Answer
$\frac{{{x^2} - 1}}{{2x}}$
Work Step by Step
$$\eqalign{
& \sinh \left( {\ln x} \right) \cr
& {\text{Use the definition of hyperbolic functions }} \cr
& \sinh t = \frac{{{e^t} - {e^{ - t}}}}{2},{\text{ Let }}t = \ln x \cr
& \sinh \left( {\ln x} \right) = \frac{{{e^{\ln x}} - {e^{ - \ln x}}}}{2} \cr
& {\text{Using logarithmic properties }}n\ln x = {x^{ - n}} \cr
& \sinh \left( {\ln x} \right) = \frac{{{e^{\ln x}} - {e^{\ln {x^{ - 1}}}}}}{2} \cr
& {\text{Where }}{e^{\ln t}} = t \cr
& \sinh \left( {\ln x} \right) = \frac{{x - 1/x}}{2} \cr
& {\text{Simplifying}} \cr
& \sinh \left( {\ln x} \right) = \frac{{\frac{{{x^2} - 1}}{x}}}{2} \cr
& \sinh \left( {\ln x} \right) = \frac{{{x^2} - 1}}{{2x}} \cr} $$
