= Solution
After <modified minimal subtraction>, the surviving logarithm in the given <fermion self-energy> is $\log[\mu^2/((1-x)(m^2-p^2x))]$. For a one-loop mass shift, set $p^2=m^2$ and let $\not p$ act as $m$ on an on-shell spinor inside that correction; changing these arguments by the mass shift contributes only at order $e^4$. Thus
$$
\Sigma_R\big|_{\not p=m}
=\frac{e^2m}{8\pi^2}\int_0^1(x-2)\left[\log\frac{\mu^2}{m^2}-2\log(1-x)\right]dx.
$$
The needed integrals are
$$
\int_0^1(x-2)dx=-\frac32,\qquad
\int_0^1(x-2)\log(1-x)dx=\frac54.
$$
For the second, put $u=1-x$ and use $\int_0^1\log u\,du=-1$ and $\int_0^1u\log u\,du=-1/4$. Consequently
$$
\Sigma_R\big|_{\not p=m}=-\frac{e^2m}{16\pi^2}\left(5+3\log\frac{\mu^2}{m^2}\right).
$$
The <pole mass> condition $m_{\rm phys}-m+\Sigma_R=0$ gives $m_{\rm phys}=m[1+e^2(5+3\log(\mu^2/m^2))/(16\pi^2)]+O(e^4)$. Invert this relation and replace $m$ by $m_{\rm phys}$ inside the already one-loop term to obtain
$$
\boxed{m=m_{\rm phys}\left[1-\frac{e^2}{16\pi^2}\left(5+3\log\frac{\mu^2}{m_{\rm phys}^2}\right)\right]+O(e^4).}
$$
\b[The negative sign in the running-mass conversion follows from the explicitly chosen self-energy convention.] Subtracting poles alone, instead of the overbarred combination, would leave additional $\log4\pi-\gamma_E$ terms.
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