Answer
$\ln|e^x-e^{-x}|+C$
Work Step by Step
$\int \frac{e^x+e^{-x}}{e^x-e^{-x}}dx$
let $e^x-e^{-x}=u$
$(e^x+e^{-x})dx=du$
$\int \frac{e^x+e^{-x}}{e^x-e^{-x}}dx$
$=\int \frac{1}{u}du$
$=\ln |u|+C$
$=\ln|e^x-e^{-x}|+C$
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