To solve the integral

$\int \left(\left(\frac{1}{4}\right)^x\right)dx$

Step 1: Rewrite the base using exponents

Recall that $\frac{1}{4} = 4^{-1}$, so we can rewrite the expression inside the integral as:

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

Thus, the integral becomes:

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

Step 2: Use the exponential form

Next, recognize that $4^{-x} = e^{-x \ln(4)}$. So the integral becomes:

$\int e^{-x \ln(4)} \, dx$

Step 3: Apply the basic integration rule

The integral of $e^{ax}$ is $\frac{e^{ax}}{a}$, where $a$ is a constant. In this case, $a = -\ln(4)$. So, we can integrate as follows:

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

Step 4: Substitute back the exponential form

Recall that $e^{-x \ln(4)} = 4^{-x}$, so the final result is:

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

Final Answer:

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

Would you like further clarification on any steps?

Here are some related questions you might find interesting:

1. How do you integrate exponential functions in general?
2. What are the steps to differentiate the result and check the answer?
3. Can you solve integrals with other exponential bases?
4. How does the natural logarithm function apply in this context?
5. What is the relationship between base changes in exponents and integrals?

Tip: Always check the derivative of your solution to verify its correctness!