The expressions log(x - f10) and log(f10 - x) differ by an additive
constant.
So both (%o3) and (%o4) are OK. If you are a fan of
integrate(1/x,x) = log(abs(x)), try this:
(%i38) integrate(1/x,x);
(%o38) log(x)
(%i39) integrate(1/x,x), logabs : true;
(%o39) log(abs(x))
Barton
maxima-admin@math.utexas.edu wrote on 10/13/2005 04:47:03 AM:
> Hi
>
> please take a look at the following code snippet.
> The first integration result is right, but not the second one.
>
> b + a x
> (%i2) INTEGRATE( - - - - - - -, x)
> d + c x
> a x (a d - b c) LOG(c x + d)
> (%o2) --- - ------------------------
> c 2
> c
> (%i3) EV(%, a = 1, b = F10, c = - 1, d = F10)
> (%o3) - x - 2 F10 LOG(F10 - x)
> x + F10
> (%i4) INTEGRATE(- - - - - - - , x)
> F10 - x
> (%o4) - 2 F10 LOG(x - F10) - x
>
> May be I am overlooking some details or is it simple an error?
>
> Klaus
>
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