How do you find the integral of #e^(-x)dx#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Ziad K. May 15, 2018 it's #-e^(-x) + c# Explanation: let #u = -x# then #(du)/dx = -1# which means #dx = -du# then #int (e^(-x))dx = int (-e^u)du = -int (e^u)du = -e^u + c= -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 233390 views around the world You can reuse this answer Creative Commons License