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

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.

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

Explore how to solve the integral of the exponential function (1/4)^x. This solution demonstrates key concepts in exponential functions, logarithmic properties, and the integration rule for non-e-based exponents. Ideal for high school students in Grades 11-12.

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