`int(dx)/(1+e^(-x))` is equal to :
Where c is the constant of integration.

Options

A. `1+e^(x)+c`
B. `ln(1+e^(-x))+c`
C. `ln(1+e^(x))+c`
D. `2ln(1+e^(-x))+c`

Answer

To solve the integral \( \int \frac{dx}{1 + e^{-x}} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{1 + e^{-x}} \]

### Step 2: Simplify the Denominator
To simplify the expression, we can multiply the numerator and the denominator by \( e^x \):
\[ I = \int \frac{e^x \, dx}{e^x + 1} \]

### Step 3: Use Substitution
Let \( u = e^x + 1 \). Then, the derivative \( du = e^x \, dx \). This means \( dx = \frac{du}{e^x} \). Since \( e^x = u - 1 \), we can substitute this into our integral:
\[ I = \int \frac{du}{u} \]

### Step 4: Integrate
Now we can integrate:
\[ I = \ln |u| + C \]
Substituting back for \( u \):
\[ I = \ln |e^x + 1| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{dx}{1 + e^{-x}} = \ln(1 + e^x) + C \]

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