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

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

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