`int (dx)/( e^(x) + e^(-x) +2)` is equal to

Options

A. `(1)/( e^(x) +1) +C`
B. `(1)/( 1 + e^(-x)) + C`
C. `- ( 1)/( e^(x) +1) + C`
D. none of these

Answer

To solve the integral

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

we can follow these steps:

### Step 1: Rewrite the Denominator
First, we can simplify the expression in the denominator. Notice that:

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

Thus, we can rewrite the integral as:

\[ I = \int \frac{dx}{2 \cosh(x) + 2} = \int \frac{dx}{2(\cosh(x) + 1)} \]

### Step 2: Factor Out the Constant
We can factor out the constant 2 from the integral:

\[ I = \frac{1}{2} \int \frac{dx}{\cosh(x) + 1} \]

### Step 3: Use a Substitution
Next, we can use the substitution \( t = e^x \). Then, \( dx = \frac{dt}{t} \) and \( \cosh(x) = \frac{t + \frac{1}{t}}{2} \). This gives us:

\[ \cosh(x) + 1 = \frac{t + \frac{1}{t}}{2} + 1 = \frac{t + \frac{1}{t} + 2}{2} = \frac{(t + 1)^2}{2t} \]

Now substituting this back into the integral:

\[ I = \frac{1}{2} \int \frac{dt/t}{(t + 1)^2/2t} = \int \frac{dt}{(t + 1)^2} \]

### Step 4: Integrate
The integral

\[ \int \frac{dt}{(t + 1)^2} \]

can be solved using the formula for the integral of \( \frac{1}{x^n} \):

\[ \int x^{-n} dx = \frac{x^{-n+1}}{-n+1} + C \]

Thus, we have:

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

### Step 5: Substitute Back
Now we need to substitute back \( t = e^x \):

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

### Final Answer
Thus, the final result of the integral is:

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