![enter image description here](https://i.sstatic.net/9k8fg.j Sakurai just mentions that if operator is changed but state kets are kept frozen but he left mathematics for students.

The following section is from Modern Quantum Mechanics by Sakurai; can any one help me finding out how this is done?

In contrast, if we follow approach 2, we obtain \begin{align} \vert\alpha\rangle&\to\vert\alpha\rangle \\ \mathbf{x}&\to\left(1+\frac{i\mathbf p\cdot d\mathbf{x}'}{\hbar}\right)\mathbf x\left(1-\frac{i\mathbf p\cdot d\mathbf{x}'}{\hbar}\right) \\ &=\mathbf x + \left(\frac{i}{\hbar}\right)\left[\mathbf p\cdot d\mathbf x',\,\mathbf x\right]\tag{this line} \\ &=\mathbf x +d\mathbf x' \end{align} We leave it as an exercise for the reader to show that both approaces lead to the same result for the expectation value of $\mathbf x$: $$\langle\mathbf x\rangle\to\langle\mathbf x\rangle+\langle d\mathbf x'\rangle$$

Sakurai just mentions that if operator is changed but state kets are kept frozen but he left mathematics for students.