Following my earlier question in this Phys.SE post I have another question regarding the derivation I am struggling through!
Considering the variation in the Lagrange density for $x'=x+\delta x$ and $\phi'(x')=\phi(x)+\delta \phi (x)$.
\begin{equation}
\mathcal L'(x')=\mathcal L'(x) +\frac {d \mathcal L}{d x^i}\delta x^i=\mathcal L(x)+\underbrace{\mathcal L'(x)-\mathcal L (x)}_{\bar \delta \mathcal L(x)}+\frac {d \mathcal L}{d x^i}\delta x^i
\end{equation}
Such that the variation in the action is given by,
\begin{equation}
\delta S=\int [\mathcal L(x)+\bar \delta \mathcal L(x)+\frac {d \mathcal L}{d x^i}\delta x^i ]dx'-\int \mathcal L(x) dx
\end{equation}
How then do we obtain the following,
\begin{equation}
\delta S=\int \big[\bar \delta \mathcal L(x)+\frac{d }{d x^i}(\mathcal L\delta x^i)\big]dx
\end{equation}
I understand it is to do with the Jacobian of the transformation that we approximate $$dx'=dx(1+\partial _i\delta x^i+\dots ),$$ but how does this follow and what exactly does this mean (again sorry if its obvious).
