By definition, if the function F(x) is the antiderivative of f(x) then we follow

the indefinite integral as `int f(x) dx = F(x)+C`

where: f(x) as the integrand

F(x) as the anti-derivative function

C as the arbitrary constant known as constant of integration

For the problem` int 8^(-x) dx,` we may apply u-substitution then basic formula for exponential function.

Using u-substitution, we let `u = -x` then` du = -1 dx` .

By dividing both sides by -1 in `du = -1 dx` , we get `-1 du = dx` .

Applying u-substitution using` -x =u ` and dx=-1 du in ` int 8^(-x) dx`

, we get: `int 8^(u) * (-1) du = -1 int 8^u du`

Applying the basic integration formula for exponential function:

`int a^u du = a^u/(ln(a)) +C` where a is a constant.

Then `(-1) int 8^u du = 8^u/(ln(8)) +C`

To express it in terms of x, we plug-in u=-x to get:

`-8^(-x)/(ln(8)) +C`

Recall `8 = 2^3` . It can be also be written as:

`-(2^3)^(-x)/(ln(2^3))+C`

Recall the logarithm property:` ln(x^n) = n ln(x)` then` ln(2^3) = 3 ln(2)`

It becomes

The final answer can be `-8^(-x)/(ln(8))+c ` or `-2^(-3x)/(3ln(2))+C` .