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

Options

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

Answer

To solve the integral

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

we can start by simplifying the integrand.

### Step 1: Rewrite the integrand
We can rewrite the integrand as follows:

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

### Step 2: Substitution
Let’s make the substitution:

\[ t = 1 + e^{-x}. \]

Now, we need to find \( dt \):

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

Since \( e^{-x} = t - 1 \), we can rewrite \( dx \):

\[ dx = -\frac{dt}{t - 1}. \]

### Step 3: Substitute in the integral
Now substitute \( e^{-x} \) and \( dx \) into the integral:

\[ \int \frac{e^{-x}}{1 + e^{-x}} \, dx = \int \frac{(t - 1)}{t} \left(-\frac{dt}{t - 1}\right). \]

This simplifies to:

\[ -\int \frac{1}{t} \, dt. \]

### Step 4: Integrate
The integral of \( -\frac{1}{t} \) is:

\[ -\log |t| + C. \]

### Step 5: Substitute back
Now we substitute back \( t = 1 + e^{-x} \):

\[ -\log |1 + e^{-x}| + C. \]

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

\[ -\log(1 + e^{-x}) + C. \]

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