The problem you're presenting involves evaluating a line integral over the unit circle $C'$, oriented counterclockwise. The integral is given by:

$\int_C \left( f_x(x, y) - 2y \right) dx + \left( f_y(x, y) + x \right) dy$

where $f(x, y)$ has continuous second-order partial derivatives.

Step 1: Apply Green's Theorem

To solve this, we can use Green's Theorem, which states that:

$\int_C P(x, y) \, dx + Q(x, y) \, dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA$

where $P(x, y) = f_x(x, y) - 2y$ and $Q(x, y) = f_y(x, y) + x$.

Step 2: Compute the Derivatives

Now, calculate the partial derivatives of $P$ and $Q$:

$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} \left( f_y(x, y) + x \right) = f_{xy}(x, y) + 1$

$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} \left( f_x(x, y) - 2y \right) = f_{xy}(x, y) - 2$

Step 3: Simplify the Expression

Substitute these into Green's Theorem:

$\iint_R \left( \left( f_{xy}(x, y) + 1 \right) - \left( f_{xy}(x, y) - 2 \right) \right) dA$

This simplifies to:

$\iint_R (1 + 2) \, dA = 3 \iint_R 1 \, dA$

The area of the region $R$ is the area of the unit circle, which is $\pi$.

Step 4: Final Calculation

Thus, the integral becomes:

$3 \times \pi = 3\pi$

Conclusion:

The value of the line integral is $3\pi$. Therefore, the correct answer is:

c. 3π

Would you like more details or explanations on any specific part of this solution?

Here are some additional questions for you to explore:

1. How does Green's Theorem relate to vector fields and circulation?
2. What are the conditions for applying Green's Theorem?
3. How do second-order partial derivatives play a role in this problem?
4. What is the interpretation of the unit circle in this problem?
5. How would the solution change if the curve was oriented clockwise?

Tip: When dealing with line integrals over closed curves, always check if Green's Theorem can simplify your work. It often reduces the problem to a double integral over the region enclosed by the curve.