= Solution
For a <Fourier mode> $e^{i(kx-\omega t)}$, the wave equation gives
$$
-(1+k^2)^2\omega^2+k^4=0.
$$
Taking the positive-frequency branch,
$$
\boxed{\omega(k)=\frac{k^2}{1+k^2}}.
$$
For $k>0$, the <phase velocity and group velocity> are
$$
c_p(k)=\frac{k}{1+k^2},
\qquad
c_g(k)=\frac{2k}{(1+k^2)^2}.
$$
The frequency rises from zero and approaches $1$; $c_p$ reaches its maximum at $k=1$, while
$$
c_g'(k)=\frac{2(1-3k^2)}{(1+k^2)^3}
$$
shows that the front velocity is
$$
V_m=c_g(1/\sqrt3)=\frac{3\sqrt3}{8}.
$$
Thus the wave at the front has
$$
\boxed{k=1/\sqrt3}.
$$
Moreover $c_g/c_p=2/(1+k^2)$, so the packet outruns its crests for $0<k<1$, the speeds agree at $k=1$, and the crests outrun the packet for $k>1$.
The initial conditions give the exact superposition
$$
\varphi(x,t)=\int_{-\infty}^{\infty}A(k)
\left[e^{i(kx-\omega(k)t)}+e^{i(kx+\omega(k)t)}\right]dk.
$$
The condition $A^*(-k)=A(k)$ makes this solution real. For $V=x/t$ with $0<V<V_m$, the equation
$$
\boxed{V=\omega'(\alpha)=\frac{2\alpha}{(1+\alpha^2)^2}}
$$
has two positive roots $0<\alpha_1<1/\sqrt3<\alpha_2$. The second exponential has the corresponding stationary points $-\alpha_1,-\alpha_2$. Since
$$
\omega''(k)=\frac{2(1-3k^2)}{(1+k^2)^3},
$$
the <stationary phase method> and pairing of complex-conjugate contributions give
$$
\boxed{
\varphi(Vt,t)\sim2\operatorname{Re}\sum_{j=1}^2
\left(\frac{2\pi}{t|\omega''(\alpha_j)|}\right)^{1/2}
A(\alpha_j)
\exp\left\{it[\alpha_jV-\omega(\alpha_j)]
-\frac{i\pi}{4}\operatorname{sgn}\omega''(\alpha_j)\right\}
}.
$$
For $-V_m<V<0$, the same picture is reflected and the stationary points have negative wavenumber. For $|V|>V_m$ there are no real stationary points, so a smooth localized spectrum gives a much smaller nonstationary tail outside the wave front.
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