`int(1)/(1+e^(-x))dx=`

Options

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

Answer

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

### Step 1: Rewrite the integrand
We start with the integral:
\[ \int \frac{1}{1 + e^{-x}} \, dx \]
We can rewrite \( e^{-x} \) as \( \frac{1}{e^x} \):
\[ \int \frac{1}{1 + \frac{1}{e^x}} \, dx = \int \frac{e^x}{e^x + 1} \, dx \]

### Step 2: Substitute
Let \( t = e^x + 1 \). Then, the derivative \( dt = e^x \, dx \) or \( dx = \frac{dt}{e^x} \). Since \( e^x = t - 1 \), we can substitute:
\[ dx = \frac{dt}{t - 1} \]
Now, substituting into the integral gives:
\[ \int \frac{e^x}{e^x + 1} \, dx = \int \frac{t - 1}{t} \cdot \frac{dt}{t - 1} = \int \frac{1}{t} \, dt \]

### Step 3: Integrate
The integral \( \int \frac{1}{t} \, dt \) is a standard integral:
\[ \int \frac{1}{t} \, dt = \ln |t| + C \]

### Step 4: Back-substitute
Now, we substitute back \( t = e^x + 1 \):
\[ \ln |t| + C = \ln |e^x + 1| + C \]

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