Let $W_t$ be a standard one dimension Brownian Motion with $W_0=0$ and $X_t=\int_0^t{W_sds}$. With the help of ito formula, we could get $E[(X_t)^2]=\frac{1}{3}t^3$ $E[(X_t)^3]=0$
When I try to employ the same method to calculate the general case $E[(X_t)^n]$, I got stuck. I guess $X_t$ should be normal distribution since it could be the limit of the following $\lim_{n\rightarrow \infty}{\sum_{i=0}^{n-1}{W_{t_i}(t_{i+1}-t_i)}},$
where $ W_{t_i}\sim norm(0,\sqrt{\frac{t_i}{n}}).$
If it is true, the problem would be trivial.
Update: Thanks for all the suggestions. Now I believe $X_t$ is a Gaussian process.
How about for this integral $Y_t=\int_0^t{f(W_s)ds}$ if we assume that $f$ is some good function, say polynomial or exponential, i.e $Y_t=\int_0^t{e^{W_s}ds}$ $Y_t=\int_0^t{[a_n(W_s)^n+a_{n-1}(W_s)^{n-1}+...+a_0]ds}$