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

Options

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

Answer

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

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

### Step 2: Multiply by \( e^{-x} \)
Next, we multiply the numerator and the denominator by \( e^{-x} \):
\[ \int \frac{e^{-x}}{1 + e^{-x}} \, dx \]

### Step 3: Substitution
Now, we will use substitution. Let:
\[ t = 1 + e^{-x} \]
Then, the derivative of \( t \) with respect to \( x \) is:
\[ dt = -e^{-x} \, dx \quad \Rightarrow \quad dx = -\frac{dt}{e^{-x}} = -\frac{dt}{t - 1} \]

### Step 4: Substitute in the Integral
Substituting \( t \) into the integral gives:
\[ \int \frac{e^{-x}}{t} \left(-\frac{dt}{e^{-x}}\right) = -\int \frac{1}{t} \, dt \]

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

### Step 6: Back Substitute
Now, we substitute back for \( t \):
\[ -\ln |1 + e^{-x}| + C \]

### Step 7: Simplify
We can also express this as:
\[ -\ln(1 + e^{-x}) + C \]

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

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