Question

The integral \( \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx \) is equal to:

• A. \( (\log 2) \sin^{-1} 2^x + C \)
• B. \( \frac{1}{2} \sin^{-1} 2^x + C \)
• C. \( \frac{1}{\log 2} \sin^{-1} 2^x + C \)
• D. \( 2 \log 2 \sin^{-1} 2^x + C \)

Hint:

When solving integrals involving powers of 2, use substitution and simplify the expression into a standard form.

Answer 1

The Correct Option is C

We are given the integral \( \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx \). Step 1: Use substitution Let \( u = 2^x \). Then, the differential \( du = 2^x \ln 2 \, dx \). Now, express the integral in terms of \( u \): \[ \int \frac{2^x}{\sqrt{1 - 4^x}} \, dx = \int \frac{du}{\sqrt{1 - u^2}} \] Step 2: Solve the integral We recognize the standard integral \( \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) \), so: \[ \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) \] Substitute \( u = 2^x \) to get the final result: \[ \frac{1}{\log 2} \sin^{-1}(2^x) + C \] Thus, the correct answer is \( \frac{1}{\log 2} \sin^{-1} 2^x + C \).