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

Options

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

Explanation

To solve the integral \( \int \frac{2}{(e^x + e^{-x})^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{2}{(e^x + e^{-x})^2} \, dx \]
We can simplify \( e^x + e^{-x} \) using the identity:
\[ e^x + e^{-x} = 2 \cosh(x) \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{2}{(2 \cosh(x))^2} \, dx = \int \frac{2}{4 \cosh^2(x)} \, dx = \frac{1}{2} \int \text{sech}^2(x) \, dx \]

### Step 2: Integrate
The integral of \( \text{sech}^2(x) \) is a standard integral:
\[ \int \text{sech}^2(x) \, dx = \tanh(x) + C \]
Thus, we have:
\[ I = \frac{1}{2} \left( \tanh(x) + C \right) = \frac{1}{2} \tanh(x) + C \]

### Step 3: Rewrite in Terms of Exponentials
Now, we can express \( \tanh(x) \) in terms of exponentials:
\[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}} \]
Thus, our integral becomes:
\[ I = \frac{1}{2} \cdot \frac{e^x - e^{-x}}{e^x + e^{-x}} + C \]

### Final Answer
The final result of the integral is:
\[ \int \frac{2}{(e^x + e^{-x})^2} \, dx = \frac{1}{2} \cdot \frac{e^x - e^{-x}}{e^x + e^{-x}} + C \]