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If we consider a process $$X_T=e^{\int_0^TtdW_t}$$it holds that it can be expressed by $$X_T=\mathbb{E}[X_T]+\int_0^Th(t)dW_t$$but how do we derive this $h(t)$?

I calculated $\mathbb{E}[X_T]=e^{t^3/6}$, and defined a new process $$Z_T=e^{-T^3/6}X_T \qquad Z_0=1$$with the hope that I could then use the martingale representation theorem, but $Z_T$ is not a martingale, since by Itô we get $$dZ_T=-\frac{1}{2}T^2e^{-T^3/6}X_TdT+Te^{-T^3/6}dW_T$$which contains a drift term. How could we solve this?

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Your formula is wrong you wrote $TdW_T$ which is $dlog(X_T)$, not $dX_T$

We have $$dZ_T=-\frac{1}{2}T^2e^{-T^3/6}X_TdT+e^{-T^3/6}dX_T$$

Do that instead :

$$X_T=e^{Y_T}$$ with $$dY_T=TdW_T$$

Apply Ito's lemma on $X$

We have $$dX_T=e^{Y_T}dY_T+\frac{1}{2}e^{Y_T}d=X_TdY_T+\frac{1}{2}X_Td$$

Because $$d=T^2dT$$

We have $$dX_T=X_TdY_T+\frac{1}{2}X_TT^2dT$$

Using the first equation , $$dZ_T=-\frac{1}{2}T^2e^{-T^3/6}X_TdT+e^{-T^3/6}(X_TdY_T+\frac{1}{2}X_TT^2dT)=e^{-T^3/6}X_TdY_T=TZ_TdW_T$$

which is what you want

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    You're right, what i meant to write was not $Te^{-T^3/6}dW_t$ but $Te^{-T^3/6}X_TdW_t$ since $de^{f(t)}=e^{f(t)}df(t)$, but this evidently does not hold when $f_t$ is a stochastic integral. Thanks! – 2017-01-25