How do you integrate # 3^x#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Eddie Sep 13, 2016 #= 1/(ln 3) 3^x + C# Explanation: We can work the derivative first #y = 3^x# #ln y = x ln 3# #1/y y' = ln 3 # #y' = ln 3 \ 3^x# #implies int 3^x \ dx# #= int d/dx(1/(ln 3) 3^x )\ dx# #= 1/(ln 3) 3^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 97070 views around the world You can reuse this answer Creative Commons License