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

Options

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

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{2}{(e^x + e^{-x})^2} \, dx \]

### Step 2: Simplify the Denominator
We know that \( e^x + e^{-x} = 2 \cosh(x) \). Therefore, 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 3: Integrate
The integral of \( \text{sech}^2(x) \) is known:
\[ \int \text{sech}^2(x) \, dx = \tanh(x) + C \]
Thus, we have:
\[ I = \frac{1}{2} \left( \tanh(x) + C \right) = \frac{1}{2} \tanh(x) + C \]

### Step 4: Final Result
The final result of the integral is:
\[ I = \frac{1}{2} \tanh(x) + C \]