How do you integrate $int (e^x-e^-x)/(e^x+e^-x)dx$ ?

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Answer 1 · 2016-11-29T09:32:32

$log(e^x+e^(-x))+C$

Note that $d/(dx)(e^x+e^(-x))=e^x-e^(-x)$ so

$(e^x-e^(-x))/(e^x+e^(-x))= d/(dx)log(e^x+e^(-x))$ and finally

$int (e^x-e^-x)/(e^x+e^-x)dx = int d/(dx)log(e^x+e^(-x))dx=log(e^x+e^(-x))+C$

How do you integrate int (e^x-e^-x)/(e^x+e^-x)dxint (e^x-e^-x)/(e^x+e^-x)dx?