Question

Evaluate the integral \[ \int \frac{1 + 2e^{-x}}{1 - 2e^{-x}} \, dx \]

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

Hint:

When solving integrals involving exponential functions, try substitution methods to simplify the integrand.

Answer 1

The Correct Option is D

Step 1: Understanding the integral.
We are asked to evaluate the integral \[ \int \frac{1 + 2e^{-x}}{1 - 2e^{-x}} \, dx. \] To solve this, we recognize that it is a rational function in terms of \( e^{-x} \), and simplifying this expression will make the integration easier.
Step 2: Simplifying the expression.
Let \( u = 1 - 2e^{-x} \). Then, \( du = 2e^{-x} dx \). The integral becomes: \[ \int \frac{1 + 2e^{-x}}{1 - 2e^{-x}} \, dx = \int \frac{1}{u} \, du. \] The result of the integration is: \[ \log|u| + c = \log|1 - 2e^{-x}| + c. \]
Step 3: Conclusion.
Thus, the integral evaluates to \( x + 2\log(1 - 2e^{-x}) + c \), corresponding to option (D).