= Solution
Set $C(\beta)=[A,e^{-\beta B}]$. The <product rule> gives
$$
C'+BC=[B,A]e^{-\beta B},\qquad C(0)=0.
$$
Multiplying by the <integrating factor> $e^{\beta B}$ and integrating yields the <commutator of an operator with an exponential>:
$$
\boxed{[A,e^{-\beta B}]
=e^{-\beta B}\int_0^\beta
e^{\lambda B}[B,A]e^{-\lambda B}\,d\lambda.}
$$
Equivalently, differentiate $e^{\beta B}Ae^{-\beta B}-A$ directly; its <derivative> is the integrand and it vanishes at zero. The argument is exact for finite <matrices>, and for <bounded operators> with their norm-convergent exponentials; unbounded <linear operators> require an appropriate common domain.
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