The value of ${\int }_{\frac{1}{2}}^2 e^{\left|x-\frac{1}{x}\right|}dx$  is equal to [Note: e denotes napier’s constant] 

Let ${\int }_{\frac{1}{2}}^2 e^{\left|x-\frac{1}{x}\right|}dx$…..(1)
Put x=1/y $2I={\int }_{1/2}^2{e}^{\left|x-\frac{{\displaystyle 1}}{{\displaystyle x}}\right|}\left(1+\frac{{\displaystyle 1}}{{\displaystyle x^2}}\right)dx$

${\int }_{1/2}^2 e^{-\left(x-\frac{1}{x}\right)}\left(1+\frac{1}{x^2}\right)dx+{\int }_1^2 e^{\left(x-\frac{1}{x}\right)}\left(1+\frac{1}{x^2}\right)dx\text{ put }\left(x-\frac{1}{x}\right)=t$
And adding,we get