Choose the correct answer of the given question
`int (e^x+1)/(e^x-1) dx = ___ +c`

Options

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

Answer

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

### Step 1: Simplify the integrand
We start by rewriting the integrand:
\[ \frac{e^x + 1}{e^x - 1} = \frac{(e^x - 1) + 2}{e^x - 1} = 1 + \frac{2}{e^x - 1} \]
Thus, we can express the integral as:
\[ \int \frac{e^x + 1}{e^x - 1} \, dx = \int \left(1 + \frac{2}{e^x - 1}\right) \, dx \]

### Step 2: Separate the integral
Now we can separate the integral into two parts:
\[ \int \left(1 + \frac{2}{e^x - 1}\right) \, dx = \int 1 \, dx + 2 \int \frac{1}{e^x - 1} \, dx \]

### Step 3: Integrate the first part
The integral of 1 is straightforward:
\[ \int 1 \, dx = x \]

### Step 4: Integrate the second part
Now we need to focus on the integral \( \int \frac{1}{e^x - 1} \, dx \). We can use substitution for this integral. Let:
\[ u = e^x - 1 \quad \Rightarrow \quad du = e^x \, dx \quad \Rightarrow \quad dx = \frac{du}{u + 1} \]
Thus, we rewrite the integral:
\[ \int \frac{1}{u} \cdot \frac{du}{u + 1} \]
This integral can be solved using partial fractions:
\[ \frac{1}{u(u + 1)} = \frac{A}{u} + \frac{B}{u + 1} \]
Multiplying through by \(u(u + 1)\) and solving for \(A\) and \(B\), we find:
\[ 1 = A(u + 1) + Bu \quad \Rightarrow \quad 1 = (A + B)u + A \]
Setting coefficients equal gives:
1. \(A + B = 0\)
2. \(A = 1\)

Thus, \(A = 1\) and \(B = -1\). Therefore:
\[ \int \frac{1}{u(u + 1)} \, du = \int \left(\frac{1}{u} - \frac{1}{u + 1}\right) \, du = \log |u| - \log |u + 1| + C \]
Substituting back \(u = e^x - 1\):
\[ \int \frac{1}{e^x - 1} \, dx = \log |e^x - 1| - \log |e^x| + C = \log \left|\frac{e^x - 1}{e^x}\right| + C \]

### Step 5: Combine the results
Now, we combine the results of both integrals:
\[ \int \frac{e^x + 1}{e^x - 1} \, dx = x + 2\left(\log \left|\frac{e^x - 1}{e^x}\right| + C\right) \]
This simplifies to:
\[ x + 2\log |e^x - 1| - 2\log |e^x| + C \]
Thus, the final result can be expressed as:
\[ x + 2\log |e^x - 1| - 2x + C = -x + 2\log |e^x - 1| + C \]

### Final Answer:
\[ \int \frac{e^x + 1}{e^x - 1} \, dx = -x + 2\log |e^x - 1| + C \]