Why this integral $\int_0^te^{41s^2+40s}ds$ is bounded by $\frac{e^{41t^2+40t}}{t}$?
This is a bound without proof I find in a published paper. 
Let $g(t)= e^{41t^2+40t}n^{-\frac{1}{6}}$ when $t\le 1$, and $g(t)=\frac{e^{41t^2+40t}}{t}n^{-\frac{1}{6}}$ when $t>1$. $a\pm b$ means $[a-b,a+b]$. How to show $\pm n^{\frac{11}{6}}\int_0^te^{41s^2+40s}ds\pm n^{\frac{5}{3}}\subseteq \pm n^2g(t)$ when $0\le t\le \sqrt{\log n}$? And since this an asymptotic bound, we may assume $n$ is large enough. 
For $t\le 1$ part, it seems we can use $n^{\frac{11}{6}}\int_0^te^{41s^2+40s}ds\le n^{\frac{11}{6}}e^{41t^2+40t}\cdot1=n^2g(t)$. But I have no idea for larger $t$.
