The integral `int(2)/(e^(2x)-1)dx` is equal to (Here C is a constant of integration)

Options

A. `x+log_(e)|e^(x)+e^(-x)|+C`
B. `-x+log_(e)|e^(x)-e^(-x)|+C`
C. `-x+log_(e)|e^(x)+e^(-x)|+C`
D. `x+log_(e)|e^(x)-e^(-x)|+C`

Answer

To solve the integral \( I = \int \frac{2}{e^{2x} - 1} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int \frac{2}{e^{2x} - 1} \, dx \]
We can factor the denominator:
\[ e^{2x} - 1 = (e^x - 1)(e^x + 1) \]
Thus, we have:
\[ I = \int \frac{2}{(e^x - 1)(e^x + 1)} \, dx \]

### Step 2: Partial Fraction Decomposition
Next, we can use partial fraction decomposition:
\[ \frac{2}{(e^x - 1)(e^x + 1)} = \frac{A}{e^x - 1} + \frac{B}{e^x + 1} \]
Multiplying through by the denominator \((e^x - 1)(e^x + 1)\) gives:
\[ 2 = A(e^x + 1) + B(e^x - 1) \]
Expanding this:
\[ 2 = (A + B)e^x + (A - B) \]
Setting coefficients equal, we have:
1. \( A + B = 0 \)
2. \( A - B = 2 \)

### Step 3: Solve for A and B
From \( A + B = 0 \), we can express \( B = -A \). Substituting into the second equation:
\[ A - (-A) = 2 \implies 2A = 2 \implies A = 1 \]
Thus, \( B = -1 \). Therefore, we can rewrite the integral:
\[ I = \int \left( \frac{1}{e^x - 1} - \frac{1}{e^x + 1} \right) \, dx \]

### Step 4: Integrate Each Term
Now we can integrate each term separately:
\[ I = \int \frac{1}{e^x - 1} \, dx - \int \frac{1}{e^x + 1} \, dx \]

### Step 5: Substitution for Each Integral
For the first integral, let \( u = e^x \), then \( du = e^x \, dx \) or \( dx = \frac{du}{u} \):
\[ \int \frac{1}{e^x - 1} \, dx = \int \frac{1}{u - 1} \cdot \frac{du}{u} = \int \frac{1}{u(u - 1)} \, du \]
For the second integral:
\[ \int \frac{1}{e^x + 1} \, dx = \int \frac{1}{u + 1} \cdot \frac{du}{u} = \int \frac{1}{u(u + 1)} \, du \]

### Step 6: Combine Results
After integrating both terms, we combine the results:
\[ I = \log |e^x - 1| - \log |e^x + 1| + C \]
Using properties of logarithms:
\[ I = \log \left( \frac{e^x - 1}{e^x + 1} \right) + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = \log \left( \frac{e^x - 1}{e^x + 1} \right) + C \]