To solve the integral $\int \left( \left( \frac{1}{4} \right)^x \right) dx$, let's first rewrite the expression using exponents.

We can express $\left( \frac{1}{4} \right)^x$ as:

$\left( \frac{1}{4} \right)^x = 4^{-x}$

So the integral becomes:

$\int 4^{-x} \, dx$

Next, apply the rule for integrating exponential functions of the form $a^x$, which states:

$\int a^{x} \, dx = \frac{a^x}{\ln(a)} + C$

Here, $a = 4^{-1} = \frac{1}{4}$, so we can directly apply this rule. Using this formula, the integral is:

$\int 4^{-x} \, dx = \frac{4^{-x}}{\ln(4^{-1})} + C$

Since $\ln(4^{-1}) = -\ln(4)$, we can simplify further:

$\frac{4^{-x}}{-\ln(4)} + C$

Thus, the solution to the integral is:

$\boxed{\frac{-4^{-x}}{\ln(4)} + C}$

Would you like more details on any step or have any further questions?

Here are some related questions you might want to explore:

1. How do you integrate exponential functions in general?
2. How do you derive the formula for integrating $a^x$?
3. How can you simplify logarithms in integrals?
4. What is the derivative of $4^{-x}$ and how does it relate to the integral?
5. How do you handle integrals involving base $e$ instead of other bases?

Tip: When integrating exponential functions, always remember to account for the logarithmic factor that comes with non-$e$-based exponentials.