To find the derivative of the function $f(x) = \frac{x^3 + 2x - 1}{x^2 + x}$, we can use the quotient rule. The quotient rule states that for a function $f(x) = \frac{g(x)}{h(x)}$, the derivative $f'(x)$ is given by:

$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$

Here, $g(x) = x^3 + 2x - 1$ and $h(x) = x^2 + x$. We need to find $g'(x)$ and $h'(x)$.

First, we find the derivatives of $g(x)$ and $h(x)$:

1. $g(x) = x^3 + 2x - 1$ $g'(x) = 3x^2 + 2$

2. $h(x) = x^2 + x$ $h'(x) = 2x + 1$

Now, applying the quotient rule:

$f'(x) = \frac{(3x^2 + 2)(x^2 + x) - (x^3 + 2x - 1)(2x + 1)}{(x^2 + x)^2}$

Let's simplify the numerator step-by-step.

1. Expand $(3x^2 + 2)(x^2 + x)$: $(3x^2 + 2)(x^2 + x) = 3x^4 + 3x^3 + 2x^2 + 2x$

2. Expand $(x^3 + 2x - 1)(2x + 1)$: $(x^3 + 2x - 1)(2x + 1) = 2x^4 + x^3 + 4x^2 + 2x - 2x - 1 = 2x^4 + x^3 + 4x^2 - 1$

3. Subtract the two expressions: $(3x^4 + 3x^3 + 2x^2 + 2x) - (2x^4 + x^3 + 4x^2 - 1)$ $= 3x^4 + 3x^3 + 2x^2 + 2x - 2x^4 - x^3 - 4x^2 + 1$ $= (3x^4 - 2x^4) + (3x^3 - x^3) + (2x^2 - 4x^2) + 2x + 1$ $= x^4 + 2x^3 - 2x^2 + 2x + 1$

Therefore, the derivative $f'(x)$ is:

$f'(x) = \frac{x^4 + 2x^3 - 2x^2 + 2x + 1}{(x^2 + x)^2}$

Comparing with the given options, the correct answer is:

$\boxed{(C)}$

Do you want more details or have any questions?

Here are 5 related questions you might find helpful:

1. How do you find the derivative using the product rule?
2. What is the chain rule for differentiation?
3. How do you simplify complex fractions?
4. What are common mistakes to avoid when applying the quotient rule?
5. How do you find the second derivative of a function?

Tip: Always double-check your algebraic simplifications to ensure accuracy in your final answer.