## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{ x^2- x^{-2}}{2}$
Use formula: $\sinh x= \dfrac{e^x-e^{-x}}{2}$ Thus, $\sinh ( 2 \ln x)= \dfrac{e^{2 \ln x}-e^{- 2 \ln x}}{2}=\dfrac{e^{\ln x^2}-e^{\ln x^{-2}}}{2}$ Hence, $\sinh ( 2 \ln x)=\dfrac{ x^2- x^{-2}}{2}$