Doob-Meyer Decomposition for $(W_{t}^{2}-t)^{2}$ Given a stochastic process $M=(M_{t})_{t\geq0} ,M_{t}=W_{t}^{2}-t,(W_{t}$ is Βrownian Motion).Find the Doob-Meyer decomposition of the $M^{2}$.
Attemp:
I firstly proved that $M^{2}$ is a submartingale
for $s\leq t$
$\mathbb{E}[M_{t}^{2}|\mathcal{F}_{s}]\overset{Jensen's\,inequality}{\geq}\left(\mathbb{E}[M_{t}|\mathcal{F}_{s}]\right)^{2}=M_{s}^{2}$  because $f(X)=X^{2}$  is a curve function and $W_{t}^{2}-t$  is a Martingale
Since $M^{2}$ is a submartingale there is a unique Doob-Meyer decomposition 
$M_{t}^{2}=X_{t}+A_{t}$
$X_{t}$ must be a Martingale and $A_{t}$ an increasing predictable process
We already Know that $W_{t}^{2}-t$ is a Martingale and I tried to use this for my proof but I didn't find the solution.
 A: Using the It$\hat{o}$  formula for the function $f(X_{t},t)=(W_{t}^{2}-t)^{2}$ we have: 
$f(x,t)=(x^{2}-t)^{2}$
$f_{x}(x,t)=4x(x^{2}-t)$
$f_{xx}(x,t)=4(3x^{2}-t)$
$f_{t}(x,t)=-2(x^{2}-t)$
$f(t,W_{t})=f(0,W_{0})+\int_{0}^{t}f_{t}(u,W_{u})du+\int_{0}^{t}f_{x}(u,W_{u})dW_{u}+\frac{1}{2}\int_{0}^{t}f_{xx}(u,W_{u})d[W]_{u}$
$f(t,W_{t})=0+\int_{0}^{t}-2(W_{u}^{2}-u)du+\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}+\frac{1}{2}\int_{0}^{t}4W_{u}(3W_{u}^{2}-u)d[W]_{u}$
but $[W]_{u}=u$  so 
$f(t,W_{t})=\int_{0}^{t}-2(W_{u}^{2}-u)du+\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}+\frac{1}{2}\int_{0}^{t}4(3W_{u}^{2}-u)du$
$f(t,W_{t})=\int_{0}^{t}\left(-2(W_{u}^{2}-u)+2W_{u}(3W_{u}^{2}-u)\right)du+\int_{0}^{t}2(W_{u}^{2}-u)dW_{u}$
$f(t,W_{t})=M_{t}^{2}=\int_{0}^{t}4W_{u}^{2}du+\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}$
But the process $\int_{0}^{t}4W_{u}^{2}du$ is increasing.In addition is predictable.
$\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}$ is a Martingale because for  $s\leq t$
$\mathbb{E}\left[\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]=\mathbb{E}\left[\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]+\mathbb{E}\left[\int_{s}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]$
$\mathbb{E}\left[\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]=\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}$because $\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}$ $\mathcal{F}_{s}$-measurable
$ dW_{u}=W_{u+du}-W_{u} , W_{u+du}-W_{u}\sim N\left(0,du\right)$ and 
$4W_{u}(W_{u}^{2}-u)$ independent of $W_{u+du}-W_{u} $
$\mathbb{E}\left[\int_{s}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]=\int_{s}^{t}\left(\mathbb{E}\left[4W_{u}(W_{u}^{2}-u)\right]\mathbb{E}\left[dW_{u}\right]\right)|\mathcal{F}_{s}=0$
so $\mathbb{E}\left[\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]=\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}$ is Martingale 
