Evaluate $\int \frac{(x^2 + 2) dx}{x + 1}$.

$\frac{x^2}{2} - x + 3 \ln|x + 1| + C$

The correct answer is Option (3) → $\frac{x^2}{2} - x + 3 \ln|x + 1| + C$

Let $I = \int \frac{x^2 + 2}{x + 1} dx$

Divide $(x^2 + 2)$ by $(x + 1)$

Then, $\frac{x^2 + 2}{x + 1} = x - 1 + \frac{3}{x + 1}$

Now, $I = \int \left( x - 1 + \frac{3}{x + 1} \right) dx = \int (x - 1) dx + 3 \int \frac{1}{x + 1} dx$

$= \frac{x^2}{2} - x + 3 \log |(x + 1)| + C$