= Solution
The stopped representation from part (a)(ii) and the <Itô isometry> give the exact second-moment identity
$$
\boxed{\mathbb E[M_{t\wedge T}^2]
=1+\mathbb E\int_0^{t\wedge T}\frac{e^{2X_s}}{(1+X_s^2)^2}\,ds.}
$$
The cross term with the initial value is zero because the stopped <stochastic integral> has zero mean. This is the useful finite-time computation; evaluating the remaining occupation integral explicitly is unnecessary for proving integrability of $T$.
For $|x|\leq L$,
$$
\frac{e^{2x}}{(1+x^2)^2}\geq c_L,
\qquad c_L=\frac{e^{-2L}}{(1+L^2)^2}>0.
$$
Combining the second-moment identity with part (i) yields
$$
1+c_L\mathbb E[t\wedge T]\leq\mathbb E[M_{t\wedge T}^2]\leq e^{2L}.
$$
Let $t\to\infty$ and use <monotone convergence>:
$$
\boxed{\mathbb ET\leq e^{2L}(1+L^2)^2(e^{2L}-1)<\infty.}
$$
In particular $T<\infty$ <almost surely>, a conclusion obtained without assuming finite exit in the stopping argument.
One can also evaluate the limiting second moment exactly. At exit $M_T$ is either $e^{-L}$ or $e^L$, so
$$
M_T^2=(e^L+e^{-L})M_T-1.
$$
By <bounded convergence theorem> and part (i), $\mathbb EM_T=1$. Therefore
$$
\boxed{\lim_{t\to\infty}\mathbb E[M_{t\wedge T}^2]
=\mathbb EM_T^2=e^L+e^{-L}-1.}
$$
The same bound also holds at every finite time, by applying $(y-e^{-L})(y-e^L)\leq0$ to $y=M_{t\wedge T}$ and taking <expected values>. It improves the preceding bound to $\mathbb ET\leq c_L^{-1}(e^L+e^{-L}-2)$ if desired.
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