= Solution
First remove the <Projection operator>[projector] denominator explicitly:
$$
i\Pi_{\mathrm{ex}}=4e^2\mu^\epsilon\int\frac{d^dl}{(2\pi)^d}\frac{l^2k^2-(l\cdot k)^2}{(l^2)^2[(l+k)^2+m^2]}.
$$
Let $A=(l+k)^2+m^2$ and $B=l^2$. The <Feynman parameter for a simple and a squared propagator> follows by differentiating the elementary two-denominator <Feynman parameter>[Feynman formula] with respect to $B$ gives
$$
\boxed{\frac1{AB^2}=2\int_0^1dx\frac{1-x}{[xA+(1-x)B]^3}.}
$$
Shift $q=l+xk$. Then $xA+(1-x)B=q^2+D$, with $D=x(1-x)k^2+xm^2$. The numerator is unchanged by this parallel shift:
$$
l^2k^2-(l\cdot k)^2=q^2k^2-(q\cdot k)^2.
$$
Therefore the correctly combined integral is
$$
\boxed{i\Pi_{\mathrm{ex}}=8e^2\mu^\epsilon\int_0^1(1-x)dx\int\frac{d^dq}{(2\pi)^d}\frac{q^2k^2-(q\cdot k)^2}{(q^2+D)^3}.}
$$
\b[The printed weight $x/2$ is not compatible with this denominator and prefactor.] Exchanging $x$ with $1-x$ changes the <Mass>[mass term] in $D$ as well and cannot supply the missing <normalization>. A concrete counterexample is $m=0$ with nonzero spacelike <Four-momentum>[external momentum] and $2<d<4$: after <Wick rotation>, the remaining integral is nonzero and symmetric under $x\leftrightarrow1-x$. The source's $4e^2\int(x/2)dx$ expression is then exactly one quarter of the correct expression. Thus the claimed equality cannot be imposed when computing the pole.
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