= Solution
The <momentum operator> in the position representation is $\widehat p=-i\hbar\,d/dx$. Differentiating the given <Gaussian wave packet> gives
$$
\frac{d\Psi}{dx}=(-2\alpha x+ik)\Psi,
\qquad
\Psi^*\widehat p\Psi=(2i\hbar\alpha x+\hbar k)|\Psi|^2.
$$
The probability density $|\Psi|^2$ is an <even function>, so the term proportional to the <odd function> $x$ has integral zero. Normalization supplies $\int_{-\infty}^{\infty}|\Psi|^2dx=1$, and hence
$$
\boxed{\langle p\rangle=\int_{-\infty}^{\infty}\Psi^*\widehat p\Psi\,dx=\hbar k}.
$$
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