Let $B^a_t$ and $B^b_t$ for any $t\in[0,T]$ where $T>0$ be Brownian motions under probability measure $P$ with correlation $\rho\geq 0$.
The task is to calculate \begin{equation} \mathbb{E}\left[\frac{\mathcal{E}(B^a_T)}{1+\mathcal{E}(B^b_T)}\right] \end{equation} where $\mathcal{E}$ is the operator for stochastic exponential. Note that $\mathcal{E}(B^i_t)=\exp(-t/2+B_t^i)$ which is a $P$-martingale for any $i\in\{a,b\}$. Let $f$ be the density function of the standard normal distribution.
Approach 1: Define an equivalent probability measure $Q$ by $\frac{dQ}{dP}=\mathcal{E}(B^a_T)$, then by Girsanov's theorem $\tilde{B}^b_t:=B^b_t-\rho t$ is a Brownian motion under $Q$ and by $\tilde{B}^b_T\sim \mathcal{N}(0,T)$ it follows that $$\begin{align} \mathbb{E}\left[\frac{\mathcal{E}(B^a_T)}{1+\mathcal{E}(B^b_T)}\right] &=\mathbb{E}^Q\left[\frac{1}{1+\mathcal{E}(\tilde{B}^b_T+\rho T)}\right]\\\\ &=\int_{\mathbb{R}}\frac{f(x)}{1+\exp(\rho T-T/2+T x)}dx=:A \end{align}$$
Approach 2: Let $B^a_t=\rho B^b_t+\sqrt{1-\rho^2}B_t$ where $B_t$ is independent from $B^b$. Then $$\begin{align} \mathbb{E}\left[\frac{\mathcal{E}(B^a_T)}{1+\mathcal{E}(B^b_T)}\right] &=\mathbb{E}\left[\frac{\mathcal{E}(\rho B^b_T+\sqrt{1-\rho^2}B_T)}{1+\mathcal{E}(B^b_T)}\right]\\\\ &=\mathbb{E}[\mathcal{E}(\sqrt{1-\rho^2}B_T)]\mathbb{E}\left[\frac{\mathcal{E}(\rho B^b_T)}{1+\mathcal{E}(B^b_T)}\right]\\\\ &=\mathbb{E}\left[\frac{\mathcal{E}(\rho B^b_T)}{1+\mathcal{E}(B^b_T)}\right]\\\\ &=\int_{\mathbb{R}}\frac{\exp(-T\rho^2/2+T\rho x)f(x)}{1+\exp(-T/2+Tx)}dx=:B \end{align}$$ However, the integrands in A and B do not equal. I wonder if I made a mistake somewhere.