$\int(x^4+x^2+1)d(x^2)$ is equal to: (where c is an integration constant)

$\frac{x^6}{3} + \frac{x^4}{2} + x^2 + C$

The correct answer is Option (3) → $\frac{x^6}{3} + \frac{x^4}{2} + x^2 + C$

Given integral: $\int (x^4 + x^2 + 1) \, d(x^2)$

Substitute $u = x^2 \Rightarrow du = d(x^2)$

$x^4 + x^2 + 1 = (x^2)^2 + x^2 + 1 = u^2 + u + 1$

Integral becomes: $\int (u^2 + u + 1) \, du = \frac{u^3}{3} + \frac{u^2}{2} + u + c$

Substitute back $u = x^2$:

$\int (x^4 + x^2 + 1) \, d(x^2) = \frac{(x^2)^3}{3} + \frac{(x^2)^2}{2} + x^2 + c = \frac{x^6}{3} + \frac{x^4}{2} + x^2 + c$