I'm trying to understand Weisskopf's 1939 paper 'On the Self-Energy and the Electromagnetic Field of the Electron'. Here Weisskopf demonstrates that in Dirac's positron theory, the self-energy of an electron diverges only logarithmically (Eq. 26), because the quadratically divergent energy due to vacuum fluctuations (equation below Eq. 25) cancels with the quadratically divergent term of energy due to the spin (Eqs. 18, 20).
My question is about the integral in Eq. 18. It seems to suggest that the following holds
$$\int_0^P \frac{p^2}{\sqrt{m^2c^4+p^2c^2}} dp = \frac{1}{c}\left(P\sqrt{P^2+m^2c^2}-\frac{m^2c^2}{2} \ln \frac{P+\sqrt{P^2+m^2c^2}}{mc}\right).$$
However, it seems to me that in the limit of large $p$, the integrand can be approximated as $\frac{p}{c}$, whose integral would yield $\frac{P^2}{2c}$, i.e. a factor two off from Weisskopf's quadratic term. When I ask WolframAlpha or ChatGPT to calculate this integral, they also seem to divide the quadratic term by $2c$ instead of $c$. This factor of two is significant, because it determines whether the quadratic divergences cancel out or not.
So my question is: am I making a careless mistake when I expect that the quadratic term should be $\frac{P^2}{2c}$ instead of $\frac{P^2}{c}$? If not, then how do we still arrive at the result that the quadratic divergences cancel?