How do you integrate # 3^x#?

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Answer 1 · 2016-09-13T07:57:45

#= 1/(ln 3) 3^x + C#

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#