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

Answer

To solve the integral \( \int \frac{e^{2x} + e^{-2x}}{e^x} \, dx \), we can break it down step by step.

### Step 1: Simplify the integrand
We start by simplifying the expression inside the integral:
\[ \frac{e^{2x} + e^{-2x}}{e^x} = \frac{e^{2x}}{e^x} + \frac{e^{-2x}}{e^x} \]
Using the property of exponents, we can rewrite this as:
\[ e^{2x - x} + e^{-2x - x} = e^{x} + e^{-3x} \]

### Step 2: Rewrite the integral
Now we can rewrite the integral:
\[ \int \left( e^{x} + e^{-3x} \right) \, dx \]

### Step 3: Integrate each term
Now we will integrate each term separately:
1. The integral of \( e^{x} \):
\[ \int e^{x} \, dx = e^{x} + C_1 \]
2. The integral of \( e^{-3x} \):
\[ \int e^{-3x} \, dx = -\frac{1}{3} e^{-3x} + C_2 \]

### Step 4: Combine the results
Combining these results, we have:
\[ \int \left( e^{x} + e^{-3x} \right) \, dx = e^{x} - \frac{1}{3} e^{-3x} + C \]
where \( C = C_1 + C_2 \) is the constant of integration.

### Final Answer
Thus, the final answer is:
\[ \int \frac{e^{2x} + e^{-2x}}{e^x} \, dx = e^{x} - \frac{1}{3} e^{-3x} + C \]

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