This question came up in the context of a bandit problem.
Let $a: [0, \infty[ \to [0,1], t\mapsto a_t$ be a measurable function. I want to prove the following claim for $t\ge 0$: $$ \int_0^t \frac{\lambda a_z}{1+\frac{p}{1-p}e^{-\int_0^z \lambda a_s \mathrm{d}s}} \mathrm{d}z = \log(p + (1-p)e^{\int_0^t \lambda a_s \mathrm{d}s}) $$ ($\lambda >0$ and $p\in]0, 1[$ are parameters)
Here is what I did: Plugging in $t=0$ we find that both sides are zero and differentiating at any $t>0$ we find that the derivatives of both sides coincide:
LHS: $$ \frac{\lambda a_t}{1+\frac{p}{1-p}e^{-\int_0^t \lambda a_s \mathrm{d}s}} $$ RHS: $$ \frac{(1-p)e^{\int_0^t \lambda a_s \mathrm{d}s}\lambda a_t}{p + (1-p)e^{\int_0^t \lambda a_s \mathrm{d}s}} = \frac{\lambda a_t}{1+\frac{p}{1-p}e^{-\int_0^t \lambda a_s \mathrm{d}s}} $$
Hence the two sides coincide for any $t\ge 0$.
The problem with this approach is that I am implicitly assuming that $a$ is "nice" enough to allow differentiation of the two sides. But for general measurable $a$ I don't know how to proceed.