= Solution
The one-dimensional <Time-dependent Schrödinger equation> for a <wavefunction> in a real potential is
$$
i\hbar\frac{\partial\Psi}{\partial t}
=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+U\Psi.
$$
Multiplying this equation by $\Psi^*$, multiplying its <complex conjugate> by $\Psi$, and subtracting gives the <probability continuity equation>
$$
\frac{\partial |\Psi|^2}{\partial t}+\frac{\partial j}{\partial x}=0,
\qquad
j=\frac{\hbar}{2mi}\left(\Psi^*\frac{\partial\Psi}{\partial x}
-\Psi\frac{\partial\Psi^*}{\partial x}\right).
$$
If $\Psi$ and its first <derivative> decay sufficiently rapidly as $x\to\pm\infty$, then the <probability current> $j$ vanishes at both ends. Integrating the continuity equation and using the <fundamental theorem of calculus> yields
$$
\frac d{dt}\int_{-\infty}^{\infty}|\Psi(x,t)|^2\,dx
=-j(\infty,t)+j(-\infty,t)=0.
$$
This <conservation of quantum probability> is required by the <Born rule>: once a <normalizable wavefunction> has total probability one, its time evolution must preserve that normalization.
For the stated <Gaussian wave packet>, put $\beta=\hbar t/m$. Since
$$
f(t)=\frac1{\alpha+i\beta},
\qquad
|f(t)|=\frac1{\sqrt{\alpha^2+\beta^2}},
\qquad
\operatorname{Re}f(t)=\frac{\alpha}{\alpha^2+\beta^2},
$$
its squared <modulus> is
$$
|\Psi(x,t)|^2
=\frac{C^2}{\sqrt{\alpha^2+\beta^2}}
\exp\left(-\frac{\alpha x^2}{\alpha^2+\beta^2}\right).
$$
The <Gaussian integral> then gives
$$
\int_{-\infty}^{\infty}|\Psi(x,t)|^2\,dx
=\frac{C^2}{\sqrt{\alpha^2+\beta^2}}
\sqrt{\frac{\pi(\alpha^2+\beta^2)}{\alpha}}
=\boxed{C^2\sqrt{\frac\pi\alpha}},
$$
which is independent of time, as required.
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