## Thomas' Calculus 13th Edition

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