How to show that $I_tdX_t + I_tP_tX_t dt = d(I_tX_t)$ using Ito's Lemma?

Question

$dX_t +P_tX_tdt = dB_t$ and the integrating factor is $I_t = e^{\int{P_t dt}}$

Answer 1

$$d(u.v)=u.dv+v.du$$ put $I_t = e^{\int{P_t dt}}$ $$d(I_tX_t)=X_tdI_t+I_tdX_t\\=X_t d(e^{\int{P_t dt}})+I_tdX_t\\ =X_t (e^{\int{P_t dt}})d(\int{P_t dt})+I_tdX_t\\ =X_t (e^{\int{P_t dt}})P_t dt+I_tdX_t\\= X_t I_t P_t dt+I_tdX_t$$ hence by multiplying integral factor we have complete differential in Left hand side $$dX_t+P_t X_t dt=dB_t \ to \\d(I_tX_t)=I_t dB_t$$ apply integral both sides $$\int d(I_sX_s)=\int I_s dB_s \to \\I_tX_t-I_0X_0=\int I_s dB_s\\I_tX_t=I_0X_0+\int I_s dB_s \to \\ X_t=I_t^{-1}I_0X_0+I_t^{-1}\int I_s dB_s $$