Answer
$x+\dfrac{1}{x}$
Work Step by Step
Use the hyperbolic function formula as follows: $ \cosh x= \dfrac{e^x+e^{-x}}{2}$
Need to solve. $2 \cosh ( \ln x)$
$2 \cosh ( \ln x)=2[ \dfrac{e^{\ln x}+e^{-\ln x}}{2}] \\ =e^{\ln x}+e^{(-\ln x)} \\=e^{\ln x}+e^{(\ln x^{-1})} \\
=x+x^{(-1)} \\=x+\dfrac{1}{x}$
