Let $B$ denotes a Brownian motion in a filtration $\mathcal{F}$ and we define $X$ as the following process:$$X_{t}=\int_{0}^{t}h\left(B_{s}\right)ds,$$ where $h:\mathbb{R}\mapsto\left[1,\infty\right)$ is a smooth function. $M$ is definied as the following: $$M_{t}=\int_{0}^{t}e^{-X_{s}}ds+v\left(B_{t}\right)e^{-X_{t}},$$ where $v:\mathbb{R}\mapsto\mathbb{R}$. What does $v$ have to satisfy if we want $M$ to be a martingale in $\mathcal{F}$?
Here is my solution so far which may contains a lot of mistakes, misunderstandings and/or inaccuracy.
If we want $M$ to be a martingale, then $M$ shouldn't contain any $\int_{0}^{t}\ldots ds$ member in its decomposition. Unfortunatelly there is $\int_{0}^{t}e^{-X_{s}}ds,$ which is a nonnegative process with finite total variation. We know it has finite total variation because we Lebesgue integrate a positive expression. Even $X$ has finite total variation because of the same reason. $X$ is nonnegative, because $h:\mathbb{R}\mapsto\left[1,\infty\right)$, so $X$ is an integration of positive values. We can make $\int_{0}^{t}e^{-X_{s}}ds$ disappear in $M$, if $M$ is the constant $0$ process, therefore$$\int_{0}^{t}e^{-X_{s}}ds=-v\left(B_{t}\right)e^{-X_{t}}.$$ I do hope this is the only way to make $M$ to be a martingale, but at this point I am not so sure. We want to find the Itô decomposition of $-v\left(B_{t}\right)e^{-X_{t}}$ in the form of $g\left(X_{t},B_{t}\right)$. I got the following result for this:$$v\left(B_{t}\right)e^{-X_{t}}=\int_{0}^{t}v\left(B_{s}\right)e^{-X_{s}}h\left(B_{s}\right)-\frac{1}{2}\frac{\partial^{2}v}{\partial y^{2}}\left(B_{s}\right)e^{-X_{s}}ds+\int_{0}^{t}-\frac{\partial v}{\partial y}\left(B_{s}\right)e^{-X_{s}}dB_{s}, $$ which should be equal with $\int_{0}^{t}e^{-X_{s}}ds$ for all $t$. As far as I know there is a statement, (I think we called it Doob-lemma, but I think I am mistaken here, because I don't really remember this crystal clean...) that the semimartingale decomposition is unique, so $$\begin{cases} e^{-x}=v\left(y\right)e^{-x}h\left(y\right)-\frac{1}{2}\frac{\partial^{2}v}{\partial y^{2}}\left(y\right)e^{-x}\\ 0=-\frac{\partial v}{\partial y}\left(y\right)e^{-x} \end{cases}\Longrightarrow\begin{cases} 1=v\left(y\right)h\left(y\right)-\frac{1}{2}\frac{\partial^{2}v}{\partial y^{2}}\left(y\right)\\ 0=\frac{\partial v}{\partial y}\left(y\right) \end{cases}.$$ We have to solve this system of differential equations. From the last equation we know $v$ is a constant function. Furthermore, $h$ has to be the reciproke of $v$, because if $v$ is constant, than $\frac{\partial^{2}v}{\partial y^{2}}\left(y\right)=0$, and $1=const\cdot h\left(y\right)$, so $h$ is a constant as well. $h$ can't be the constant $0$, because $h:\mathbb{R}\mapsto\left[1,\infty\right)$. First I thought we made this clause, because we didn't want $X$ to cross the $0$, therefore in the case of division we can use Itô's formula appropriately. This is my solution, but I don't think it is right. I feel like a made a lot of mistakes but I want to get a clear solution...I don't want this exercise to be unsolved.
Some corrections:
You have written in the comments quite well, if $v\left(B_{t}\right)e^{-X_{t}}$ has the form $-\int_{0}^{t}e^{-X_{s}}ds+M_{t}$, then I get back my own $M$ martingale which means $M$ doesn't necessary have to be the constant zero process. As I did before, the dynamic of $v\left(B_{t}\right)e^{-X_{t}}$ can be written as $$\int_{0}^{t}-v\left(B_{s}\right)e^{-X_{s}}h\left(B_{s}\right)+\frac{1}{2}\frac{d^{2}v}{dy^{2}}\left(B_{s}\right)e^{-X_{s}}ds+\int_{0}^{t}\frac{dv}{dy}\left(B_{s}\right)e^{-X_{s}}dB_{s},$$ which leads back to the differential equation: $$v\left(y\right)h\left(y\right)-\frac{1}{2}\frac{d^{2}v}{dy^{2}}\left(y\right)=1,$$ thanks to the uniqueness of Doob-Meyer decomposition and the fact, that the expressions above are measurable functions of a normal distributed random variable, which is absolute continuous to the Lebesgue measure (moreover they are equivalent). We made the $\int_{0}^{t}\ldots ds$ member disappear, but it is not enough for $M$ to be a martingale. However, I know a condition, which makes $M$ to be a martingale.
We know $M_{t}=\int_{0}^{t}\frac{dv}{dy}\left(B_{s}\right)e^{-X_{s}}dB_{s}$ from the dynamic mentioned above. If $\frac{dv}{dy}\left(B_{s}\right)e^{-X_{s}}$ is progressive measurable, and $$\mathbb{E}\left[\int_{0}^{t}\left(\frac{dv}{dy}\left(B_{s}\right)e^{-X_{s}}\right)^{2}ds\right]<\infty,$$ then $M$ is a martingale. If a process is adapted and continuous, then the process is progressive measurable, so it is enough to $\frac{dv}{dy}\left(y\right)$ be continuous, because we would get compositions of continuous processes which is also continuous. Finally, if I did everything alright, and $v$ satisfies the conditions below, than $M$ is martingale.
1. $v$ is a solution of the differential equation:$$v\left(y\right)\cdot h\left(y\right)-\frac{1}{2}\frac{d^{2}v}{dy^{2}}\left(y\right)=1.$$
2. $\frac{dv}{dy}\left(y\right)$ is continuous.
3. $\mathbb{E}\left[\left(\frac{dv}{dy}\left(B_{s}\right)\right)^{2}e^{-2\int_{0}^{s}h\left(B_{u}\right)du}\right]<\infty$ for all $s$, thanks to the Fubini theorem.
The conditions above perhaps can be generalized, is it a better solution?