Evaluate the following integrals:
`int frac{(x^2+2)}{x+1}dx`

Answer

To evaluate the integral \( \int \frac{x^2 + 2}{x + 1} \, dx \), we can start by simplifying the integrand through polynomial long division since the degree of the numerator is higher than that of the denominator.

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

1. Divide the leading term: \( \frac{x^2}{x} = x \).
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 rewrite the integral:
\[ \int \frac{x^2 + 2}{x + 1} \, dx = \int \left( x + \frac{2 - x}{x + 1} \right) \, dx. \]

### Step 2: Split the Integral
Now 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 3: Evaluate the First Integral
The first integral is straightforward:
\[ \int x \, dx = \frac{x^2}{2} + C_1. \]

### Step 4: Simplify the Second Integral
For the second integral \( \int \frac{2 - x}{x + 1} \, dx \), we can separate it:
\[ \int \frac{2 - x}{x + 1} \, dx = \int \left( \frac{2}{x + 1} - \frac{x}{x + 1} \right) \, dx = \int \frac{2}{x + 1} \, dx - \int 1 \, dx. \]

### Step 5: Evaluate the Second Integral
1. The integral \( \int \frac{2}{x + 1} \, dx = 2 \ln |x + 1| + C_2 \).
2. The integral \( \int 1 \, dx = x + C_3 \).

Combining these results:
\[ \int \frac{2 - x}{x + 1} \, dx = 2 \ln |x + 1| - x + C_4. \]

### Step 6: Combine All Parts
Now we combine all parts:
\[ \int \frac{x^2 + 2}{x + 1} \, dx = \frac{x^2}{2} + 2 \ln |x + 1| - x + C, \]
where \( C = C_1 + C_2 + C_3 + C_4 \).

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