after, he restores the presence of the counterterms, and states that the effective potential changes to equation (15):

$$V_{\textrm{eff}}(\phi_c)=V_0(\phi_c)+\color{red}{B\phi_c^2+C\phi_c^4}+\frac{\hbar}{2}\int \frac{d^4k}{(2\pi)^4}\log \left[\frac{k^2+V_0''(\phi_c)}{k^2}\right]\tag{15}$$

My question: Why don't the (red) counterterms show up in the logarithm? Why is the answer not given by:

$$V_{\textrm{eff}}(\phi_c)=V_0(\phi_c)+\color{red}{B\phi_c^2+C\phi_c^4}+\frac{\hbar}{2}\int \frac{d^4k}{(2\pi)^4}\log \left[\frac{k^2+V_0''(\phi_c)+\color{red}{2B+12C\phi_c^2}}{k^2}\right]\tag{15*}$$

The first obvious answer is that we are interested in the counterterms to leading order in $$\hbar$$. Including the counterterms in the logarithm would give us an $$\mathcal{O}(\hbar^2)$$ contribution to them. If we really wanted the correct $$\mathcal{O}(\hbar^2)$$ counterterms we should also include the 2-loop contribution to the effective action, $$\frac{\delta^3 S[\phi]}{\delta \phi^3}$$. However, I would like to avoid arguments that rely on expansions in the dimensionful parameter $$\hbar$$.