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

1 Answer
Feb 20, 2017

Answer:

#int (e^x + e^-x)/(e^x - e^-x)dx = ln|sinhx| + C = ln|1/2(e^x -e^-x)| + C#

Explanation:

We know that

#•coshx = (e^x + e^-x)/2#
#•sinhx = (e^x- e^-x)/2#

This integral can be rewritten as

#int coshx/sinhx dx#, where #coshx# and #sinhx# represent the hyperbolic trigonometric functions

Now use a substitution to solve. Let #u = sinhx#. Just like with regular trigonometric functions, #du = coshx dx# and #dx= (du)/coshx#.

#int coshx/u * (du)/coshx #

#int 1/u du#

#ln|u| + C#

#ln|sinhx| + C#

If you wish, the answer can be written as

#ln|1/2(e^x - e^-x)| + C#

Hopefully this helps!