How do you integrate #int e^(3x)dx#?

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Answer 1 · 2017-03-12T11:07:59

The answer is #=1/3e^(3x)+C#

We need

#inte^xdx=e^x+C#

We do this integral by substitution

Let #u=3x#

#du=3dx#

#dx=(du)/3#

Therefore,

#inte^(3x)dx=int1/3e^(u)du#

#=1/3e^(u)#

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

Answer 2 · 2017-03-12T11:09:33

#int e^(3x)dx =e^(3x)/3 +C#

#I = int e^(3x)dx#

Let #u=3x -> (du)/dx= 3#

#I = int e^u* 1/3du#

#= 1/3*e^u +C#

Undo substitution:

#I= e^(3x)/3 +C#