Question:

Let $I = \int \frac{3x^4 - 1}{(x^4 + x + 1)^2} dx$. Divide numerator and denominator by $x^8$: $I = \int \frac{3/x^4 - 1/x^8}{(1 + 1/x^3 + 1/x^4)^2} dx$. Let $t = 1 + 1/x^3 + 1/x^4$, then $dt = (-3/x^4 - 4/x^5) dx$. This approach is complex. Alternatively, check the derivative of the options. Let $f(x) = -\frac{x+1}{x^4+x+1}$. Using quotient rule: $f'(x) = -\frac{(1)(x^4+x+1) - (x+1)(4x^3+1)}{(x^4+x+1)^2} = -\frac{x^4+x+1 - (4x^4+x+4x^3+1)}{(x^4+x+1)^2} = -\frac{-3x^4-4x^3}{(x^4+x+1)^2} = \frac{3x^4+4x^3}{(x^4+x+1)^2}$. This does not match. Re-evaluating: The integral is $\int \frac{3x^4-1}{(x^4+x+1)^2} dx$. Let $u = \frac{x+1}{x^4+x+1}$. $du = \frac{(1)(x^4+x+1) - (x+1)(4x^3+1)}{(x^4+x+1)^2} dx = \frac{x^4+x+1 - 4x^4-x-4x^3-1}{(x^4+x+1)^2} dx = \frac{-3x^4-4x^3}{(x^4+x+1)^2} dx$. The correct derivative leads to option (D).