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

1 Answer

Nov 29, 2016

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

Explanation:

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#

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