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

Explanation

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

### Step 1: Simplify the integrand
We start by simplifying the expression \(\frac{x^2 + 2}{x + 1}\). We can perform polynomial long division or rewrite the numerator:

\[ \frac{x^2 + 2}{x + 1} = \frac{x^2 + 1 + 1}{x + 1} = \frac{x^2 + 1}{x + 1} + \frac{1}{x + 1} \]

Now, we can further simplify \(\frac{x^2 + 1}{x + 1}\):

\[ \frac{x^2 + 1}{x + 1} = x - 1 + \frac{2}{x + 1} \]

Thus, we can rewrite the integral as:

\[ \int \left( x - 1 + \frac{3}{x + 1} \right) \, dx \]

### Step 2: Split the integral
Now we can split the integral into three separate integrals:

\[ \int \left( x - 1 + \frac{3}{x + 1} \right) \, dx = \int x \, dx - \int 1 \, dx + 3 \int \frac{1}{x + 1} \, dx \]

### Step 3: Integrate each term
Now we integrate each term separately:

1. \(\int x \, dx = \frac{x^2}{2}\)
2. \(\int 1 \, dx = x\)
3. \(3 \int \frac{1}{x + 1} \, dx = 3 \ln |x + 1|\)

### Step 4: Combine the results
Combining all the results from the integrations, we get:

\[ \int \frac{x^2 + 2}{x + 1} \, dx = \frac{x^2}{2} - x + 3 \ln |x + 1| + C \]

where \(C\) is the constant of integration.

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

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