#int(2e^x - 2e^-x) / (e^x+e^-x)^2 dx =# ? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Cesareo R. Jan 31, 2017 #-sechx+C# Explanation: #sinhx=(e^x-e^(-x))/2# and #coshx=(e^x+e^(-x))/2# then #int(2e^x - 2e^-x) / (e^x+e^-x)^2 dx = int tanhx/coshx dx = -sechx+C# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1504 views around the world You can reuse this answer Creative Commons License