Is $ X(t) = t^2 B(t) - 2 \int_0^t sB(s) ds$ a martingale? This is part of my homework assignment, so I don't expect a complete solution.
$B(t)$ is standard Brownian motion and $(\mathcal{F}_s^B)$ is the natural filtration of $B(t)$.
I've tried to show that $\mathbb{E}(X(t) | \mathcal{F}_s^B) = X(s)$.
So far I have
\begin{align*}
    \mathbb{E} \big( X(t) | \mathcal{F}_s^B \big) &=
    \mathbb{E} \Big( t^2 B(t) - 2 \int_0^t u B(u) du \Big| \mathcal{F}_s^B \Big) = \\ &=
    \mathbb{E} \Big( \big( t^2 B(t) - s^2 B(s) \big) + s^2 B(s) -2 \int_0^s u B(u) du - 2 \int_s^t u B(u) du \Big| \mathcal{F}_s^B \Big) = \\ &=
    \mathbb{E} \Big( t^2 B(t) - s^2 B(s) - 2 \int_s^t u B(u) du \Big| \mathcal{F}_s^B \Big) + \mathbb{E} \Big( s^2 B(s) - 2 \int_0^s uB(u) du \Big| \mathcal{F}_s^B \Big) = \\ &=
    \mathbb{E} \Big( t^2 B(t) - s^2 B(s) - 2 \int_s^t u B(u) du \Big| \mathcal{F}_s^B \Big) + X(s) = \\ &=
    \mathbb{E} \Big( t^2 \big( B(t) - B(s) \big) + t^2 B(s) - s^2 B(s) \big| \mathcal{F}_s^B \Big) + \mathbb{E} \Big( -2 \int_s^t uB(u) du \Big| \mathcal{F}_s^B \Big) + X(s) = \\ &=
    t^2 \mathbb{E} \big( B(t) - B(s) \big| \mathcal{F}_s^B \big) + B(s) \mathbb{E}( t^2 - s^2 | \mathcal{F}_s^B) + \mathbb{E} \Big( -2 \int_s^t uB(u) du \Big| \mathcal{F}_s^B \Big) + X(s) = \\ &=
    (t^2 - s^2) B(s) + \mathbb{E} \Big( -2 \int_s^t uB(u) du \Big| \mathcal{F}_s^B \Big) + X(s)
\end{align*}
Now, I don't know if this is the 'correct' way to show this or if I made any mistakes. So please tell me if I did.
What I don't know is how to compute the remaining expectation.
 A: By Time-depdendent Ito:
$$\mathrm{d}f\left(t,B_{t}\right)=f_{t}\left(t,B_{t}\right)\mathrm{d}t+f_{x}\left(t,B_{t}\right)\mathrm{d}B_{t}+\frac{1}{2}f_{xx}\left(t,B_{t}\right)\mathrm{d}t$$
We have:
$$\mathrm{d}t^{2}B_{t}=2tB_{t}\mathrm{d}t+t^{2}\mathrm{d}B_{t}$$
Thus:
$$t^{2}B_{t}-2\int_{0}^{t}sB_{s}\mathrm{d}s=\mathrm{\int_{0}^{t}}t^{2}\mathrm{d}B_{t}+\int_{0}^{t}2sB_{s}\mathrm{d}s-\int_{0}^{t}2sB_{s}\mathrm{d}s=\mathrm{\int_{0}^{t}}t^{2}\mathrm{d}B_{t}$$
Which is a martingale, as the drift is 0.
Alternatively:
$$
X(t_{2})-X(t_{1})=\left(t_{2}^{2}B_{t_{2}}-2\int_{0}^{t_{2}}sB_{s}\mathrm{d}s\right)-\left(t_{1}^{2}B_{t_{1}}-2\int_{0}^{t_{1}}sB_{s}\mathrm{d}s\right)=
$$
$$
=t_{2}^{2}\left(B_{t_{2}}-B_{t_{1}}\right)+\left(t_{2}^{2}-t_{1}^{2}\right)B_{t_{1}}-2\int_{t_{1}}^{t_{2}}s\left(B_{s}-B_{t_{1}}\right)\mathrm{d}s-2B_{t_{1}}\underbrace{\int_{t_{1}}^{t_{2}}s}_{\frac{1}{2}\left(t_{2}^{2}-t_{1}^{2}\right)}\mathrm{d}s=
$$
$$
=t_{2}^{2}\left(B_{t_{2}}-B_{t_{1}}\right)-2\int_{t_{1}}^{t_{2}}s\left(B_{s}-B_{t_{1}}\right)\mathrm{d}s
$$
Now Fubini+etc to change integrals + Markov property:
$$
\mathbb{E}\left(X(t_{2})-X(t_{1})|\mathcal{F}_{t_{1}}\right)=\mathbb{E}\left(t_{2}^{2}\left(B_{t_{2}}-B_{t_{1}}\right)-2\int_{t_{1}}^{t_{2}}s\left(B_{s}-B_{t_{1}}\right)\mathrm{d}s|B_{t_{1}}\right)=0
$$
