`int(dx)/(1+e^(-x))` is equal to

Options

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

Explanation

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

### Step 1: Simplify the Integral
We start with the integral:
\[ \int \frac{dx}{1 + e^{-x}} \]
To simplify this, we can multiply the numerator and denominator by \( e^{x} \):
\[ \int \frac{e^{x} \, dx}{e^{x}(1 + e^{-x})} = \int \frac{e^{x} \, dx}{e^{x} + 1} \]

### Step 2: Substitution
Now, we can use the substitution:
\[ u = e^{x} + 1 \]
Then, we differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = e^{x} \implies du = e^{x} \, dx \]
This means we can replace \( e^{x} \, dx \) with \( du \).

### Step 3: Rewrite the Integral
Substituting into the integral gives us:
\[ \int \frac{du}{u} \]

### Step 4: Integrate
The integral of \( \frac{1}{u} \) is:
\[ \ln |u| + C \]

### Step 5: Substitute Back
Now we substitute back for \( u \):
\[ \ln |e^{x} + 1| + C \]

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