Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 354: 104

$\frac{2}{(e^x+e^{-x})}+C$

$\int \frac{2e^x-2e^{-x}}{(e^x+e^{-x})^2}dx$ let $(e^x+e^{-x})=u$ $(e^x-e^{-x})dx=du$ $\int \frac{2e^x-2e^{-x}}{(e^x+e^{-x})^2}dx$ $=2 \int \frac{1}{u^2}du$ $=2u^{-1}+C$ $=\frac{2}{(e^x+e^{-x})}+C$