How do you integrate #int (e^x+e^-x)/(e^x-e^-x)dx#?
1 Answer
Feb 20, 2017
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
#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!