Answer
The integral $\int_{1}^{\infty} e^{-x} dx$ converges to $e^{-1}$. Therefore, by the comparison test, the integral $\int_{1}^{\infty} e^{-(x+x^{-1})} \ dx$ converges as well.
Work Step by Step
We are given the function
$f(x)=\int_{1}^{\infty} e^{-(x+x^{-1})} \ dx$
Since, $\dfrac{1}{x}\geq 0$ and $x+\dfrac{1}{x} \geq x$
This yields:
$e^{-(x+x^{-1})} \leq e^{-x} $
Consider the integral
$\int_{1}^{\infty} e^{-x} dx=\lim\limits_{R \to \infty} \int_{1}^{R} e^{-x} dx\\=\lim\limits_{R \to \infty}-e^{-R}+e^{-1}\\=e^{-1}$
Thus, the integral $\int_{1}^{\infty} e^{-x} dx$ converges to $e^{-1}$. Therefore, by the comparison test, the integral $\int_{1}^{\infty} e^{-(x+x^{-1})} \ dx$ converges as well.
