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

Answer

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

### Step 1: Rewrite the integrand
First, we can rewrite the integrand by performing polynomial long division. We divide \(x^2 + 1\) by \(x + 1\).

\[ x^2 + 1 = (x + 1)(x - 1) + 2 \]

This means we can express the integrand as:

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

### Step 2: Split the integral
Now we can split the integral into two parts:

\[ \int \frac{x^2 + 1}{x + 1} \, dx = \int (x - 1) \, dx + \int \frac{2}{x + 1} \, dx \]

### Step 3: Integrate each part
Now we will integrate each part separately.

1. For the first integral \(\int (x - 1) \, dx\):

\[ \int (x - 1) \, dx = \frac{x^2}{2} - x \]

2. For the second integral \(\int \frac{2}{x + 1} \, dx\):

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

### Step 4: Combine the results
Now we combine the results of the two integrals:

\[ \int \frac{x^2 + 1}{x + 1} \, dx = \left(\frac{x^2}{2} - x\right) + 2 \ln |x + 1| + C \]

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

### Final Answer
Thus, the final answer is:

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