Let $C$ be parametrization $\mathbf{r}=\space5\cos(t) \mathbf{i}+\space4\sin(t) \mathbf{j}\space, t \in [0, 2\pi]$. Calculate $\oint_C \mathbf{F}\cdot d \mathbf{r}$ where $F$ is vector field
$$F(x,y)= \frac{-4x^3y}{(x^4+y^4)^2} \mathbf{i}+\frac{x^4-3y^4}{(x^4+y^4)^2} \mathbf{j}$$
Using Green's theorem:
\begin{align} &\oint_C \bigg(\frac{-4x^3y}{(x^4+y^4)^2} \mathbf{i}+\frac{x^4-3y^4}{(x^4+y^4)^2} \mathbf{j} \bigg) \cdot d \mathbf{r} \\\\ &= \iint_R \bigg[\frac{\partial}{\partial x} \bigg(\frac{x^4-3y^4}{(x^4+y^4)^2 } \bigg)- \frac{\partial}{\partial y} \bigg(\frac{-4x^3y}{(x^4+y^4)^2} \bigg)\bigg] dA \end{align}
But the vector field presents a singularity on $(0,0)$ so I cannot directly apply Green's. I use Extended Green’s Theorem instead.
$$\oint_CF-\oint_LF=\iint_R (\nabla×F) ·k\space dA = 0$$
$$\oint_CF=\oint_LF$$ Which $L$ will let me calculate the line integral in the easiest way possible?
PD: I've already tried with a square $(|x|=1\space |y|=1)$ but it just gives me more problems.
I don't know what problems the square gave you. It works excellently:
Define $$\mathbf r(t) = \begin{cases}(t+3)\mathbf{i} - \mathbf{j}& t \in [-4,-2)\\ (\mathbf{i} + (t+1)\mathbf{j}& t\in [-2, 0)\\ (1 - t)\mathbf{i} + \mathbf{j}& t \in [0,2)\\ -\mathbf{i} + (3-t)\mathbf{j}& t\in [2, 4)\end{cases}$$ Then $$\dot{\mathbf r}(t) = \begin{cases}\mathbf{i}& t \in [-4,-2)\\ \mathbf{j}& t\in [-2, 0)\\ -\mathbf{i}& t \in [0,2)\\ -\mathbf{j}& t\in [2, 4)\end{cases}$$ So $$\begin{align}\oint_LF\cdot d\mathbf r = &\int_{-4}^{-2} \frac{4(t+3)^3}{((t+3)^4+1)^2} dt\\ &+\int_{-2}^{0}\frac{1-3(t+1)^4}{(1+(t+1)^4)^2}dt\\ &+\int_{0}^{2} \frac{4(1-t)^3}{((1-t)^4+1)^2} dt\\ &+\int_{2}^{4}\frac{3(3-t)^4-1}{(1+(3-t)^4)^2}dt\\=&\int_{-1}^{1} \frac{4t^3}{(t^4+1)^2} dt\\ &+\int_{-1}^{1}\frac{1-3t^4}{(1+t^4)^2}dt\\ &+\int_{-1}^{1} \frac{4t^3}{(t^4+1)^2} dt\\ &+\int_{-1}^{1}\frac{3t^4-1}{(1+t^4)^2}dt\end{align}$$
For the first and third integrals, the integrand is odd, so the integrals are $0$. The second and fourth integrals are opposites, so their sum is $0$.