The value of `int(x^(2)+1)/(x^(2)-1)dx` is

Options

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

Explanation

To solve the integral \(\int \frac{x^2 + 1}{x^2 - 1} \, dx\), we can break it down into simpler parts. Here’s a step-by-step solution:

### Step 1: Simplify the integrand
We start by rewriting the integrand:
\[ \frac{x^2 + 1}{x^2 - 1} = \frac{x^2 - 1 + 2}{x^2 - 1} = 1 + \frac{2}{x^2 - 1} \]
This allows us to separate the integral into two parts:
\[ \int \frac{x^2 + 1}{x^2 - 1} \, dx = \int 1 \, dx + \int \frac{2}{x^2 - 1} \, dx \]

### Step 2: Integrate the first term
The integral of 1 is straightforward:
\[ \int 1 \, dx = x \]

### Step 3: Integrate the second term
Now we need to integrate \(\int \frac{2}{x^2 - 1} \, dx\). We can factor the denominator:
\[ x^2 - 1 = (x - 1)(x + 1) \]
We can use partial fraction decomposition for \(\frac{2}{x^2 - 1}\):
\[ \frac{2}{x^2 - 1} = \frac{A}{x - 1} + \frac{B}{x + 1} \]
Multiplying through by the denominator \(x^2 - 1\) gives:
\[ 2 = A(x + 1) + B(x - 1) \]
Expanding this, we have:
\[ 2 = Ax + A + Bx - B = (A + B)x + (A - B) \]
Setting up the equations:
1. \(A + B = 0\)
2. \(A - B = 2\)

From the first equation, we can express \(B\) in terms of \(A\):
\[ B = -A \]
Substituting into the second equation:
\[ A - (-A) = 2 \implies 2A = 2 \implies A = 1 \]
Then substituting back to find \(B\):
\[ B = -1 \]
Thus, we have:
\[ \frac{2}{x^2 - 1} = \frac{1}{x - 1} - \frac{1}{x + 1} \]

### Step 4: Integrate using the partial fractions
Now we can integrate:
\[ \int \frac{2}{x^2 - 1} \, dx = \int \left( \frac{1}{x - 1} - \frac{1}{x + 1} \right) \, dx \]
This results in:
\[ \int \frac{1}{x - 1} \, dx - \int \frac{1}{x + 1} \, dx = \ln |x - 1| - \ln |x + 1| \]
Using properties of logarithms, we can combine these:
\[ \ln \left| \frac{x - 1}{x + 1} \right| \]

### Step 5: Combine the results
Putting everything together, we have:
\[ \int \frac{x^2 + 1}{x^2 - 1} \, dx = x + \ln \left| \frac{x - 1}{x + 1} \right| + C \]
where \(C\) is the constant of integration.

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