Evaluate `int(x^(2)+x+1)/(x^(2)-1)dx`

Explanation

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

### Step 1: Long Division
Since the degree of the numerator \(x^2 + x + 1\) is equal to the degree of the denominator \(x^2 - 1\), we perform polynomial long division.

1. Divide \(x^2\) by \(x^2\) to get \(1\).
2. Multiply \(1\) by \(x^2 - 1\) to get \(x^2 - 1\).
3. Subtract \((x^2 - 1)\) from \((x^2 + x + 1)\):
\[ (x^2 + x + 1) - (x^2 - 1) = x + 2 \]

So, we can rewrite the integral as:
\[ \int \left(1 + \frac{x + 2}{x^2 - 1}\right) \, dx \]

### Step 2: Separate the Integral
Now we can separate the integral:
\[ \int 1 \, dx + \int \frac{x + 2}{x^2 - 1} \, dx \]

### Step 3: Evaluate the First Integral
The first integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 4: Evaluate the Second Integral
Now we need to evaluate \(\int \frac{x + 2}{x^2 - 1} \, dx\). We can split this into two parts:
\[ \int \frac{x}{x^2 - 1} \, dx + \int \frac{2}{x^2 - 1} \, dx \]

#### Part A: Evaluate \(\int \frac{x}{x^2 - 1} \, dx\)
For this integral, we can use substitution:
Let \(u = x^2 - 1\), then \(du = 2x \, dx\) or \(\frac{1}{2} du = x \, dx\).

Thus, we have:
\[ \int \frac{x}{x^2 - 1} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |x^2 - 1| + C \]

#### Part B: Evaluate \(\int \frac{2}{x^2 - 1} \, dx\)
The integral \(\int \frac{2}{x^2 - 1} \, dx\) can be simplified using partial fractions:
\[ \frac{2}{x^2 - 1} = \frac{2}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} \]
Multiplying through by the denominator \((x - 1)(x + 1)\) gives:
\[ 2 = A(x + 1) + B(x - 1) \]
Setting \(x = 1\):
\[ 2 = A(2) \Rightarrow A = 1 \]
Setting \(x = -1\):
\[ 2 = B(-2) \Rightarrow B = -1 \]
Thus, we have:
\[ \frac{2}{x^2 - 1} = \frac{1}{x - 1} - \frac{1}{x + 1} \]
Now we can integrate:
\[ \int \frac{2}{x^2 - 1} \, dx = \int \left(\frac{1}{x - 1} - \frac{1}{x + 1}\right) \, dx = \ln |x - 1| - \ln |x + 1| + C = \ln \left|\frac{x - 1}{x + 1}\right| + C \]

### Step 5: Combine All Parts
Combining all parts, we have:
\[ \int \frac{x^2 + x + 1}{x^2 - 1} \, dx = x + \frac{1}{2} \ln |x^2 - 1| + \ln \left|\frac{x - 1}{x + 1}\right| + C \]

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