Question

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

Answer 1

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

Explanation:

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`