The integral $$\int \frac{x^4 + 2}{x^2 + 1} \, dx$$ is equal to

Correct option is A

We have

$$\int \frac{x^4 + 2}{x^2 + 1} \, dx \\= \int \frac{x^4 - 1 + 3}{x^2 + 1} \, dx \\= \int \frac{(x^2 + 1)(x^2 - 1)}{x^2 + 1} \, dx + \int \frac{3}{x^2 + 1} \, dx\\= \int (x^2 - 1) \, dx + 3 \int \frac{dx}{x^2 + 1} \\= \frac{x^3}{3} - x + 3 \tan^{-1}x + C$$​