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

The integral $\int (x^4 + x^2 + 1)d(x^2)$ has $d(x^2)$ instead of $dx$. This means $x^2$ is treated as the variable of integration.

Let $u = x^2$

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The integral becomes:

$\int (x^4 + x^2 + 1)d(x^2)$

Substituting $u = x^2$:

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

$x^2 = u$

$d(x^2) = du$

The integral transforms to:

$\int (u^2 + u + 1)du$

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Integrating term by term:

$\int (u^2 + u + 1)du$

$= \dfrac{u^3}{3} + \dfrac{u^2}{2} + u + C$

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Substituting back $u = x^2$:

$= \dfrac{(x^2)^3}{3} + \dfrac{(x^2)^2}{2} + x^2 + C$

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

Therefore, $\int (x^4 + x^2 + 1)d(x^2) = \dfrac{x^6}{3} + \dfrac{x^4}{2} + x^2 + C$