The value of the integral `int(dx)/((e^(x)+e^(-x))^2)` is

Options

A. `(1)/(2)(e^(2x)+1)+C`
B. `(1)/(2)(e^(-2x)+1)+C`
C. `-(1)/(2)(e^(2x)+1)^(-1)+C`
D. `(1)/(4)(e^(2x)-1)+C`

Explanation

To solve the integral

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

we will follow these steps:

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

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

Thus,

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

Now, substituting this back into the integral gives:

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

### Step 2: Substitution
Let us make the substitution:

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

Then, differentiating both sides gives:

\[ dt = 2e^{2x} \, dx \quad \Rightarrow \quad dx = \frac{dt}{2e^{2x}} \]

From our substitution, we have:

\[ e^{2x} = t - 1 \]

Substituting \(dx\) into the integral:

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

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

\[ I = \frac{1}{2} \left(-\frac{1}{t}\right) + C = -\frac{1}{2t} + C \]

### Step 4: Substitute Back
Now we substitute back \(t = e^{2x} + 1\):

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

### Final Result
Thus, the value of the integral is:

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