Given some probability space, assume $X_t$ is a square integrable continuous process adapted to the filtration $\mathcal{F}_{t}$ generated by the standard Brownian process $B_t$. I denote by $(X.B)_t=\int_0^t X_sdB_s$ the Ito integral. Then it is not very hard to show that $E((X.B)_t) = 0 $ and $E((X.B)^2_t) = E(\int_0^t X^2_s ds)$. I want to compute $E((X.B)_t\int_0^tX_sds)$.
For each $n$ choosing a partition $[t^n_i, t^n_{i+1}]$ going to $0$. Then I define $$ (X.B)^n_t = \sum_{i=0}^{K_n} X_{t^n_i}(B_{t^n_{i+1}}-B_{t^n_i}) $$ which converges in $L^2$ norm to $(X.B)^n_t$ as the mesh size goes to $0$. Similiarly $(X)^n_t = \sum_{i=0}^{K_n} X_{t^n_i}({t^n_{i+1}}-{t^n_i})$ converges to $\int_0^tX_sds$. So $E((X.B)^n_t(X)^n_t)$ converges to $E((X.B)_t\int_0^tX_sds)$ (Actually I am not %100 percent sure of this when the random variables are not independent). To compute the terms $E((X.B)^n_t(X)^n_t)$, we write $$ E((X.B)^n_t(X)^n_t)=\sum_{i,j=0}^{K_n} E(X_{t^n_j}X_{t^n_i}(B_{t^n_{i+1}}-B_{t^n_i})({t^n_{j+1}}-{t^n_j})) $$ One can see that conditioning with respect to $\mathcal{F}_{t^n_i}$ if $t^n_{j}<t^n_i$ then the terms are zero so the sum reduces to only terms with $j>i$. But I am not sure what to after here. Is it possible to anything more without knowing anything about $X_t$.