I attempted to solve this on my own...please let me know if I am correct $\displaystyle \frac{e^x+e^{-x}}{e^x-e^{-x}} $ my answer: $\displaystyle e^x-e^{-x}+c $ Thank you.
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Hello, Note that the numerator is the derivative of the denominator. So you have something in the form u'/u. Thus the integral would give $\displaystyle \ln(e^x-e^{-x})+c$
The following is completely unnecessary, but true: $\displaystyle \int \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} \ dx = \int \coth{x} \ dx = \ln ({\sinh {x}}) + C = \ln \Big(\frac{e^{x}-e^{-x}}{2}\Big) + C = \ln (e^{x}-e^{-x}) + B$
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