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

Explanation

To solve the integral \(\int \frac{x^2 + 2}{x + 1} \, dx\), we can use polynomial long division to simplify the integrand. Here’s a step-by-step solution:

### Step 1: Polynomial Long Division
We divide \(x^2 + 2\) by \(x + 1\).

1. Divide the leading term: \(x^2 \div x = x\).
2. Multiply \(x\) by \(x + 1\): \(x(x + 1) = x^2 + x\).
3. Subtract:
\[ (x^2 + 2) - (x^2 + x) = 2 - x \]

So, we can express the integrand as:
\[ \frac{x^2 + 2}{x + 1} = x + \frac{2 - x}{x + 1} \]

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

### Step 3: Split the Integral
We can split the integral into two parts:
\[ \int \left( x + \frac{2 - x}{x + 1} \right) \, dx = \int x \, dx + \int \frac{2 - x}{x + 1} \, dx \]

### Step 4: Integrate the First Part
The first integral is straightforward:
\[ \int x \, dx = \frac{x^2}{2} \]

### Step 5: Simplify the Second Integral
Now, we need to integrate \(\frac{2 - x}{x + 1}\). We can split this into two separate fractions:
\[ \frac{2 - x}{x + 1} = \frac{2}{x + 1} - \frac{x}{x + 1} \]

### Step 6: Integrate the Second Part
Now we can integrate each term:
1. \(\int \frac{2}{x + 1} \, dx = 2 \ln |x + 1|\)
2. \(\int \frac{x}{x + 1} \, dx\): We can use substitution or recognize that \(\frac{x}{x + 1} = 1 - \frac{1}{x + 1}\):
\[ \int \frac{x}{x + 1} \, dx = \int (1 - \frac{1}{x + 1}) \, dx = x - \ln |x + 1| \]

### Step 7: Combine All Parts
Now we can combine all the parts together:
\[ \int \frac{x^2 + 2}{x + 1} \, dx = \frac{x^2}{2} + \left(2 \ln |x + 1| - (x - \ln |x + 1|)\right) + C \]
This simplifies to:
\[ \int \frac{x^2 + 2}{x + 1} \, dx = \frac{x^2}{2} + 3 \ln |x + 1| - x + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{x^2 + 2}{x + 1} \, dx = \frac{x^2}{2} - x + 3 \ln |x + 1| + C \]