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7

The "trick" which works frequently when Mathematica refuses to calculate a definite integral is to calculate first the indefinite integral, then take the limits at the ends of the integration interval and subtract the results. Here we go. Let the integrand be f = -((cA Log[1 + mgl^2/sa1])/(-1 + x)) + (cA Log[-(sa1/(-mgl^2 - sa1))])/(1 - x) + (cA cF Log[-1 ...


3

seems to work: expr /. Times[xx_?(! FreeQ[#, x] &) , p_plusd, rest___] :> Times[xx /. x -> 1, p, rest] and terms like: 2 Log[(sa1 (-2 + x))/(-mgl^2 + sa1 (-2 + x))] plusd[1/(1 - x)] and Log[2 + mgl^2/sa1 - x] plusd[-(2/(-1 + x))] were reduced to 2 Log[-(sa1/(-mgl^2 - sa1))] plusd[1/(1 - x)] and Log[1 + mgl^2/sa1] plusd[-(2/(-1 + x))] ...


1

booleanTable := Module[ {var, iter}, f = Input["Enter Boolean function"]; var = BooleanVariables@f; iter = Sequence @@ ({#, {True, False}} & /@ var); Column[{Table[ Evaluate[{var, f} // Flatten], Evaluate[iter]] // Flatten[#, Length[var] - 1] & // Prepend[#, {var, f} // Flatten] & // ...



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