NIntegrate of a convergent integral working with large integration limits, but not with infinite integration limits

Numerical integration can be without any problems by splitting the integration into two parts:

f1 = FullSimplify[Integrand[x], Assumptions -> x > 0];
f2 = FullSimplify[Integrand[-x], Assumptions -> x > 0];
NIntegrate[f1 + f2, {x, 0, ∞}]
(*2.81307*)

Higher accuracy can be reached (again without any warnings as follows)

NIntegrate[f1 + f2, {x, 0, 1, ∞}, WorkingPrecision -> 100]
 (*2.81306842544297991357319331946335010348754523474644575568568457957304\
   5154609418329813658075049081527*)

Update

Assuming that the principal value is wanted, the method below is probably the best way, which computes the PV at infinity by reflecting the integration over the negative axis onto the positive axis.

NIntegrate[Integrand[x] + Integrand[-x], {x, 0, Infinity}]

(*  2.8130684254425558`  *)

@user64494 pointed out that it's a divergent integral (at infinity) and that Method -> "PrincipalValue" computes the principal value only at finite points and not at infinity. Add the singularity at 1 to the interval to get higher precision:

    NIntegrate[Integrand[x] + Integrand[-x], {x, 0, 1, Infinity},...]

(See older part below for discussion of the singularities.)

The OP's NIntegrate[Integrand[x], {x, -10000000, 10000000}] in essence approximates the PV.

---

I can get them to agree in this way:

sing = SortBy[
   x /. Solve[FunctionSingularities[Integrand[x], x], x, Reals], N];
N@sing

(*  {-1., -0.56984, 0., 2.1479}  *)

NIntegrate[integrand[x], 
 Evaluate@{x, -10000000, Sequence @@ sing, 10000000}]
NIntegrate[integrand[x],
 Evaluate@{x, -Infinity, Sequence @@ sing, Infinity}, 
 WorkingPrecision -> 32, PrecisionGoal -> 6, AccuracyGoal -> 20, 
 Method -> "PrincipalValue"]

(*
  2.813065409383123       (no warnings)
  2.81306842613267712727  (warnings omitted)
*)

At -1 and 0, there seem to be infinite derivatives, so-called "weak" singularities. At the other two "singularities" there might be numerical difficulties from subtractive cancellation (e.g. they're where the denominator (1 + x)^2 - Abs[x]^3 is zero). Weak singularities tend to make convergence harder.