Answer
From Eq.(1.29),
\begin{eqnarray}
\label{1}
\frac{d\langle x\rangle}{dt}=\int x \frac{\partial}{\partial t} \mid \Psi \mid ^{2} dx
\end{eqnarray}
In Eq.\eqref{1}, the L.H.S. shows the variation of $\langle x \rangle$ with respect to time because of the time dependence of probability density function $\mid \Psi\left(x,t\right)\mid^{2}$.
From the chain rule, R.H.S. of Eq.\eqref{1} can be written in following manner:
\begin{eqnarray}
\label{2}
\frac{d\langle x\rangle}{dt}&=&\int x \left( \frac{\partial \mid \Psi \mid ^{2}}{\partial x}\frac{\partial x}{\partial t}\right) dx \nonumber \\
\end{eqnarray}
From integration by parts,
\begin{eqnarray}
\label{3}
\frac{d\langle x\rangle}{dt}&=& x\frac{\partial x}{\partial t}\int \frac{\partial \mid \Psi \mid ^{2}}{\partial x} dx-\int \frac{d}{dx}\left(x\frac{\partial x}{\partial t}\right) \mid \Psi \mid ^{2} dx,
\end{eqnarray}
In Eq.\eqref{3}, the term $\frac{\partial x}{\partial t}$ vanishes as $x$ is independent of time '$t$'. Therefore, it implies $\frac{d\langle x\rangle}{dt}=0$. So it would not be appropriate to apply integration by parts for the solution of Eq.\eqref{1}.
Work Step by Step
From Eq.(1.29),
\begin{eqnarray}
\label{1}
\frac{d\langle x\rangle}{dt}=\int x \frac{\partial}{\partial t} \mid \Psi \mid ^{2} dx
\end{eqnarray}
In Eq.\eqref{1}, the L.H.S. shows the variation of $\langle x \rangle$ with respect to time because of the time dependence of probability density function $\mid \Psi\left(x,t\right)\mid^{2}$.
From the chain rule, R.H.S. of Eq.\eqref{1} can be written in following manner:
\begin{eqnarray}
\label{2}
\frac{d\langle x\rangle}{dt}&=&\int x \left( \frac{\partial \mid \Psi \mid ^{2}}{\partial x}\frac{\partial x}{\partial t}\right) dx \nonumber \\
\end{eqnarray}
From integration by parts,
\begin{eqnarray}
\label{3}
\frac{d\langle x\rangle}{dt}&=& x\frac{\partial x}{\partial t}\int \frac{\partial \mid \Psi \mid ^{2}}{\partial x} dx-\int \frac{d}{dx}\left(x\frac{\partial x}{\partial t}\right) \mid \Psi \mid ^{2} dx,
\end{eqnarray}
In Eq.\eqref{3}, the term $\frac{\partial x}{\partial t}$ vanishes as $x$ is independent of time '$t$'. Therefore, it implies $\frac{d\langle x\rangle}{dt}=0$. So it would not be appropriate to apply integration by parts for the solution of Eq.\eqref{1}.
