`int ( x^(2) + 1)/( x^(2) - 1)dx` is equal to

Options

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

Answer

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

### Step 1: Rewrite the integrand
We start by rewriting the integrand:
\[ \frac{x^2 + 1}{x^2 - 1} = \frac{x^2 - 1 + 2}{x^2 - 1} = \frac{x^2 - 1}{x^2 - 1} + \frac{2}{x^2 - 1} \]
This simplifies to:
\[ 1 + \frac{2}{x^2 - 1} \]

### Step 2: Set up the integral
Now we can express the integral as:
\[ \int \left( 1 + \frac{2}{x^2 - 1} \right) \, dx = \int 1 \, dx + \int \frac{2}{x^2 - 1} \, dx \]

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

### Step 4: Integrate the second term
Next, we need to integrate \( \frac{2}{x^2 - 1} \). We can factor \( x^2 - 1 \) as \( (x - 1)(x + 1) \). We will use partial fraction decomposition:
\[ \frac{2}{x^2 - 1} = \frac{2}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} \]
Multiplying through by \( (x - 1)(x + 1) \) gives:
\[ 2 = A(x + 1) + B(x - 1) \]
Setting \( x = 1 \):
\[ 2 = A(2) + B(0) \implies A = 1 \]
Setting \( x = -1 \):
\[ 2 = A(0) + B(-2) \implies B = -1 \]
Thus, we have:
\[ \frac{2}{x^2 - 1} = \frac{1}{x - 1} - \frac{1}{x + 1} \]

### Step 5: Integrate the partial fractions
Now we can integrate:
\[ \int \left( \frac{1}{x - 1} - \frac{1}{x + 1} \right) \, dx = \int \frac{1}{x - 1} \, dx - \int \frac{1}{x + 1} \, dx \]
This results in:
\[ \log |x - 1| - \log |x + 1| = \log \left| \frac{x - 1}{x + 1} \right| \]

### Step 6: Combine the results
Combining all parts, we have:
\[ \int \frac{x^2 + 1}{x^2 - 1} \, dx = x + 2 \log \left| \frac{x - 1}{x + 1} \right| + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{x^2 + 1}{x^2 - 1} \, dx = x + 2 \log \left| \frac{x - 1}{x + 1} \right| + C \]