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 $A$. So, \begin{equation*} EX_1\,1_A\le\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u E(\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}\,dt\Big)\,P(A). \tag{2} \end{equation*} Because the distribution of $B_t^u$ is symmetric and $(a+b) e^{(a+b)^2}+(a-b)e^{(a-b)^2}$ is convex in real $b$ for each real $a\ge0$, we can use Jensen's inequality to get \begin{align*} &2E(a+B_t^u) \exp\{(a+B_t^u)^2\} \\ & =E[(a+B_t^u) \exp\{(a+B_t^u)^2\}+(a-B_t^u) \exp\{(a-B_t^u)^2\}] \\ & \ge2ae^{a^2} \end{align*} for $a\ge0$. So, by (2), \begin{align*} EX_1\,1_A&\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(A) \\ &=\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\frac u{2b} \, (e^{b^2}-1)\Big)\,P(A) \\ &=-e^{b^2(1+o(1))}\,P(A). \tag{3} \end{align*} On the other hand, \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&\le-e^{b^2(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.