If $I = \int \frac{x^4 + x^2 + 1}{x^2 - x + 1} \, dx = \alpha x + \beta x^2 + \gamma x^3 + \delta$, $\delta$ is constant of integration, then $(\alpha + 2\beta + 3\gamma)$ equals:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
Given integral: $I = \int \frac{x^4 + x^2 +1}{x^2 - x +1} dx$
Upon Performing long division: $x^4 + x^2 +1$ divided by $x^2 - x +1$
Thus, $\frac{x^4 + x^2 +1}{x^2 - x +1} = x^2 + x + 1$
Integrate: $I = \int (x^2 + x +1) dx = \frac{x^3}{3} + \frac{x^2}{2} + x + \delta$
Compare with $I = \alpha x + \beta x^2 + \gamma x^3 + \delta$:
$\alpha = 1,\ \beta = \frac{1}{2},\ \gamma = \frac{1}{3}$
Compute: $\alpha + 2\beta + 3\gamma = 1 + 1 + 1 = 3$