`int1/((e^(2x)+e^(-2x))^(2))dx=`

Options

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

Answer

To solve the integral

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

we can follow these steps:

### Step 1: Simplify the expression inside the integral
We start by rewriting \( e^{2x} + e^{-2x} \). We can express this in terms of hyperbolic functions:

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

Thus, we have:

\[ (e^{2x} + e^{-2x})^2 = (2 \cosh(2x))^2 = 4 \cosh^2(2x). \]

So, we can rewrite the integral as:

\[ \int \frac{1}{(e^{2x} + e^{-2x})^2} \, dx = \int \frac{1}{4 \cosh^2(2x)} \, dx = \frac{1}{4} \int \frac{1}{\cosh^2(2x)} \, dx. \]

### Step 2: Use the identity for hyperbolic secant
Recall that

\[ \frac{1}{\cosh^2(u)} = \text{sech}^2(u). \]

Thus, we can rewrite the integral as:

\[ \frac{1}{4} \int \text{sech}^2(2x) \, dx. \]

### Step 3: Integrate using the derivative of hyperbolic tangent
The integral of \(\text{sech}^2(u)\) is \(\tanh(u)\). Therefore, we have:

\[ \int \text{sech}^2(2x) \, dx = \frac{1}{2} \tanh(2x) + C, \]

where \(C\) is the constant of integration.

### Step 4: Substitute back into the integral
Now we can substitute back into our expression:

\[ \frac{1}{4} \int \text{sech}^2(2x) \, dx = \frac{1}{4} \left( \frac{1}{2} \tanh(2x) + C \right) = \frac{1}{8} \tanh(2x) + \frac{C}{4}. \]

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

\[ \int \frac{1}{(e^{2x} + e^{-2x})^2} \, dx = \frac{1}{8} \tanh(2x) + C, \]

where \(C\) is the constant of integration.

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