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Re: Please explain this weird Integrate result
Jose Reckoner wrote: > Expand[Integrate[(2*x)/(2*x - 1), x]] > > gives > > -1/2 + x + Log[-1 + 2*x]/2 > > But, > > Expand[Integrate[1 + 1/(-1 + 2*x), x]] > > gives > > x + Log[-1 + 2*x]/2 > > even though > > (2*x)/(2*x - 1) == 1 + 1/(-1 + 2*x) > > Where did the extra -1/2 in the first result come from? > Hi Jose, the constant term -1/2 is a constant of integration. Say f(x) is any real function, F(x) is the primitive of f(x), then any function F(x) + C, C being an arbitrary constant, is an indefinite integral of f(x); that is (F(x) + C)' = f(x). Now, *Integrate* does not return any symbolic constant term such as C (so we would get x + (1/2)*Log[-1 + 2*x] + C for instance), nor does it return, usually, any numeric constant term (or you may think of the value of the constant to be zero). However, a constant term might be returned sometimes, like in your case. Two different, although equivalent, expressions may lead to different representations of the solution since the algorithms used by *Integrate* might be different in each case. As long as you work with indefinite integrals, you may "discard" the constant term -1/2 and be sure that both indefinite integrals represent the same family of functions. In:= int1 = Expand[Integrate[(2*x)/(2*x - 1), x]] Out= -(1/2) + x + (1/2)*Log[-1 + 2*x] In:= int2 = Expand[Integrate[1 + 1/(-1 + 2*x), x]] Out= x + (1/2)*Log[-1 + 2*x] In:= int1 - int2 Out= -(1/2) In:= D[int1, x] Out= 1 + 1/(-1 + 2*x) In:= D[int2, x] Out= 1 + 1/(-1 + 2*x) Obviously, the second form for the integrand is the one preferred by Mathematica. Best regards, /J.M.