What is the antiderivative of #e^(8x)#?

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Answer 1 · 2016-01-22T14:21:46

#1/8 e^(8x) +C#

We can go through the steps of integrating by substitution, but some find the following more clear:

We know that #d/dx(e^(8x)) = 8e^(8x)#

That is #8# time more that we want the derivative to be. So, we'll multiply by #1/8# (divide by #8#).
#d/dx(1/8e^(8x)) = 1/8*8e^(8x) = e^(8x)#

The general antiderivative is, therefore, #1/8e^(8x)+C#.

Here is the substitution solution

#inte^(8x) dx#

Let #u=8x#, so we get #du = 8 dx# and #dx = 1/8 du#

The integral becomes:

#int e^u * 1/8 du = 1/8 int e^u du = 1/8 e^u +C#

Reversing the substitution gives

#inte^(8x) dx = 1/8e^(8x)+C#