How do you integrate #int (5-e^x)/(e^(2x))dx#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer sjc Dec 3, 2016 #-5/2e^(-2x)+e^-x+C# Explanation: rearrange as folows. #int((5-e^x)/e^(2x))dx# #=int(5/e^(2x)-e^x/e^(2x))dx# #int(5e^(-2x)-e^(-x))dx# #=-5/2e^(-2x)+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 10607 views around the world You can reuse this answer Creative Commons License