Assuming that $W(t)$ is a Brownian motion, and considering two integrals
$$ X :=\int_{0}^{T} W(t) d W,\quad\text{and}\quad Y :=\int_{0}^{T}(W(t)+t)^{2} d W $$
I'm looking for the covariance $Cov(X, Y)$.
Here is my try:
$$ \operatorname{Cov}(X, Y)=E[(X(t)-E[X(t)]) \cdot (Y(t)-E[Y(t)])] $$
$$ X(t)=\int_{0}^{t} W(s) d W=\frac{1}{2} W^{2}(t)-\frac{1}{2} t $$
$$ E\left[X_{t}\right]=E\left[\frac{1}{2} W_{t}^{2}-\frac{1}{2} t\right]=E\left[\frac{1}{2} W_{t}^{2} \right]-\frac{1}{2}t=\frac{1}{2}E\left[ W_{t}^{2} \right]-\frac{1}{2}t = 0 $$
$$ \operatorname{Cov}(X, Y)=E[(\int_{0}^{T} W(t) d W) \cdot (\int_{0}^{T}(W(t)+t)^{2}-E[Y(t)])] $$
Here is what things get messy. First, Im struggling with finding expectation of Y(t). Second, multiplication of integrals get messy:
$$=E\left[\int_{0}^{T} W^{3}(t) d W+\int_{0}^{T} t^{2} W(t) d W+\int_{0}^{T} 2 t W^{2}(t) d W-(\int_{0}^{T} W(t) d W)\cdot E[Y(t)]\right]$$
Due to linearity of expectation, I can further write that:
$$=E\left[\int_{0}^{T} W^{3}(t) d W\right] + E\left[\int_{0}^{T} t^{2} W(t) d W\right] + E\left[\int_{0}^{T} 2 t W^{2}(t) d W\right] - E\left[(\int_{0}^{T} W(t) d W)\cdot E[Y(t)]\right]$$
$$=\int_{0}^{T}E\left[ W^{3}(t) \right]d W + \int_{0}^{T} E\left[t^{2} W(t)\right] d W + \int_{0}^{T} E\left[2 t W^{2}(t)\right] d W - E\left[(\int_{0}^{T} W(t) d W)\cdot E[Y(t)]\right]$$
Now I'm wondering if I can continue as $$=\int_{0}^{T}E\left[ W^{3}(t) \right]d W + \int_{0}^{T} E\left[t^{2} W(t)\right] d W + \int_{0}^{T} E\left[2 t W^{2}(t)\right] d W - E\left[(\int_{0}^{T} W(t) d W)\cdot E[Y(t)]\right]$$
I would appreciate your comments on how far this is correct and how to get things straight. Thanks.