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:

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

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