Let $X_t$ be a standard Brownian. Put $Y_t := \exp (\int _0^t udX_u)$ and determine, whether $Y$ is a martingale.
We know $M_t :=\int _0^t u dX_u$ is a martingale, therefore $\langle M \rangle _t = \int _0^t u^2 du $. By Ito's formula we have $$ Y_t = 1 + \int _0^t Y_u dM_u + \frac{1}{2} \int _0^t Y_u d\langle M \rangle _u, $$ where the $dM_u$ term is a martingale. Therefore, $Y$ is martingale if and only if $$ E \left ( \int _0^t Y_u d\langle M \rangle _u \ \Bigg\vert\ s \right ) = 0 \tag{1} $$ for every $s< t$ and this is where I'm stuck. How do we make sense of quantities like (1) and why (not) it would be zero?
Well unfortunately it is never the case. This comes from the fact that $\int _0^t u dX_u$ follows a Gaussian law (as this is a Wiener integral) and so $Y_t$ is a log-normal random variable for all $t$ with expectation equal to $e^{\mu_t+\sigma_t^2/2}$ where $\mu_t=0=E[\int _0^t u dX_u]$ and $\sigma_t^2=E[(\int _0^t u dX_u)^2]=\int_0^t u^2du=t^3/3$ so the expectation of $Y_t$ is a strictly increasing function of time and it's not a martingale.