Question

The value of the integral \(\int \frac{dx}{1 + e^x}\) is equal to:

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

Hint:

Exam Tip:
For integrals of the form \(\int \frac{dx}{1 + e^x}\):

• Multiply numerator and denominator by \(e^{-x}\).
• Or use the substitution \(u = 1 + e^{-x}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to evaluate an indefinite integral involving an exponential function. A common substitution is to set \(u = e^x\) or to multiply numerator and denominator by \(e^{-x}\).

Step 2: Key Formula or Approach:
Multiply numerator and denominator by \(e^{-x}\): \[ \int \frac{dx}{1 + e^x} = \int \frac{e^{-x}}{e^{-x} + 1} \, dx = \int \frac{e^{-x}}{1 + e^{-x}} \, dx \]

Step 3: Detailed Explanation:
Let \(u = 1 + e^{-x}\). Then \(du = -e^{-x} dx\).
So, \(e^{-x} dx = -du\).
Thus, \[ \int \frac{e^{-x}}{1 + e^{-x}} \, dx = \int \frac{-du}{u} = -\ln|u| + c = -\ln(1 + e^{-x}) + c \] Wait, this gives option (C).
Let's check:
If \(I = \int \frac{dx}{1 + e^x}\), let \(u = 1 + e^x\), then \(du = e^x dx\).
This is not directly useful.
Alternatively, let \(u = e^x\), \(dx = \frac{du}{u}\): \[ I = \int \frac{1}{1 + u} \cdot \frac{du}{u} = \int \frac{du}{u(1 + u)} = \int \left( \frac{1}{u} - \frac{1}{1 + u} \right) du = \ln|u| - \ln|1 + u| + c \] \[ = \ln\left( \frac{u}{1 + u} \right) + c = \ln\left( \frac{e^x}{1 + e^x} \right) + c = \ln\left( \frac{1}{e^{-x} + 1} \right) + c = -\ln(1 + e^{-x}) + c \] So, the answer is \(-\ln(1 + e^{-x}) + c\), which is option (C).
But the answer key says option (B).
Let's re-evaluate:
\(\frac{e^x}{1 + e^x} = \frac{1}{1 + e^{-x}}\).
So, \(\ln\left( \frac{e^x}{1 + e^x} \right) = \ln\left( \frac{1}{1 + e^{-x}} \right) = -\ln(1 + e^{-x})\).
Thus, the integral is \(-\ln(1 + e^{-x}) + c\).
Option (C) is \(-\ln(1 + e^{-x}) + c\).
Option (B) is \(\ln(1 + e^{-x}) + c\).
The correct answer is option (C).

Step 4: Final Answer:
Therefore, option (C) is correct.