In Landau & Lifshitz's book, Classical theory of fields, the action for a free particle is defined as:
$$\tag{8.1} S= \int ^b _a {-mc \ \text d s}=0,$$
where $$\text d s=c\,\text d t\sqrt{1-\frac {v^2}{c^2}}$$ is the the invariant space-time interval beetween points along the particles worldline. From the latter expression of the Lagrangian, it's easy to obtain the expressions of the momentum and the energy of the particle.
However, after a "classical" derivation, the author repeats the calculations with a different notation (I'll post the derivation from the "particle in a electromagnetic field" case omitting the 4-potential terms, because I think that my book contains an error in the free particle case):
Since $\text d s= \sqrt {\text dx^i \text dx_i}$: $$\delta S = -mc \int ^b _a\dfrac{\text dx_i \text d\delta x_i}{\text ds}.$$ Integrating by part, introducing the 4-velocity $u_i =dx_i/ds$, we get:
$$\tag{9.10}\delta S =-mcu_i\delta x^i|^b _a+\int_a ^bmc\dfrac{\text d u_i}{\text d s} \delta x^i \,\text ds.$$
I'm very puzzled about the meaning of these two lines; I have three questions:
1) What does the notation $\text d \delta x_i$ mean?
2) How does he obtain the $\delta S$ expression?
3) How does he pass from the first integral to the second?
If someone could explain in detail I'd be very grateful.
Note, I have no problem in getting the result, I can obtain it by replacing $\text d s =c\text d t \sqrt {...}$ and doing the variation on $v$.