Evaluate the indefinite integral. 
${\int }_{}^{}-{\left(6\right)}^{x}dx$

• A. $-x{6}^{x-1}+C$
• B. $-\frac{1}{x+1}{6}^{x+1}+C$
• C. $-\frac{{\left(6\right)}^x}{\ln 6}+C$
• D. $-\frac{{\left(6\right)}^{x+1}}{\ln x}+C$

Answer: (C)

Verified step by step guidance

1. Step 1: Recognize that the integral involves an exponential function of the form \( 6^x \). The general formula for integrating \( a^x \) is \( \int a^x dx = \frac{a^x}{\ln a} + C \), where \( a > 0 \) and \( a \neq 1 \).
2. Step 2: Apply the formula to the given integral \( \int -6^x dx \). The negative sign can be factored out, so the integral becomes \( -\int 6^x dx \).
3. Step 3: Using the formula \( \int a^x dx = \frac{a^x}{\ln a} + C \), substitute \( a = 6 \). This gives \( \int 6^x dx = \frac{6^x}{\ln 6} + C \).
4. Step 4: Multiply the result by \( -1 \) to account for the negative sign. This gives \( -\frac{6^x}{\ln 6} + C \).
5. Step 5: Write the final expression for the indefinite integral as \( -\frac{6^x}{\ln 6} + C \), where \( C \) is the constant of integration.