What is the integral ∫x(t)dt

Question

I am learning classical physics using the Lagrangian and came across a problem when solving an example.

Suppose you have a spring system, where:

$KE = \frac{1}{2}m(\dot{x})^{2}$
$\ \ \ \ \ and$
$PE = \frac{1}{2}kx^{2}$

The Lagrangian of the system is therefore:

$L \equiv T-U$
$L = \frac{1}{2}m(\dot{x})^{2}-\frac{1}{2}kx^{2}$

Using the Euler-Lagrange equation we can calculate the equation of motion,
$\frac{d}{dt}(\frac{∂L}{∂\dot{x}})=\frac{∂L}{∂x}$

We get,
$m\ddot{x}=-kx$

Therefore, acceleration $\ddot{x}$ must be -$\frac{kx}{m}$

If we integrate the differential equation to solve for the velocity, we get

$\dot{x}=-\frac{k}{m}\int{x}dt$

Here arises my confusion:
How would one integrate $x(t)$ in respect to $t$, if $x(t) =\int{\dot{x}dt}$ or $\ddot{x}$ integrated twice in respect to time

$or$

$-\frac{k}{m}\int{x}dt = -\frac{k}{m}xt$

Answer 1

$m\ddot{x}=-kx\implies \int m(\dot{x}\ddot{x})dt=\int -kx\dot{x}dt$

$m\dot{x}^2/2=-kx^2/2+C$

Worth noting $C$ is a constant of motion, the total energy.

$\dot{x}=\sqrt{2C-kx^2}/\sqrt{m}$

$\frac{dx}{\sqrt{(2C-kx^2)/m}}=dt$

Integrate both sides, the left using trig substitution. Can you take it from there?