Question

Evaluate the integral \( 3 \int \frac{3^x}{\sqrt{1 - 9^x}} \, dx \)

• A. \( (\log 3)\sin^{-1}(3^x) + C \)
• B. \( \frac{1}{3} \sin^{-1}(3^x) + C \)
• C. \( \frac{1}{\log 3} \sin^{-1}(3^x) + C \)
• D. \( 3 \log 3 \sin^{-1}(3^x) + C \)

Hint:

When encountering integrals with square roots of exponential functions, use substitution to simplify the terms.

Answer 1

The Correct Option is C

We need to evaluate the integral: \[ 3 \int \frac{3^x}{\sqrt{1 - 9^x}} \, dx \] The expression \( 9^x = (3^2)^x = 3^{2x} \), and we can use substitution. Let \( u = 3^x \), hence \( du = 3^x \ln 3 \, dx \). Thus, the integral simplifies as: \[ 3 \int \frac{du}{\sqrt{1 - u^2}} = 3 \sin^{-1}(u) + C = \frac{1}{\log 3} \sin^{-1}(3^x) + C \] Thus, the correct answer is (C).