I'm trying to solve a problem similar to Stochastic Integral. I have to evaluate $$ \mathbb{Var}\left(\int_{0}^t ((B_s)^2 + s) \mathrm{d}B_s \right)$$
I have split the problem in two parts:
1) $ \mathbb{Var}\left(\int_{0}^t (B_s)^2 \mathrm{d}B_s \right) = \mathbb{E}(\left(\int_{0}^t (B_s)^2 \mathrm{d}B_s \right)^2) - \mathbb{E}^2\left(\int_{0}^t (B_s)^2 \mathrm{d}B_s \right) = \left(\int_{0}^t \mathbb{E}(B_s)^4 \mathrm{d}s \right) - 0 = \left(\int_{0}^t 3t^2 \mathrm{d}s \right)= t^3$
Is this right?
2) $ \mathbb{Var}\left(\int_{0}^t s \mathrm{d}B_s \right) = \mathbb{E}(\left(\int_{0}^t s \mathrm{d}B_s \right)^2) - \mathbb{E}^2\left(\int_{0}^t s \mathrm{d}B_s \right) =\left(\int_{0}^t \mathbb{E}(s^2) \mathrm{d}s \right) - 0 = \left(\int_{0}^t s^2 \mathrm{d}s \right) = t^3$
And is this right?
So $$ \mathbb{Var}\left(\int_{0}^t ((B_s)^2 + s) \mathrm{d}B_s \right) = t^3 + t^3 = 2t^3$$
Hint Recall that for any progressively measurable function $f$
$$M_t := \int_0^t f(s) \, dB_s, \qquad t \leq T$$
is a martingale whenever
$$\mathbb{E} \left( \int_0^T |f(s)|^2 \, ds \right)<\infty$$
In particular, $$\mathbb{E}(M_t) = 0. \tag{1}$$ Moreover, Itô's isometry states
$$\mathbb{E} \left[ \left( \int_0^t f(s) \, dB_s \right)^2 \right] = \mathbb{E} \left( \int_0^t |f(s)|^2 \, ds \right) \tag{2}$$
Using $(1)$, $(2)$ for the function $f(s) := s+B_s^2$ allows you to compute the variance you are looking for.