How do integrate this #int (e^x + e^-x)/(e^x - e^-x)dx# without hyperbolic indentities, preferebly with subsititution?

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Answer 1 · 2018-06-05T23:55:59

#=ln(e^x-e^(-x))+C#

#int(e^x+e^-x)/(e^x-e^(-x))dx#

Substitute:

#u=e^x-e^(-x)#

From this it will follow that:

#du=e^x+e^(-x)dx#

So #du# is just the top of the fraction, now substituting in:

#int(e^x+e^-x)/(e^x-e^(-x))dx=int(du)/u#

#=ln(u)+C#

Reverse the substitution:

#=ln(e^x-e^(-x))+C#