Not sure how to solve the SDE $d X_t = X_t dt + d B_t,$ for $t > 0$, and $X_0 = 1$ Where $B_t$ is Brownian Motion
Not sure if I'm doing it right: $dX_t/X_t$ = $B_t$ dt + $B_t$ d Bt Set $R_t = log(X_t)$ and apply Itos formula
$R-t - R_0$ = $∫_{t_0}^t 1/(X_s) dX_s - 1/2∫_{t_0}^t (1/(X_s)^2) dX_s*$ ; where $dX_s=B_sX_sds + B_sX_sdB_s$ and $dX_s* B_s^2 X_s^2$
so combined: $R_t - R_0 = ∫_{t_0}^t B_sdB_s + ∫_{t_0}^t (B_s-1/2B_s^2)ds$
Not sure if this is right or where to go from here.
Let $Y_t=X_te^{-t}$. Using Ito's lemma
$$dY_t=e^{-t}dX_t-X_te^{-t}dt=e^{-t}(X_tdt+dB_t)-X_te^{-t}dt=e^{-t}dB_t$$
Hence
$$Y_t=Y_0+\int_0^te^{-s}dB_s$$
and
$$X_t=e^{t}X_0+e^{t}\int_0^te^{-s}dB_s$$
Note that here $\int_0^te^{-s}dB_s \sim N(0,\frac{1}{2}(e^{2t}-1))$.