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Question #19f25

1 Answer

Aug 7, 2016

ln|(e^x-e^-x)|+C#.

Explanation:

Let #e^x-e^-x=t rArr {e^x-(-e^-x)}dx=dt#.

# rArr (e^x+e^-x)dx=dt#

Hence, #I=int(e^x+e^-x)/(e^x-e^-x)dx=intdt/t=ln|t|=ln|(e^x-e^-x)|+C#.

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