Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 72

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 .

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