`int1/(1+e^(x))dx` is equal to

Options

A. `log_(e)((e^(x)+1)/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^(x)/(e^(x)-1))+c`

Answer

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

### Step 1: Set up the integral
Let \( I = \int \frac{1}{1 + e^x} \, dx \).

### Step 2: Rewrite the integral
We can rewrite the integrand by factoring out \( e^x \):
\[ I = \int \frac{1}{1 + e^x} \, dx = \int \frac{e^{-x}}{e^{-x} + 1} \, dx \]

### Step 3: Substitution
Let \( u = 1 + e^{-x} \). Then, differentiate \( u \):
\[ du = -e^{-x} \, dx \quad \Rightarrow \quad dx = -\frac{du}{e^{-x}} = -\frac{du}{u - 1} \]

### Step 4: Substitute in the integral
Now, substituting \( u \) into the integral:
\[ I = \int \frac{-1}{u} \, du = -\ln |u| + C \]

### Step 5: Back-substitute \( u \)
Substituting back for \( u \):
\[ I = -\ln |1 + e^{-x}| + C \]

### Step 6: Simplify the expression
We can simplify the expression further:
\[ I = \ln |1 + e^{-x}|^{-1} + C = \ln \left( \frac{1}{1 + e^{-x}} \right) + C \]

### Step 7: Final simplification
Using the property of logarithms:
\[ I = \ln \left( \frac{e^x}{e^x + 1} \right) + C \]

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

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