I am absolute newbie to stochastic calculus and have to solve a weighted hazard rates integral, where the hazard rates are stochastic, their logarithm governed by arithmetic Ornstein-Uhlenbeck (OU) process (if it simplifies, asymptotically for large times so that all transients are gone).

I.e. I am looking for the distribution at time $t$ of the stochastic process

$$A(t) = \int_0^t q(s) e^{x_s} ds $$

where $q(s)$ is the weight function of time (let's for simplicity think it polynomial or even constant),

and $x_s = x_0 e^{-as} + (1-e^{-as})\mu + \sigma e^{as} \int_0^s e^{-as} dW_s$ being the OU solution with mean $\mu$ and mean-reversion strength parameter $a$.

After transients, $$ x_s \approx \mu + \sigma e^{as} \int_0^s e^{-as} dW_s = \mu + \beta W_s $$

with $\beta^2 = \frac{\sigma^2}{2a}(1 - e^{- 2 a s}) \approx \frac{\sigma^2}{2a}$

So the simplified version of the initial problem: $$A(t) = \int_0^t q(s) e^{\mu + \beta W_s} ds $$ with constant $\mu$ and $\beta$ and a Wiener process $W_s$ (to be more precise, a random variable with density N(0,1))

**Now my problem:**

tried to solve $A(t)$ integral by parts and got stuck $\Rightarrow$ need any possible hint top get any further:

With $Q(t) = \int q(s) ds$

$$ A(t) = Q(t)e^{\mu + \beta W_t} - Q(0)e^{\mu} - \beta \int_0^t Q(s) e^{\mu + \beta W_s} dW_s $$

**If I am correct so far, how to solve the right integral?**

Must be a one-liner well described in the literature but I could not find any information. Maybe solvable by Taylor expansion of $e^W_s$?

Would be very grateful for any hint!!!