`l=int(dx)/(1+e^(x))` is equal to

Options

A. `log_(e)((1+e^(x))/(e^(x)))+c`
B. `log_(e)((e^(x))/(1+e^(x)))+c`
C. `log_(e)(e^(x))(e^(x)+1)+c`
D. `log_(e)(e^(2x)+1)+c`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{1 + e^x} \]
To simplify this expression, we can multiply and divide by \( e^{-x} \):
\[ I = \int \frac{e^{-x} \, dx}{e^{-x} + 1} \]

### Step 2: Substitute
Now, we will use the substitution:
\[ t = e^{-x} + 1 \]
Differentiating both sides gives:
\[ dt = -e^{-x} \, dx \quad \Rightarrow \quad dx = -\frac{dt}{e^{-x}} = -\frac{dt}{t - 1} \]
Substituting \( e^{-x} = t - 1 \) into our integral:
\[ I = \int \frac{-dt}{t - 1 + 1} = \int \frac{-dt}{t} = -\int \frac{dt}{t} \]

### Step 3: Integrate
The integral of \( \frac{1}{t} \) is:
\[ -\int \frac{dt}{t} = -\ln |t| + C \]
where \( C \) is the constant of integration.

### Step 4: Substitute Back
Now we substitute back for \( t \):
\[ I = -\ln |t| + C = -\ln |e^{-x} + 1| + C \]
Using the property of logarithms:
\[ I = \ln \left| \frac{1}{e^{-x} + 1} \right| + C \]

### Step 5: Simplify
We can rewrite this as:
\[ I = \ln \left( \frac{1}{1 + e^{-x}} \right) + C \]
To express it in a more standard form, we can multiply and divide by \( e^x \):
\[ I = \ln \left( \frac{e^x}{e^x + 1} \right) + C \]

### Final Answer
Thus, the integral evaluates to:
\[ I = \ln \left( \frac{e^x}{e^x + 1} \right) + C \]

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