Answering my own question.
Comes trivially from the BCH-D integral expression; "$ln(e^Ke^L)-K-L$" being equal to the iterated integral of ${(M(x)-1)y}/{(M(x)(1-y)+y)}*L$, $M(x)=Ad(e^Ke^{xL})$, on $[0,1]²$.
Kanghun Kim
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