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

Options

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

Answer

To solve the integral

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

we will follow a series of steps.

### Step 1: Rewrite the integral
We can rewrite the terms with negative exponents in reciprocal form:

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

### Step 2: Simplify the expression
Now, we can multiply the numerator and the denominator by \(e^{2x}\):

\[ I = \int \frac{e^{4x} + 1}{e^{3x} + e^{x}} \, dx. \]

### Step 3: Factor the denominator
The denominator can be factored as follows:

\[ e^{3x} + e^{x} = e^{x}(e^{2x} + 1). \]

Thus, we can rewrite the integral:

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

### Step 4: Split the integral
Now, we can split the integral:

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

This simplifies to:

\[ I = \int \frac{e^{3x}}{e^{2x} + 1} \, dx + \int \frac{e^{-x}}{e^{2x} + 1} \, dx. \]

### Step 5: Substitution
Let \(t = e^{x}\). Then, \(dt = e^{x} \, dx\) or \(dx = \frac{dt}{t}\). The limits change accordingly, but since this is an indefinite integral, we will not worry about limits.

Substituting \(t\) into the integral gives:

\[ I = \int \frac{t^3}{t^2 + 1} \cdot \frac{dt}{t} + \int \frac{1}{t(t^2 + 1)} \, dt. \]

This simplifies to:

\[ I = \int \frac{t^2}{t^2 + 1} \, dt + \int \frac{1}{t(t^2 + 1)} \, dt. \]

### Step 6: Solve the integrals
1. For the first integral:

\[ \int \frac{t^2}{t^2 + 1} \, dt = \int \left(1 - \frac{1}{t^2 + 1}\right) dt = t - \tan^{-1}(t) + C_1. \]

2. For the second integral:

\[ \int \frac{1}{t(t^2 + 1)} \, dt. \]

Using partial fractions:

\[ \frac{1}{t(t^2 + 1)} = \frac{A}{t} + \frac{Bt + C}{t^2 + 1}. \]

Solving for \(A\), \(B\), and \(C\) gives:

\[ \int \left(\frac{1}{t} - \frac{t}{t^2 + 1} + \frac{1}{t^2 + 1}\right) dt = \ln|t| - \frac{1}{2} \ln|t^2 + 1| + \tan^{-1}(t) + C_2. \]

### Step 7: Combine results
Combining both integrals:

\[ I = \left(t - \tan^{-1}(t) + C_1\right) + \left(\ln|t| - \frac{1}{2} \ln|t^2 + 1| + \tan^{-1}(t) + C_2\right). \]

### Step 8: Substitute back
Substituting back \(t = e^{x}\):

\[ I = e^{x} + \ln|e^{x}| - \frac{1}{2} \ln|e^{2x} + 1| + C. \]

### Final Result
Thus, the final result is:

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