By what I know, there are 3 expression for the acceleration:
$$a=\frac{dv}{dt} \Longleftrightarrow dv=a\,dt$$ $$a=\frac{d^2x}{dt^2} \Longleftrightarrow a\,dt^2=d^2x$$ $$a=v\frac{dv}{dx} \Longleftrightarrow a\,dx=v\,dv$$
The reason of my question is to obtain the expressions of the movement with constant acceleration, which I don't really know the name in English.
In pdf of the class, they have 3 expression already (methods to obtain them included), which are:
Integrating the 2 sides of the first expression, they obtained $v=v_0+at$; Equaling this to $\frac{dx}{dt}$, they obtained $x=x_0+v_0+\frac 12at^2$; Lastly, they integrated the 2 sides of the last expression to obtain $v^2=v_0^2+2a\Delta x$;
What I was curious about was if it possible to do anything with the second expression, so what I did, was as follows:
Put the 2 sides of the expression on integrals, obtaining the equation $$a\int_0^t \int_0^t dt\,dt=\int_{x_0}^{x} \int_{x_0}^{x} dx\,dx.$$ Integrating the first time to obtain the expression $$\int_{x_0}^{x}(x-x_0) dx=a\int_0^t t\,dt,$$ but don't know what to do from here. Does the fact that I obtained $x_0$ on the development of the equation make it impossible to move forward?
Say you have a prescribed acceleration $a(t)$.
The first integration yields velocity
$$ v(t) = \int a(t)\,{\rm d}t + C_1 $$
The second integration yields displacement
$$ x(t) = \int v(t)\,{\rm d}t + C_2 =\int \left( \int a(t)\,{\rm d}t \right) {\rm d}t + C_1 t + C_2 $$
So for example, with constant acceleration, starting $t = 0$ with $x_0$ and $v_0$ you have
$$ \begin{cases} v(t) = a t + C_1 \\ x(t) = \frac{1}{2} a t^2 + C_1 t + C_2 \\ x(0) = x_0 \\ v(0) = v_0 \end{cases} $$
Use the last two equations to find $C_1$ and $C_2$.
$$ x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 $$