Question:

Step 1: Identify the function to be integrated and evaluate the given options. We need to find the primitive (antiderivative) of the function $\frac{3x^4 - 1}{(x^4 + x + 1)^2}$. For multiple-choice questions involving primitives, a common strategy is to differentiate the given options and check which one yields the original function. Let's consider Option B: $y = -\frac{x}{x^4 + x + 1}$. Step 2: Apply the quotient rule to differentiate $y$ with respect to $x$. Let $y = -\frac{x}{x^4 + x + 1}$. We can write this as $y = \frac{-x}{x^4 + x + 1}$. Using the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = -x$ and $v = x^4 + x + 1$. First, find the derivatives of $u$ and $v$: $u' = \frac{d}{dx}(-x) = -1$ $v' = \frac{d}{dx}(x^4 + x + 1) = 4x^3 + 1$ Step 3: Substitute $u, v, u'$, and $v'$ into the quotient rule formula. $$ \frac{dy}{dx} = \frac{(-1)(x^4 + x + 1) - (-x)(4x^3 + 1)}{(x^4 + x + 1)^2} $$ Step 4: Simplify the expression for the derivative. Expand the terms in the numerator: $$ \frac{dy}{dx} = \frac{-x^4 - x - 1 - (-4x^4 - x)}{(x^4 + x + 1)^2} $$ Distribute the negative sign in the numerator: $$ \frac{dy}{dx} = \frac{-x^4 - x - 1 + 4x^4 + x}{(x^4 + x + 1)^2} $$ Combine like terms in the numerator: $$ \frac{dy}{dx} = \frac{(-x^4 + 4x^4) + (-x + x) - 1}{(x^4 + x + 1)^2} $$ $$ \frac{dy}{dx} = \frac{3x^4 - 1}{(x^4 + x + 1)^2} $$ Step 5: Conclude the result. The derivative we obtained, $\frac{dy}{dx} = \frac{3x^4 - 1}{(x^4 + x + 1)^2}$, is exactly the function given in the problem statement whose primitive we need to find. Therefore, the primitive of $\frac{3x^4 - 1}{(x^4 + x + 1)^2}$ is $y = -\frac{x}{x^4 + x + 1} + c$. The final answer is $\boxed{\text{(B)} \frac{x}{x^4 + x + 1} + c}$.