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 $$X_1=\int_0^1 f'(B_t)\,dB_t =f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt =f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt.$$ From here, it should be not too hard to see that $EX_1$ does not exist.
One way to do this may be to recall that $B_1$ and the Brownian bridge $(Y_t):=(B_t-tB_1)$ are independent and write this for a real $c>0$: $$EX_1\,I\{Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1)\}\le P(Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1))\ Q,$$ $$Q:=EF(B_1),$$ $$F(b):=f(b)-\int_0^1 (tb+c\sqrt{(1-t)t})\, \exp((tb+c\sqrt{(1-t)t})^2)\,dt. $$ Then $P(Y_t>c\sqrt{(1-t)t}\ \ \forall t\in(0,1))$ should be $>0$ and hopefully $Q=-\infty$.
Here is the graph $\{(b,\ln(-F(b))/b^2)\colon1\le b\le10\}$ for $c=1$, which strongly suggests that $F(b)=-e^{b^2(1+o(1))}$ as $b\to\infty$, which would indeed imply $Q=-\infty$.
