Integral with respect to the integral itself

Question

If you had a velocity function $v(s)=s\cdot k+1$ where s is the stretch and k is a constant, and you wanted to find the stretch from $t_0$ to $t$, witch would be an integral of the velocity function with respect to the stretch: $$s=\int_{t_0}^{t}{v(s)ds}$$ How would you represent this in maths? I made a visualization of the integral in excel, screenshot here. I realize that the function can be written as: $$s(t)=\sum^t_i\Delta s_i+1=\sum^t_i=(s(i-1)+1)+1$$ Where $$\Delta s_t=v(s_t)=s_{t-1}+1=s({t-1})+1$$ If that brings more clarity.

Answer 1

This is a differential equation and must be processed as such, because the derivative of the stretch function is a function of the stretch. You write

$$\frac{ds}{dt}=ks+1$$ or $$\frac{ds}{ks+1}=dt$$ and you can integrate both members,

$$\int_{s_0}^s\frac{d\sigma}{k\sigma+1}=\int_{t_0}^t\tau\,d\tau.$$ (Greek letters used to avoid confusion with the dummy variables.)

Hence,

$$\frac1k\log\frac{ks+1}{ks_0+1}=t-t_0$$ from which you draw $s(t)$.