How to compute $\int^{x} _{1/x} \frac{e^{-(t+\frac{1}{t})}}{t}dt$ 
How to compute $$I = \int^{x} _{1/x} \frac{e^{-(t+\frac{1}{t})}}{t}dt\ ?$$

My attempt:
Let $t=1/y$ so $dt=\frac{-1}{y^2}dy$
Now just substitute so we get :(Limits will be also changed )
$I = \int^{1/x} _{x} e^{-(\frac{1}{y} + y)}y(\frac{-1}{y^2}dy)$
So regardless of variable we get the same form again
$I = -\int^{1/x} _{x} \frac{e^{-(y+\frac{1}{y})}}{y}dy$
Removing the negative sign by changing the limits we get : $I = \int^{x} _{1/x} \frac{e^{-(y+\frac{1}{y})}}{y}dy$
I am stuck here How to find the integration? Hint are too appreciated.
Thanks in advance.
 A: Assume $x>1$.
By splitting the integration range, then applying the substitution $t+\frac{1}{t}=z$, we get:
$$I=\int_{1/x}^{x}\exp\left(-t-\frac{1}{t}\right)\frac{dt}{t}=2\int_{1}^{x}\exp\left(-t-\frac{1}{t}\right)\frac{dt}{t}=2\int_{2}^{x+\frac{1}{x}}\frac{e^{-z}}{\sqrt{z^2-4}}\,dz $$
or, applying successively the substitutions $z=2u$ and $u=\cosh\theta$:
$$ I = 2\int_{1}^{\frac{1}{2}\left(x+\frac{1}{x}\right)}\frac{e^{-2u}}{\sqrt{u^2-1}}\,du = 2\int_{0}^{\log x}\exp\left(-2\cosh \theta\right)\,d\theta $$
that is a non-elementary integral related with Bessel functions. For instance:
$$\int_{2}^{+\infty}\frac{e^{-z}}{\sqrt{z^2-4}}\,dz = e^2\,K_0(2).$$
However, $\exp(-2\cosh\theta)$ behaves like $\exp\left(-2-\theta^2\right)$, hence it is not difficult to approximate such integral with an error function or through the Cauchy-Schwarz inequality:
$$ \int_{x+\frac{1}{x}}^{+\infty}\frac{e^{-z}}{\sqrt{z^2-4}}\,dz\leq \sqrt{\frac{1}{2} e^{-2x-2/x}\int_{x+\frac{1}{x}}^{+\infty}\frac{dz}{z^2-4}}$$
