Equivalent of $\int_1^x e^{-\sqrt{\ln(t)}} \mathrm dt$ when $x \rightarrow \infty$

How can I prove that:

$$\int_1^x e^{-\sqrt{\ln(t)}} \mathrm dt \sim_{x \rightarrow \infty} xe^{-\sqrt{\ln(x)}}$$

without using l'Hôpital's rule ?

Integration by parts:

$$\int_1^x e^{-\sqrt{\ln(t)}} \mathrm dt= xe^{-\sqrt{\ln(x)}}-1+\frac{1}{2}\int_1^x \frac{e^{-\sqrt{\ln(t)}}}{\sqrt{\ln(t)}} \mathrm dt=xe^{-\sqrt{\ln(x)}}+o(xe^{-\sqrt{\ln(x)}})+\frac{1}{2}\int_1^x \frac{e^{-\sqrt{\ln(t)}}}{\sqrt{\ln(t)}} \mathrm dt$$

So how can I show that $$\int_1^x \frac{e^{-\sqrt{\ln(t)}}}{\sqrt{\ln(t)}} \mathrm dt =o(xe^{-\sqrt{\ln(x)}})$$ ?

-
Using the change of variable $t=\mathrm e^{s^2}$, one sees that one must show that $I(u)\ll K(u)$ when $u\to+\infty$, with $$K(u)=\mathrm e^{u^2-u},\qquad I(u)=\int_0^uK(s)\mathrm ds.$$ Assume that $u\gt1$ and pick $v$ in $(1,u)$. Since $K\leqslant K(v)$ on $(0,v)$ and $K\leqslant K(u)$ on $(v,u)$, $$I(u)\leqslant vK(v)+(u-v)K(u)\leqslant uK(v)+(u-v)K(u).$$ Assume that $v=u-w$ with $w\to0$ when $u\to+\infty$, then $(u-v)K(u)=wK(u)\ll K(u)$ and $$K(v)=\mathrm e^{u^2-2uw+w^2-u+w}\leqslant\mathrm e^{u^2-u-uw}=K(u)\mathrm e^{-uw},$$ for every $u$ large enough. Hence $uK(v)\leqslant u\mathrm e^{-uw}K(u)\ll K(u)$ for every choice of $w$ such that $w=o(1)$ and $u=o(\mathrm e^{uw})$, for example $w=1/\sqrt{u}$. QED.