I haven't found an *automated* way of reexpressing this with tanh.
However, but substituting atanh(tanh(x)) for x (but temporarily blocking the
simplification), you can derive the result:
(%i56) f: y = (%e^(2*x)+1)/(%e^(2*x)-1)+%c$
(%i57) subst(atanh(box(tanh(x))),x,f);
(%o57) y = (%e^(2*atanh(box(tanh(x))))+1)/(%e^(2*atanh(box(tanh(x))))-1)+%c
(%i58) expand(rembox(ratsimp(logarc(%))));
(%o58) y = 1/tanh(x)+%c
On Sun, Jan 9, 2011 at 06:08, nijso beishuizen <
nijso at numerically-related.com> wrote:
> f: y = (%e^(2*x)+1)/(%e^(2*x)-1)+%c;