How do you evaluate the integral #int e^x/(1+e^(2x))dx# from #-oo# to #oo#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Cesareo R. Aug 22, 2016 #pi# Explanation: Making #y = e^x# we have #dy = e^x dx = y dx# #int e^x/(1+e^(2x))dx equiv int dy/(1+y^2) = arctan(y)# but #lim_{x->oo}e^x=lim_{y->oo}=oo# so #lim_{a->oo} int_{-a}^a dy/(1+y^2) = pi/2-(-pi/2) = pi# 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 11939 views around the world You can reuse this answer Creative Commons License