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

The numerator has degree 4 and the denominator has degree 2. Polynomial long division is needed to simplify before integrating.

Dividing $(x^4 + x^2 + 1)$ by $(x^2 - x + 1)$:

First term: $x^4 ÷ x^2 = x^2$

$x^2(x^2 - x + 1) = x^4 - x^3 + x^2$

$(x^4 + x^2 + 1) - (x^4 - x^3 + x^2) = x^3 + 1$

Second term: $x^3 ÷ x^2 = x$

$x(x^2 - x + 1) = x^3 - x^2 + x$

$(x^3 + 1) - (x^3 - x^2 + x) = x^2 - x + 1$

Third term: $x^2 ÷ x^2 = 1$

$1(x^2 - x + 1) = x^2 - x + 1$

$(x^2 - x + 1) - (x^2 - x + 1) = 0$

Therefore:

$\frac{x^4 + x^2 + 1}{x^2 - x + 1} = x^2 + x + 1$

---

$I = \int (x^2 + x + 1) dx$

$I = \frac{x^3}{3} + \frac{x^2}{2} + x + \delta$

---

The result can be written as:

$I = x + \frac{1}{2}x^2 + \frac{1}{3}x^3 + \delta$

Comparing with $I = \alpha x + \beta x^2 + \gamma x^3 + \delta$:

$\alpha = 1$

$\beta = \frac{1}{2}$

$\gamma = \frac{1}{3}$

---

$\alpha + 2\beta + 3\gamma = 1 + 2\left(\frac{1}{2}\right) + 3\left(\frac{1}{3}\right)$

$= 1 + 1 + 1$

$= 3$

Therefore, $(\alpha + 2\beta + 3\gamma) = 3$