= Solution
Apply a <Feynman parameter> and shift the loop momentum to $\ell=p+xk$. The common denominator becomes $[\ell^2+M_x^2]^2$, where
$$
M_x^2=m^2+x(1-x)k^2.
$$
For positive $M_x^2$, the Gamma-integral representation $a^{-2}=\int_0^\infty ds\,s e^{-sa}$ and a <Gaussian integral> give
$$
\int\frac{d^d\ell}{(2\pi)^d}\frac1{(\ell^2+M_x^2)^2}
=\frac1{(4\pi)^{d/2}}\int_0^\infty ds\,s^{1-d/2}e^{-sM_x^2}
=\frac{\Gamma(2-d/2)}{(4\pi)^{d/2}}(M_x^2)^{d/2-2}.
$$
This formula initially converges for $\operatorname{Re}d<4$ and defines the <Euclidean massive loop integral> at other dimensions by <analytic continuation>. Consequently,
$$
I(k)=\frac{g^2\mu^{6-d}}{2(4\pi)^{d/2}}\Gamma(2-d/2)\int_0^1dx\,[m^2+x(1-x)k^2]^{d/2-2}.
$$
With $d=6-\epsilon$, the <Gamma function> factor is $\Gamma(-1+\epsilon/2)=-2/\epsilon+O(1)$. Only its pole matters: the other factors can be evaluated at $\epsilon=0$ when extracting that pole. Since $\int_0^1x(1-x)dx=1/6$,
$$
\boxed{I_{\mathrm{div}}(k)=-\frac{g^2}{(4\pi)^3\epsilon}\left(m^2+\frac{k^2}{6}\right).}
$$
This is the <one-loop two-point divergence in six-dimensional cubic scalar theory>. Its polynomial momentum dependence is precisely what permits subtraction by local <counterterms>.
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