Let $B_t:=B(t)$. By the Itô formula
$$f(B_1)-f(B_0)=\int_0^1 f'(B_t)\,dB_t+\frac12\,\int_0^1 f''(B_t)\,dt$$
with $f(b):=\int_0^b e^{a^2}da$ (with $\int_0^b:=-\int_b^0$ for $b<0$), we have
\begin{equation*}
X_1=\int_0^1 f'(B_t)\,dB_t
=f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt
=f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt. \tag{1}
\end{equation*}
From here, it is not hard to see that $EX_1$ does not exist.

Indeed, consider the event
\begin{equation*}
A:=\{B_u\in[b,b+1/b],B_1-B_u<-2/b,M_u>-b\},
\end{equation*}
where $b\to\infty$,
\begin{equation*}
u:=1-1/b^2,
\end{equation*}
\begin{equation*}
M_u:=\min_{t\in[u,1]}(B_t-B_u).
\end{equation*}
Note that on the event $A$ we have $B_u\ge b$, $B_1<b-1/b$ and $B_t>0$ for all $t\in[u,1]$. Therefore and because (by the l'Hospital rule) $f(b)\sim e^{b^2}/(2b)$, it follows from (1) that on $A$
\begin{align*}
X_1&\le\frac{e^{(b-1/b)^2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\
&=\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\
&\le\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u (\tfrac tu\,b+B_t^u)
\exp\{(\tfrac tu\,b+B_t^u)^2\}\,dt,
\end{align*}
where $B_t^u:=B_t-\frac tu\,B_u$; for the latter, displayed inequality, we recall that $B_u\ge b$ on $A$ and use the fact that $se^{s^2}$ is increasing in real $s$.

The Brownian bridge $(B_t^u)_{t\in[0,u]}$ is a zero-mean (Gaussian) process independent of $(B_u,B_1-B_u,M_u)$; so, $(B_t^u)_{t\in[0,u]}$ is independent of the event $A$.
Introduce also the event
$$C_x:=\{\max_{t\in[0,u]}|B_t^u|\le x\}$$
for real $x>0$, which allows us to use the Fubini theorem to get
\begin{align*}
EX_1\,1_{A\cap C_x}\le\Big(&\frac{e^{b^2-2}}{(2+o(1))b}P(C_x) \\
&-\int_0^u E(\tfrac tu\,b+B_t^u)
\exp\{(\tfrac tu\,b+B_t^u)^2\}1_{C_x}\,dt\Big)\,P(A). \tag{2}
\end{align*}
Because $g_a(b):=(a+b) e^{(a+b)^2}+(a-b)e^{(a-b)^2}$ is convex and even in real $b$ for each real $a\ge0$, we have $g_a(b)\ge g_a(0)=2ae^{a^2}$. Therefore and
because the distribution of the Brownian bridge $(B_t^u)_{t\in[0,u]}$ is symmetric,

\begin{align*}
E(a+B_t^u) \exp\{(a+B_t^u)^2\}1_{C_x} =\tfrac12\,Eg_a(B_t^u)\,1_{C_x} \ge ae^{a^2}\,P(C_x)
\end{align*}
for $a\ge0$. So, by (2),
\begin{align*}
EX_1\,1_{A\cap C_x}&\le\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u \tfrac tu\,b
\exp\{(\tfrac tu\,b)^2\}\,dt\Big)\,P(C_x)P(A) \\
&=\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\frac u{2b} \, (e^{b^2}-1)\Big)\,P(C_x)P(A) \\
&=-e^{b^2(1+o(1))}\,P(C_x)P(A).
\end{align*}

On the other hand, letting now $x\to\infty$, we have $P(C_x)\to1$
and, still with $b\to\infty$,
\begin{align*}
P(A)&=P(B_u\in[b,b+1/b])P(B_1-B_u<-2/b,M_u>-b) \\
&\ge P(B_u\in[b,b+1/b])[P(B_1-B_u<-2/b)-P(M_u\le-b)] \\
&=e^{-b^2/(2+o(1))}[P(B_1<-2)-o(1)]=e^{-b^2/(2+o(1))}.
\end{align*}
Thus,
\begin{align*}
EX_1\,1_{A\cap C_x}&\le-e^{b^2(1+o(1))}\,(1-o(1))\,e^{-b^2/(2+o(1))}=-e^{b^2/(2+o(1))}\to-\infty,
\end{align*}
which shows that indeed $EX_1$ does not exist.