Position Dependence in Equation of Motion Our lecturer gives study material which contained that Newton's second law could be written as:
$$ \begin{aligned} F &= m \ddot{x} \\ &= m \frac{d \dot{x}}{dt} \\ &= m \frac{dx}{dx} \frac{d \dot{x}}{dt} \\ &= m \frac{dx}{dt} \frac{d\dot{x}}{dx} \\ &= m \dot{x} \frac{d \dot{x}}{dx} \end{aligned} $$
The last line of that equation say that $ F = m \dot{x} \frac{d \dot{x}}{dx} $, when I trying to find it's $ \dot{x} $ equation like:
$$ \int_{a}^{b} \frac{F}{m} dx = \int_{c}^{d} \dot{x} \ d\dot{x} $$
What should I take for $a$, $b$ and $c$, $d$? Are all of these stand for x (the position) or a,b is for x and c,d for t (time) or vice-versa?
I'm confused because usually the variable x is time dependent,  $ x(t) $ so the $ \dot{x} \text{ is for } \dot{x}(t) $, isn't (?). Give me some explanation about this, please.
 A: Look at the differentials on your integrals. The left differential is a position, so the limits on the integral need to be positions. The right differential is a velocity, so the limits on the integral need to be velocities.
If you could express the velocity as a function of position  $\dot x=\dot x(x)$, then you could have
$$\mathbf \int_a^b\frac Fm\,\text dx=\int_{\dot x(a)}^{\dot x(b)}\dot x'\,\text d\dot x'$$

I'm confused because usually the variable $x$ is time dependent, $x(t)$ so the $\dot x$ is for $\dot x(t)$, isn't (?). Give me some explanation about this, please.

Yes, these can be express as functions of time. However, there are instances where you can express the velocity as a function of position as well. For example, in constant acceleration motion where the object doesn't change direction, you have
$$\dot x(x)=\sqrt{\dot x(x(t=0))^2+2\ddot xx}$$
Of course this is not always the case. For example, when you throw a ball straight in the air then it comes back down along the same path, so the velocity is not a function of the position there. In order to perform those integrals you would have to break the integrals up into pieces where the velocity could be defined as a function of the position, e.g. along the upward path and then along the downward path.
A: $x$ and $\dot x$ are both functions of time. To simplify the notation, let's write $v(t)$ instead of $\dot x(t)$.
Suppose at some time $t_0$ we have $x(t_0) =x_0$ and $v(t_0)=v_0$, and at some later time $t_1$ we have $x(t_1) = x_1$ and $v(t_1)=v_1$. These are the boundaries of the integrals. So we have
$\displaystyle \int_{x_0}^{x_1}F \space dx = m \int_{v_0}^{v_1} v \space dv
\\ \displaystyle \int_{x_0}^{x_1}F \space dx = \frac 1 2m(v_1^2-v_0^2)$
The left hand side is the work done by the force $F$ between $t=t_0$ and $t=t_1$and the right hand side is the change in kinetic energy.
A: In terms of differential quantities you have
$$ F\, {\rm d} x =m v\, {\rm d} v \tag{1} $$
where $v = \dot{x}$. When integrating consider the initial conditions $(x_0,v_0)$ and the final conditions $(x_1,v_1)$.
$$ \int \limits_{x_0}^{x_1} F\, {\rm d} x = \int \limits_{v_0}^{v_1}  m v\, {\rm d} v \tag{2} $$
which is equivalent to $\text{(Work)} = \text{(Kinetic Energy)}$.
Now if the final conditions is what we are trying to evaluate, then do the following
$$ \int \limits_{x_0}^{x} F\, {\rm d} x = \int \limits_{v_0}^{v}  m v\, {\rm d} v \tag{3} $$
For example, a spring with $F = -k x$ is attached to a mass with initial position $x_0=0$ and velocity $v_0>0$.
$$ \int \limits_{0}^{x} (-k x)\, {\rm d} x = \int \limits_{v_0}^{v}  m v\, {\rm d} v  $$
$$ -\tfrac{1}{2} k x^2 = \tfrac{1}{2} m (v^2-v_0^2) $$
or $v = \sqrt{ v_0^2 - \tfrac{k}{m} x^2} $ which is the time independent equation of simple harmonic motion.
