What is the antiderivative of #e^(-3x)#?

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Answer 1 · 2016-03-26T13:51:26

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

The question you are asking is:

#inte^(-3x)dx#

We will want to put this in the form

#inte^udu=e^u+C#

So, let #u=-3x#, then #du=-3dx#.

So that there can be a #-3dx# inside the integral, multiply by #-3#. Compensate for this by multiplying the outside of the integral by #-1//3#.

#=-1/3inte^(-3x)(-3)dx#

Substitute using the #u# and #du# values we defined earlier.

#=-1/3inte^udu#

This becomes

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