here is a derivation of equation 2.28 in the book

We’ll do that by differentiating with respect to time and show that this iszero.So, let A be proportional to the RHS of Equation 1:A =and let us compute dAdt .∫ [ ψ(x, t) ∂ψ∗ (x, t)∂x− ψ ∗ (x, t)2.1 puase, take a deep breath ...∫dA[dt = ∂ψ(x, t) ∂ψ ∗ (x, t)+ ψ(x, t) ∂2 ψ ∗ (x, t)∂t ∂x∂x∂tThe integrand can be written[∂ ∂ψ ∗ (x, t)ψ(x, t) +∂x ∂xwhich can, in turn, be written[∂ ∂ψ ∗ (x, t)ψ(x, t) +∂x ∂x]∂ψ(x, t)ψ ∗ (x, t) −2∂x]∂ψ(x, t)dx∂x]− ∂ψ∗ (x, t) ∂ψ(x, t)− ψ ∗ (x, t) ∂2 ψ(x, t)dx∂t ∂x∂x∂t[ ∂ψ ∗ (x, t)∂t∂ψ(x, t)∂x]∂ψ(x, t)ψ ∗ (x, t) − 2 ∂ [ψ ∗ (x, t)∂x∂t]+ ∂2 ψ(x, t)ψ ∗ (x, t)∂x∂t]∂ψ(x, t)∂xSo, as usual, the left part goes away when we integrate by parts, leavingdAdt = −2 ∂ ∫ [ ]ψ ∗ ∂ψ(x, t)(x, t) dx (2)∂t∂xNow, remember we’re trying to show that this is zero, since that willgive us uniform motion of the wavepacket. So, what do we know is certainlyindependent of time? For a free particle it is true that the normalization isindependent of time: ∫ψ ∗ (x, t)ψ(x, t)dx = 1.Somehow we need to introduce this normalization. Let’s try:∫∂[ ]ψ ∗ ∂ψ(x, t)(x, t) dx = ∂ ∫∂t∂x ∂t∂∂x |ψ(x, t)|2 dx− ∂ ∂t∫ [ ]ψ(x, t) ∂ψ∗ (x, t)dx∂x(3)3