Answer
$x+\dfrac{1}{x}$
Work Step by Step
Use formula: $ \cosh x= \dfrac{e^x+e^{-x}}{2}$
Thus, $2 \cosh ( \ln x)=2[ \dfrac{e^{\ln x}+e^{-\ln x}}{2}]=e^{\ln x}+e^{-\ln x}=e^{\ln x}+e^{\ln x^{-1}}$
Hence, $2 \cosh ( \ln x)=x+x^{-1}=x+\dfrac{1}{x}$
