How to realize the inverse Laplace transform?

Questions
1

(E^(-Sqrt[s] x + Sqrt[s] (2 + x))
u0 (Sqrt[s] - [Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
3/2) - (1 + E^(2 Sqrt[s])) s [Lambda]) - (
u0 (Sqrt[s] + [Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
3/2) - (1 + E^(2 Sqrt[s])) s [Lambda]) + (
E^(Sqrt[s] x) (-F + u0 [Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
3/2) - (1 + E^(2 Sqrt[s])) s [Lambda]) + (
E^(2 Sqrt[s] -
Sqrt[s] x) (-F + u0 [Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
3/2) - (1 + E^(2 Sqrt[s])) s [Lambda])

2

Please format your code as code, not text. All of the\ before [Lambda] are missing. Read this .

3

This website should be improved and compile the codes into what you see is what you get.

4

There is no reason to think it has a closed form.

One way is calculate numerically:

 $Version
 (*12.3.0 for Microsoft Windows (64-bit) (May 10, 2021)*)

f[x_, t_, F_, u0_, \[Lambda]_] := 
 InverseLaplaceTransform[(
   E^(-Sqrt[s] x + Sqrt[s] (2 + x))
     u0 (Sqrt[s] - \[Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
     3/2) - (1 + E^(2 Sqrt[s])) s \[Lambda]) - (
   u0 (Sqrt[s] + \[Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
     3/2) - (1 + E^(2 Sqrt[s])) s \[Lambda]) + (
   E^(Sqrt[s] x) (-F + u0 \[Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
     3/2) - (1 + E^(2 Sqrt[s])) s \[Lambda]) + (
   E^(2 Sqrt[s] - 
     Sqrt[s] x) (-F + u0 \[Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
     3/2) - (1 + E^(2 Sqrt[s])) s \[Lambda]), s, N@t, 
  Method -> "Crump"]

   ListLinePlot[Table[{t, f[1/2, t, 1, 2, 3]}, {t, 0.001, 4, 1/30}], 
    PlotRange -> All](*for: x=1/2,F=1,u0=2,\[Lambda]=3*)

   data = Flatten[
      Table[{x, t, f[x, t, 1, 2, 3]}, {x, -1, 1, 1/5}, {t, 0.001, 3, 
        1/5}], 1];

  ListPlot3D[data, InterpolationOrder -> 3, 
   AxesLabel -> {"x", "t"}](*for: F=1,u0=2,\[Lambda]=3*)
5

The version of Mathematica in my PC is 12 and it doesn’t run your codes.