The problem tells me to show that $X_t$ is a martingale but I am getting that it is not. Here the assumed filtration is ${\scr F}_t=\{\sigma(X_s):s \le t\}$. Here is what I tried.
$$E(X_t|{\scr F}_s) = E(e^{B(t)}|{\scr F}_s)-1-\frac{1}{2}E\left(\int_0^te^{B(k)}dk|{\scr F}_s\right) $$
For the first term, $$\begin{align} E(e^{B(t)}|{\scr F}_s)&= E(e^{B(t)-B(s)}e^{B(s)}|{\scr F}_s) \\ &= E(e^{B(t)-B(s)})e^{B(s)} \\ &=e^{\frac{t-s}{2}}e^{B(s)} \end{align}$$ Above we use the fact that $B(t)-B(s)$ follows a $N(0,t-s)$ distribution and is also independent of ${\scr F}_s$. We also used the fact that $e^{B(s)}$ is ${\scr F}_s$ measurable.
For the last term, $$\begin{align} E\left(\int_0^te^{B(k)}dk|{\scr F}_s\right) &= E\left(\int_0^se^{B(k)}dk|{\scr F}_s\right)+E\left(\int_s^te^{B(k)}dk|{\scr F}_s\right) \\ &=\int_0^se^{B(k)}dk + \int_s^tEe^{B(k)}dk \\ &=\int_0^se^{B(k)}dk + \int_s^t e^{\frac{k}{2}}dk \\ &= \int_0^se^{B(k)}dk + 2(e^{\frac{t}{2}}-e^{\frac{s}{2}}) \end{align}$$
Now putting everything together we get $$ E(X_t|{\scr F}_s)= e^{\frac{t-s}{2}}e^{B(s)}-1-\frac{1}{2}\int_0^se^{B(k)}dk-e^{\frac{t}{2}}+e^{\frac{s}{2}}$$
Any help is appreciated!
It is not ture that $B(k)$ is independent of $\mathcal F_s$ for $k >s$. Hence$E\left(\int_s^te^{B(k)}dk|{\scr F}_s\right)$ is not $\int_s^tEe^{B(k)}dk $. It is $\int_s^tE(e^{B(k)}|\mathcal F_s)dk$ and $E(e^{B(k)}|\mathcal F_s)=E(e^{B(k)-B(s)}|\mathcal F_s)e^{B(s)}=E(e^{B(k)-B(s)})e^{B(s)}$.