Gronwall's theorem is given as follows:
Assume that for $t_0\leq t \leq t_0 + a$, with $a$ a positive constant, we have the estimate $$\phi(t)\leq \delta_1\int_{t_0}^t\psi(s)\phi(s)ds + \delta_3\,\,\,\,\,(1.4)$$ in which, for $t_0\leq t \leq t_0 + a$, $\phi(t)$ and $\psi(t)$ are continuous functions, $\phi(t) \geq 0$ and $\psi(t)\geq 0$; $\delta_1$ and $\delta_3$ are positive constants. Then we have for $t_0\leq t\leq t_0 + a$ $$\phi(t) \leq \delta_3e^{\delta_1\int_{t_0}^t\psi(s)ds}$$
In my book the following proof of this theorem is proof is given:
From the estimate $(1.4)$ we derive $$\dfrac{\phi(t)}{\delta_1\int_{t_0}^t\psi(\tau)\phi(\tau)d\tau + \delta_3}\leq 1$$ multiplication with $\delta_1\psi(t)$ and integration yields $$\int_{t_0}^t\dfrac{\delta_1\psi(s)\phi(s)ds}{\delta_1\int_{t_0}^t\psi(\tau)\phi(\tau)d\tau + \delta_3}\leq \delta_1\int_{t_0}^t\psi(s)ds$$ so that $$\ln(\delta_1\int_{t_0}^t\psi(s)\phi(s)ds + \delta_3) - \ln\delta_3\leq \delta_1\int_{t_0}^t\psi(s)ds$$ which produces $$\delta_1\int_{t_0}^t\psi(s)\phi(s)ds + \delta_3 \leq \delta_3 e^{\delta_1\int_{t_0}^t\psi(s)ds}$$ Applying the estimate $(1.4)$ again, but now to the lefthand side, yields the required inequality.
Question: How does the step from $$\int_{t_0}^t\dfrac{\delta_1\psi(s)\phi(s)ds}{\delta_1\int_{t_0}^t\psi(\tau)\phi(\tau)d\tau + \delta_3}\leq \delta_1\int_{t_0}^t\psi(s)ds$$ to $$\ln(\delta_1\int_{t_0}^t\psi(s)\phi(s)ds + \delta_3) - \ln\delta_3\leq \delta_1\int_{t_0}^t\psi(s)ds$$ work? I think I understand the rest of the proof but I can't figure out the above step.
Thanks!
Ok, the thing is simple: think of the function $$F(t)=\delta_1\int_{t_0}^t\psi(\tau)\phi(\tau)d\tau + \delta_3,$$ it is in $C^1$ and $F'(t)=\delta_1\psi(t)\phi(t)$.
Then you just have to integrate $$\int_{t_0}^t\dfrac{F'(s)ds}{F(s)}=\int_{t_0}^t\dfrac{dF(s)}{F(s)},$$ and this is possible to do by variables change whenever $F(s)>0$, but that is guaranteed because $\psi(t),\phi(t),\delta_1,\delta_3$ are all positive and $t\geq{t_0}$.
Note: there are other proofs for Grönwall-Bellman's inequality, you may want to check the other ones (even Wikipedia have it's own).