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

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

The integral is taken with respect to \(x^2\), not x. Let \(t = x^2\); then \(x^4 = t^2\):

    \( \displaystyle \int (x^4 + x^2 + 1)\,d(x^2) = \int (t^2 + t + 1)\,dt \)

    \( \displaystyle = \frac{t^3}{3} + \frac{t^2}{2} + t + C \)

    \( \displaystyle = \frac{x^6}{3} + \frac{x^4}{2} + x^2 + C \)