Suppose $B_t$ is standard Brownian motion and $X_t$ satisfies
$$dX_t = X_t ^2 dt + X_t B_t$$
Given $Y_t=\exp\left\{\int_0^t (X_s^2 +1 ) ds\right\}$, how to compute $d\langle Y\rangle_t$?
My try
$dY_t = Y_t (X_t^2 + 1) dt + Y_t (\int_0^t 2X_s ds) dX_t=(\ldots)dt + X_tY_t(\int_0^t 2X_s ds)dB_t$
Therefore, we have
$d\langle Y\rangle_t = \left(X_tY_t(\int_0^t 2X_s ds)\right)^2 dt$
Am I correct?