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I want to find a process $h$ such that $$m(T) = Em(T) + \int_0^T h(t) dW(t), $$

where $m(T) = e^{ \int_0^T t dW(t)}$. Here, $T$ is some positive constant, and $W(t)$ is Brownian motion.

I get $Em(T) = \frac{1}{6}T^3$. But then I am unsure how to continue.

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    Is the exponent in the expresion for $m(T)$ to be $\int_0^T T\,dW(t)$ or $\int_0^T t\,dW(t)$? – 2017-01-07
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    See my edit. I've also got an idea which I am adding shortly, but I am unsure if it'll work out. – 2017-01-07

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Hint: Use Ito's formula to expand the martingale $e^{-t^3/6}m(t)$, $0\le t\le T$. Having done this set $t=T$ and multiply through by $e^{T^3/6}$.

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    Thanks. Any comment on why you know that this function is a martingale beforehand? – 2017-01-11
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    It has the "exponential martingale" form $\exp(Y_t-\langle Y\rangle_t/2)$, where $Y_t=\int_0^t s\,dW(s)$ is a martingale with quadratic variation $\langle Y\rangle_t=\int_0^t s^2\,ds$. – 2017-01-11