What is `int (1)/(1 + e^(x))dx` equal to ?
where c is a constant of integration

Options

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

Explanation

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

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

### Step 2: Substitution
Let’s make the substitution:
\[ t = 1 + e^{-x} \]
Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = -e^{-x} \quad \Rightarrow \quad dt = -e^{-x} \, dx \quad \Rightarrow \quad dx = -\frac{dt}{e^{-x}} = -\frac{dt}{t - 1} \]

### Step 3: Substitute in the Integral
Now we can substitute \( t \) and \( dx \) into the integral:
\[ \int \frac{e^{-x}}{e^{-x} + 1} \, dx = \int \frac{e^{-x}}{t} \left(-\frac{dt}{t - 1}\right) \]
Since \( e^{-x} = t - 1 \), we have:
\[ = \int \frac{t - 1}{t} \left(-\frac{dt}{t - 1}\right) \]
This simplifies to:
\[ = -\int \frac{1}{t} \, dt + \int \frac{1}{t} \, dt = -\int \frac{1}{t} \, dt \]

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

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

### Final Result
Thus, the integral evaluates to:
\[ -\log(1 + e^{-x}) + C \]