How do you integrate #int (e^x-e^-x)/(e^x+e^-x)dx#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Cesareo R. 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# Answer link Related questions How do you evaluate the integral #inte^(4x) dx#? How do you evaluate the integral #inte^(-x) dx#? How do you evaluate the integral #int3^(x) dx#? How do you evaluate the integral #int3e^(x)-5e^(2x) dx#? How do you evaluate the integral #int10^(-x) dx#? What is the integral of #e^(x^3)#? What is the integral of #e^(0.5x)#? What is the integral of #e^(2x)#? What is the integral of #e^(7x)#? What is the integral of #2e^(2x)#? See all questions in Integrals of Exponential Functions Impact of this question 37280 views around the world You can reuse this answer Creative Commons License