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

Options

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

Answer

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

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

### Step 2: Simplify the Denominator
Next, we can factor out \( e^{-x} \) from the denominator:
\[ = \int \frac{1}{e^{-x}(1 + e^{x})} \, dx = \int \frac{e^{x}}{1 + e^{x}} \, dx \]

### Step 3: Substitution
Now, we can use substitution. Let:
\[ t = 1 + e^{x} \implies dt = e^{x} \, dx \quad \Rightarrow \quad dx = \frac{dt}{e^{x}} \]
Substituting \( e^{x} = t - 1 \) into the equation gives:
\[ dx = \frac{dt}{t - 1} \]

### Step 4: Substitute in the Integral
Now substitute \( t \) into the integral:
\[ \int \frac{e^{x}}{1 + e^{x}} \, dx = \int \frac{(t - 1)}{t} \cdot \frac{dt}{t - 1} = \int \frac{1}{t} \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ \int \frac{1}{t} \, dt = \ln |t| + C \]

### Step 6: Substitute Back
Substituting back for \( t \):
\[ = \ln |1 + e^{x}| + C \]

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