$X_t = e^{B_t-\frac{1}{2}t^2}$
I need to find $[X]_t$, the quadratic variation process. I have tried to solve the problem and my main question is whether this approach is correct or not.
Given that
$$ \begin{equation} \begin{split} dX_t=&-tX_tdt+X_tdB_t+\frac{1}{2}X_td[B]_t \\ =&\left(\frac{1}{2}-t\right)X_tdt+X_tdB_t \end{split} \end{equation} $$.
$[X]$ is the unique process such that $X^2-[X]$ is a martingale.
$d( X^2_t)=2X_tdX_t+d[X]_t$
How do I compute $d[X]_t$ more rigorously than saying $dX_tdX_t=X_t^2dt$ because $dt\cdot dt=0$, $dt\cdot dB_t=0$, $dB_t\cdot dB_t=t$?
Now integrate from $0$ to $t$ and substitute in $dX_s$ (above)
$$ \begin{equation} \begin{split} X^2_t-X^2_0=&2\int_0^tX_sdX_s+\int_0^tX_s^2ds \\ =&\int_0^t X_s^2 \left(1-2s\right)ds+2\int_0^t X_s^2dB_s + \int_0^t X^2_s ds\\ &=2\int_0^t X_s^2 \left(1-s\right)ds + 2\int_0^t X_s^2 dB_s \end{split} \end{equation} $$
So it would seem that since the stochastic integral $2 \int_0^t X^2_s dB_s$ is a martingale, by uniqueness and since $X_0^2 = 1$, we have that $[X]_t=1+2\int_0^t X_s^2\left( 1-s \right) ds.$ Is this correct?