integrate(x/(1-x^2),x)?



>>>>> "Richard" == Richard Fateman <fateman at cs.berkeley.edu> writes:

    Richard> Commercial Macsyma and Mathematica give the same results as you show.
    Richard> The answers look OK to me.

    Richard> Raymond Toy wrote:

    >> Maxima currently says
    >> 
    >> integrate(x/(1-x^2),x) ->
    >> -log(1-x^2)/2
    >> 
    >> and
    >> 
    >> integrate(x/(x^2-1),x) ->
    >> log(x^2-1)/2
    >> 

I think the result are correct, but information is missing.

Here is what's happening in bug 1374700.

integrate((1+tan(x)^2)/tan(x),x) is converted to
integrate(cos(x)/sin(x)*(1+sin(x)^2/cos(x)^2),x) by replacing tan with
sin/cos.

Maxima recognizes this form and uses the substitution y = sin(x) to
get a new integral integrate(1/y*(1+y^2/(1-y^2)),y), which maxima
figures is

    log(y)-log(y^2-1)/2.

Substitute back to get the final answer

    log(sin(x)) - log(sin(x)^2-1)/2.

So the result isn't a real number for real x.

If, however, maxima chose the different integral for y, by doing
integrate(expand(1/y*(1+y^2/(1-y^2)),y)), we would have gotten

    log(y)-log(1-y^2)/2

with a final result of 

    log(sin(x)) - log(1-sin(x)^2)/2.

(Of course, this isn't quite right either because we're assuming
sin(x) > 0.)

Perhaps this isn't an issue, and we should just live with the result.

Ray