Question

Evaluate the integral \[ \int \frac{e^x}{2^x} dx. \]

• A. \( \frac{e^x}{(\log_2 2) 2^x} + C \)
• B. \( \frac{e^x}{2(2^x)} + C \)
• C. \( \frac{2 \left(\frac{e}{2} \right)^{x-1}}{e} + C \)
• D. \( \frac{e^x}{(1 - \log_2 2)2^x} + C \)
• E. \( \frac{e^x}{2^x} + C \)

Hint:

Convert exponentials of different bases into the natural exponential \( e^x \) for easier integration.

Answer 1

The Correct Option is D

Rewriting the given integral: \[ I = \int \frac{e^x}{2^x} dx. \] Since \( 2^x = e^{x \ln 2} \), we rewrite: \[ I = \int e^x e^{-x \ln 2} dx = \int e^{x(1 - \ln 2)} dx. \] Integrating, \[ I = \frac{e^{x(1 - \ln 2)}}{1 - \ln 2} + C. \] Thus, the correct answer is: \[ \frac{e^x}{(1 - \log_2 2)2^x} + C. \]