Let's assume that we have two correlated Brownian motions $Z_{t}$ and $W_{t}$ with correlation $\rho$.
Let $0\leq t \leq T<+\infty$.
How to compute:
$$E\left[exp\left(\int_{0}^{T}f(s)dZ_{s}+\int_{0}^{t}g(s)dW_{s}\right)\right]$$
We can assume that $f(s)=s$ and $g(s)=s$ for simplicity.
If $Z:=\{Z_t; 0 \leq t \leq T\}$ and $W:=\{W_t; 0 \leq t \leq T\}$ have correlation $\rho,$ we can express $Z$ as
$$Z_t = \rho W_t + \sqrt{1-\rho^2}B_t, \quad 0 \leq t \leq T,$$ where $(W,B)$ is a (standard) two-dimensional Brownian motion (i.e. $W$ and $B$ are two one-dimensional independent B.M.). Then \begin{align} E\left[ \exp \left(\int_0^t sdZ_s + \int_0^t sdW_s \right)\right]&=E\left[\exp \left(\int_0^t s(\rho dW_s + \sqrt{1-\rho^2}dB_s) + \int_0^tsdW_s\right)\right] \\ &= E\left[ \exp \left(\sqrt{1-\rho^2}\int_0^t sdB_s + (\rho+1)\int_0^tsdW_s \right)\right]. \end{align}
If $\lambda$ is a constant, and since $f(s)=s$ is a deterministic function, it is well known that $$\lambda \int_0^t sdW_s \sim N \left(0,\lambda^2 \int_0^t s^2ds=\frac{\lambda^2 t^3}{3} \right).$$ We also know that if $X \sim N(\mu, \sigma^2)$, then $E[e^X] = e^{\mu+\frac{\sigma^2}{2}}.$ Thus \begin{align} E\left[ \exp \left(\sqrt{1-\rho^2}\int_0^t sdB_s + (\rho+1)\int_0^tsdW_s \right)\right] &=E\left[\exp \left(\sqrt{1-\rho^2}\int_0^t sdB_s \right) \right] \times \\ &E \left[ \exp \left((\rho +1)\int_0^tsdW_s\right)\right] \\ &= e^{\frac{2 t^3(\rho+1)}{6}}. \end{align}