int 8^(-x) dx Find the indefinite integral

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 .