@Kitonum I would do that as

sin(-(1/6)*Pi+(1/2)*arccos(1/3)): 
convert(%, exp):        # your (essential) suggestion 
expand(%): simplify(%): # resolves the nested radicals 
combine(%);             # beautify it

                                1/2
                               6
                             - ---- + 1/2
                                6

If I understand the question correctly: x in IN s.th. 1 = (1+x)*exp(-x)

By Lindemann's theorem exp is transcendent except for x=0, hence ...
(as you already know that Naturals are algebraic)

For short: I would provide that as Maple document, *.wm as format. Or pdf.

I am not ware that Maple can manipulate Word documents. And would guess that additionally you would need VBA code (or even more, say VB script).

Note that there are (at least) 2 formats for word: the old one, *.doc.

And the new one, *.docx. That is a zipped directory of XML files and their description. If you rename to *.zip then you can extract them, obeying the directory structure.

The images are locates in a subdirectory named 'media', enumerated files. Those can be replaced by desired images (but you have to regard the names).

After zipping again you have the desired modification. Then rename the zip.

Initially you may start with a Word document having only 1- bit pictures.

Just played with that, seems to work. But certainly that means to do some serious work. So providing 'intrinsic' solutions is better..

Workaround for 'abs' over the Reals (which works in other cases as well)

eval(%, abs='t -> sqrt(t^2)'); # for Reals

Make the input for 'frem' more simple by a linear shift

  Int(exp(f(x+1)-1),x=-5..5);
  ``=expand(%);
  IntegrationTools[Change](%, x+1=t, t);

                         5
                        /
                       |
                       |   exp(f(x + 1) - 1) dx
                       |
                      /
                        -5

                                   6
                                  /
                                 |
                      = exp(-1)  |   exp(f(t)) dt
                                 |
                                /
                                  -4

Plot this situation to get some idea

  ''frem''(x+1*0,3)-1*0;
  plot(%,x=-4..6, color = red, tickmarks=[12, 5]);

One can 'see' the jumps and that frem(x,3) = x for x in -3/2 ... 3/2

Now split the integral and use period = 3 to succeed:

  Int(exp(F(t)),t = -4 .. 6);
  Int(F(t),t= -4 .. -3/2) + Int(F(t),t= -3/2 .. 3/2) + 
    Int(F(t),t= 3/2 .. 3/2+3) + Int(F(t),t= 3/2+3 .. 6);
  Int(F(t),t= -4      +3 .. -3/2    +3) + 
    Int(F(t),t= -3/2 .. 3/2) + 
    Int(F(t),t= 3/2   -3 .. 3/2+3   -3) + 
    Int(F(t),t= 3/2+3 -6   .. 6     -6);
 
  eval(%, F=exp); # note: in that range frem(x,3) = x
  value(%);

                             6
                            /
                           |
                           |   exp(F(t)) dt
                           |
                          /
                            -4


          3/2                  3/2                0
         /                    /                  /
        |                    |                  |
        |     exp(t) dt + 2  |     exp(t) dt +  |     exp(t) dt
        |                    |                  |
       /                    /                  /
         -1                   -3/2               -3/2


               -exp(-1) + 3 exp(3/2) - 3 exp(-3/2) + 1


After regarding the factor from above you are done:

  exp(-1)*%;
  evalf(%);



          exp(-1) (-exp(-1) + 3 exp(3/2) - 3 exp(-3/2) + 1)


                           4.93245297416353



PS: acer's trick for me in Maple 18 needs a method, for example
evalf(Int('x ->  exp(frem(x+1,3)-1)', -5..5, method = _d01ajc));

                           4.93245297416352

Edited: not acer, it was posted by Carl Love. Sorry :-(

Usually one may set Digits:= trunc(evalhf(Digits)) for the NAG integrators like method = _d01ajc

I use shift+enter

I doubt that it is possible in general (?), for topological reasons, whatever you mean by 'parametric'.

I guess that is through your IP adress assigned by your provider, done by tables (off your control)

There are ways to check that, http://browserspy.dk/ip.php will show your IP, http://www.utrace.de can show your location and other sites can provide you with such informations as well, http://www.gulli.com/hard-und-software/tools/gullitools/anonymitaetscheck or http://ip-check.info/index.php?lang=en

Mostly on desire of the advertising industry

consider eqn2+eqn3 - eqn1; giving 200 = 4845.859874

Use the library which already knows some solutions:

  restart; with(inttrans):
  'ln(1/sqrt(1+(tau*w)^2))';
  ``=%;
  g:=unapply(rhs(%), w);
                                         2  2
                    g := w -> -1/2 ln(tau  w  + 1)

  hilbert(f(t), t, w): convert(%, Int); # f = 'any' function
  eval(%, f=g);

                    2    2
   1            ln(t  tau  + 1)
  ---- Int(-1/2 ---------------, t = -infinity .. infinity, CauchyPrincipalValue)
   Pi                t - w

  hilbert(g(t), t, w):
  R:=evalc(%) assuming 0 < tau:          # that silently assumes real variables
 
  simplify(R) assuming 0<tau, w < 0;
  #simplify(R) assuming 0<tau, 0 < w;

                          Pi             1
                         ---- + arctan(-----)
                          2            tau w

On Win 7 mine is maple.ini at C:\programs_x86\Maple18\lib From which I call the actual settings, so I have the same for all versions and no duplicates

If you feel unsure you may insert some trivial statement, which prompts into opened Maple file. Like Digits:=15; (having a semi colon at the end)

Try to get used to write exp(z), not e^z (or use exp(1)^z if you really need it)

May be you have an error in coding?

If I take Denom(s, 2*M0*m, g, m) from your poleR for your data

'Denom(s, 2*M0*m, g, m)';
#subs(M0=M0, g=g, m=mK, %); indets(%, symbol);
subs(M0=M0, g=.935, m=mK, %); 
H:=unapply(%, M0, s);
indets(%%, symbol);

then you want to solve for s, if a specific M0 = 0.93 is given.

'H(0.93,s)';
plot(%, s=0.9 .. 1);

indicates that there is no real solution for your expected s ~ 0.95.

'H(0.93,x+y*I)';
Re(%):
plot3d((%), x=0.9 .. 0.98, y=-0.1 .. 0.1, axes=boxed);

indicates that there is no complex solution.

0.1*num/den; subs(x[3]=1,x[4]=2,%); simplify(%);
                                                            
                          0.0999999999999999

Note that using floating point numbers you may not get exact
results, they are approximations.

But you may use 1/10 instead of 0.1

Edited. Or using other wordings: for numerical calculations
the usual mathematical rules are only valid up to an error
(for example the 'route' of calculation or input matters).
And that happens for any (computer) system