SDE solution using Itô formula

I'd like to solve the Langevin SDE $$dX(t)=-bX(t)dt+\sigma dW(t),\\X(0)=X_0,$$ $W(t)$ being a standard Brownian motion, using the Itô formula $$du(t,X(t)) = \frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dX+\frac{1}{2}\frac{\partial^2 u}{\partial x^2}\sigma^2 dt.$$ However, I don't know where to start: if I try to divide by $X(t)$ in the fist formula and integrate I find $$\frac{dX(t)}{X(t)} = -bdt+\frac{\sigma}{X(t)}dW(t) \Longleftrightarrow \ln(X(t)) = -bt+\sigma\int_0^t\frac{1}{X(s)}dW(s).$$ I could apply Itô on $\ln(X(t))$ but I wouldn't still know the right hand side.

Apply Ito to $$X_t\mathrm{e}^{bt}\implies \mathrm{e}^{bt}dX_t + b\mathrm{e}^{bt}X_tdt$$ so multiply your original SDE by $\mathrm{e}^{bt}$ we find $$\mathrm{e}^{bt}dX_t = -b\mathrm{e}^{bt}X_tdt+\sigma \mathrm{e}^{bt}dW_t$$ rearrange and using the first equation I derived $$d\left(X_t\mathrm{e}^{bt}\right) = \sigma \mathrm{e}^{bt}dW_t$$ then integrate we find $$X_t\mathrm{e}^{bt} = X_0 +\int_0^t\sigma \mathrm{e}^{bs}dW_s$$ or (and using your notation) $$X(t) = X(0)\mathrm{e}^{-bt} + \sigma\int_0^t\mathrm{e}^{b(s-t)}dW(s)$$