`int (dx)/((1+e^x)(1+e^(-x)))=`

Options

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

Answer

To solve the integral

\[ I = \int \frac{dx}{(1 + e^x)(1 + e^{-x})} \]

we can simplify the integrand step by step.

### Step 1: Simplify the Denominator
We start by rewriting the denominator:

\[ (1 + e^x)(1 + e^{-x}) = 1 + e^x + e^{-x} + 1 = 2 + e^x + e^{-x} \]

Using the identity \( e^{-x} = \frac{1}{e^x} \), we can express this as:

\[ = 2 + e^x + \frac{1}{e^x} = 2 + \frac{e^{2x} + 1}{e^x} \]

Thus, we can rewrite \( I \) as:

\[ I = \int \frac{e^x \, dx}{e^x(2 + e^x + e^{-x})} = \int \frac{e^x \, dx}{2e^x + e^{2x} + 1} \]

### Step 2: Substitution
Let \( t = 1 + e^x \). Then, we differentiate \( t \):

\[ dt = e^x \, dx \quad \Rightarrow \quad dx = \frac{dt}{e^x} \]

Substituting \( e^x = t - 1 \) into the integral gives:

\[ I = \int \frac{dt}{(t)(t)} = \int \frac{dt}{t^2} \]

### Step 3: Integrate
Now we can integrate:

\[ I = \int t^{-2} \, dt = -\frac{1}{t} + C \]

### Step 4: Back Substitute
We substitute back \( t = 1 + e^x \):

\[ I = -\frac{1}{1 + e^x} + C \]

### Final Answer
Thus, the final answer for the integral is:

\[ I = -\frac{1}{1 + e^x} + C \]

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