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

Options

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

Answer

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

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form:
\[ \int \frac{e^{-x}}{1 + e^{x}} \, dx = \int \frac{1}{e^{x}(1 + e^{x})} \, dx \]

### Step 2: Split the Fraction
Next, we can manipulate the integrand by adding and subtracting \( e^{x} \):
\[ \int \left( \frac{1 + e^{x}}{e^{x}(1 + e^{x})} - \frac{e^{x}}{e^{x}(1 + e^{x})} \right) \, dx \]
This simplifies to:
\[ \int \left( \frac{1}{e^{x}} - \frac{1}{1 + e^{x}} \right) \, dx \]

### Step 3: Separate the Integrals
Now we can separate the integral into two parts:
\[ \int \frac{1}{e^{x}} \, dx - \int \frac{1}{1 + e^{x}} \, dx \]

### Step 4: Integrate the First Part
The first integral can be computed as follows:
\[ \int \frac{1}{e^{x}} \, dx = \int e^{-x} \, dx = -e^{-x} + C_1 \]

### Step 5: Integrate the Second Part
For the second integral, we can use a substitution. Let:
\[ t = e^{-x} + 1 \quad \Rightarrow \quad dt = -e^{-x} \, dx \quad \Rightarrow \quad dx = -\frac{dt}{t - 1} \]
Then, we have:
\[ \int \frac{1}{1 + e^{x}} \, dx = \int \frac{1}{t} \left(-\frac{dt}{t - 1}\right) \]
This integral simplifies to:
\[ -\int \frac{1}{t} \, dt = -\ln|t| + C_2 \]

### Step 6: Combine the Results
Combining both parts, we have:
\[ -e^{-x} - \ln|e^{-x} + 1| + C \]

### Step 7: Final Simplification
We can write the final answer as:
\[ -e^{-x} + \ln|1 + e^{-x}| + C \]

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