Thx - I like it, anyway: here expand is a good idea to avoid mess

My feeling is: avoid any complicated syntax for 'newbies'. Essentially
you already have done the following (by intuition)

  Mg(xi) = Mf(x)/(F*L):
  subs(x=xi*L, %);

which displays what you show in your picture as problem.

To reduce to your desired result you need 0 < L (else you can not cancel
it out [and implicitly all are Reals]). 

And you want to have F*L within the 'piecewise' and have it canceled.

So your desired result is some expression, which is 'piecewise' on xi.

For that you can continue quite 'naturally' by 

  convert(%, piecewise, xi) assuming 0 < L;
  simplify(%);

                        {      xi
                        {     ----            xi <= 1/2
                        {      2
               Mg(xi) = {
                        {    xi
                        { - ---- + 1/2        1/2 < xi
                        {    2

The previous works as well (omitting to collect for C), just continue by
 
 'V1 = -1140'; fsolve(%, C= 0 .. infinity);

@Markiyan Hirnyk 

Uploading is broken and only works with pain for me - an if it works
then the links are broken. So I avoid that mess. What a shame.

Here are the essential commands (Digits=15, Maple 18):

  EQ1 := convert(eq1, rational);

  st:=time():
  V1:=value( lhs(EQ1) ) assuming 0<C:
  `time needed`[sec]=time() - st;

                      time needed[sec] = 10.951

  V1:=convert(V1, piecewise, C):
  V1:=map(collect, V1, C);

  'V1 = -1140'; fsolve(%, C= 0 .. infinity);

                          0.100907264887074

It may be 'better' to look at 

  'C*(V1+1140)';
  convert(%, piecewise, C): expand(%): sort(%);

showing that is a list of polynomial equations, convert/pwlist
should allow to re-formulate for symbolics (note: conditions are
'cascading', not exclusively), should give a formal solution.

  P:=% assuming 0.1 < C, C < 0.11;

gives a polynom of degree = 3. One can see that all roots are real, so

  [solve(P)]:
  map(evalc@Re, %): simplify(%);
  evalf(%);



Do the same with C < 0, naming it V2. Plot, and guess a symmetry

  plot(V1, C= 0 .. +1, color=red):
  plot(V2, C= -1 .. 0, color = blue):
  plots[display](%, %%);

Confirm that 'symbolically'

  0 = 'V1 - eval(V2, C=-C)';
  convert(%, piecewise, C): simplify(%); is(%);

                                 true

But (of course/here ?) numerically it is much faster.

t*Int(exp(-t*tan(x)), x = 0 .. (1/2)*Pi);
Change(%, x=arctan(y), y);

                           infinity
                          /
                         |          exp(-t y)
                      t  |          --------- dy
                         |            2
                        /            y  + 1
                          0

with(inttrans);

t*laplace(f(y), y, t);
convert(%, Int);
eval(%, f = 'z -> 1/(z^2+1)'); 

                        t laplace(f(y), y, t)


                        infinity
                       /
                      |
                   t  |          f(y) exp(-t y) dy
                      |
                     /
                       0


                           infinity
                          /
                         |          exp(-t y)
                      t  |          --------- dy
                         |            2
                        /            y  + 1
                          0

So by the initial value theorem (for example your link)
the value is f(0), f = 'z -> 1/(z^2+1)'

PS: which neither needs to trust the explicit result of integration
nor needs it to believe in the validity of Maple's limit.

@Stephen Forrest 

Thank you, I like that (and sorry that I have not looked into the help, it already states that in short)

@Stephen Forrest 

Thank you. So s/he is exposed to be a Maple user (with known IP, of course). Would it be possible to use some more anonymous 'standard browser data'?

@DSkoog 

Sounds good.

Out of curiosity: what is sent as header data for such a request ?

yes, I reply at Preben Alsholm's answer

Writing a = b^3 (and a possible choice in general) the change b*x = t
reduces that to someFactor * Int(1/sqrt(x^3+1), x = 0 ... upper).

Maple solves it in terms of EllipticF, diff w.r.t. X gives the correct
value, 1/sqrt(X^3*a + 1) and plotting is 'correct' (but very slow).

Have not checked the books seriously for a 'simple' answer.

Using value(INT) gives some 2F1 and a complex value for a=1/10, X=-1,
which either is seen directly to be not correct or by numerical check.

I dont have Maple here - but for your plot the assumptions for INT2 and INT3 are violated (you only can have X on one of each sides with that sign assumptions)

why that complicated input? Here 'simplify' already does the job

So it remains to explain how to find the explicite zero in Math for that special situation. How does one find it here?

@Markiyan Hirnyk 

P(x0)^7: simplify(%); # works for your solution as well

                                  0

and using "sturmseq" shows there is only 1 real root

Edited: or even more simple for the candites x0,x1

x0^7 - x1^7: simplify(%);

                                  0

a) Purely with Maple? I do not expect that. Why should I, even G&R refers to old work.

b) Feeding and enter? I do not trust it: use "expand" inbetween.

Ignore the "1500" and read r as below/above 1 and look at G&R 6.512.3,
the answer depends on that and may be zero or not. But wanted to see it.