I'm self-studying Bruijn (1961)'s Asymptotic Methods in Analysis. Below is the first exercise of Chapter 1.
Show that $$ \int_1^x(1+t^{-1})^tdt=ex-\frac{1}{2}e\log x+O(1)\quad(x>1). $$
The given hint is: first, show that $e^{-1}(1+t^{-1})^t=1-\frac{1}{2}t^{-1}+O(t^{-2}),\quad (t\geq 1)$.
Attempt: the book gives a rule for term-by-term integration of an asymptotic series but only for nonnegative powers. But I do that anyway for negative powers: the hint gives $$ \int_1^x(1+t^{-1})^tdt=e\int_1^x1-\frac{1}{2}t^{-1}+O(t^{-2})dt=e\left(x-1-\frac{\log(x)}{2}+O(1-x^{-1})\right) $$ which does simplify to the RHS of the claim. So if I can show the hint, then I can arrive at the claim in a heuristic manner.
But what is a proper solution to this problem please?
Rewrite the integral as
$$\int_1^x dt \, \exp{\left [t \ln{\left ( 1+\frac1{t} \right )}\right ]} $$
Because $t \gt 1$ we can Taylor expand the log term to get an approximate expression for the integral:
$$\int_1^x dt \, e^{\displaystyle 1-\frac1{2 t} + \frac1{3 t^2}-\cdots} = e \int_1^x dt \, e^{\displaystyle -\frac1{2 t} + \frac1{3 t^2}-\cdots} $$
Again, because of the integration region we may Taylor expand the integrand and get that the integral is approximately
$$ e (x-1) - \frac{e}{2} \ln{x} + \frac{11 e}{24} \frac1{x} + O \left (\frac1{x^2} \right ) $$