Problem:
Find the value of $p$ such that:
$$\int_{0}^{t_{end}} \frac{d}{dt}e^{-p \cdot t}+(e^{-p \cdot t})^2 dt$$
is minimum.
Attempt:
I tried determine the derivative $\frac{d}{dp}$, directly expand the integral and equate to zero,i.e.:
$$\frac{d}{dp}\left(\int_{0}^{t_{end}} \frac{d}{dt}e^{-p \cdot t}+(e^{-p \cdot t})^2 dt\right)=0$$
but got an undecidable equation:
$$\frac{e^{-2 p t_{\text{end}}} -1}{2 p^2}-t_{\text{end}} e^{-p t_{\text{end}}}+\frac{t_{\text{end}} e^{-2 p t_{\text{end}}}}{p}=0$$
but I am not sure if the above is a valid step. Even if it is, I don't know how to proceed from here.
Can't you just get the integral, given $\frac{d}{dt}\left(e^{-pt}\right) = -pe^{-pt}$ and $\left(e^{-pt}\right)^2 = e^{-2pt}$, thus the integral is:
$$ \left.\left(\frac{-pe^{-pt}}{-p} - \frac{-pe^{-2pt}}{-2p}\right)\right|_0^{t_{\text{end}}} = e^{-pt_\text{end}} - \frac{1}{2}e^{-2pt_\text{end}} - \left(1 - \frac{1}{2}\right) $$
You can factor out $e^{-pt_\text{end}}$ out of that equation, and, I suspect, solve a for a minumum.
Hint:
\begin{align*} f(x) =&\ e^{-x} - e^{-2x} \\ f'(x) = &\ -e^{-x} + 2e^{-2x} = -y + 2y^2&\text{where } y = e^{-x} \end{align*}
So $y = 0$ is a solution but $e^\zeta \neq 0$, so no $x$ there, but what about $y\left(-1 +2y\right)= 0 \leadsto -1 + 2y = 0 \leadsto y = \frac{1}{2}$?
Can you find a solution to $y = e^{-x} = \frac{1}{2}$?