compute $\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}$ 
The question is:
$$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}
\quad \gamma:r(t)=(t,2t^2-2), \quad -1\leq t\leq 1
$$

I have tried to solve it like this:
since $Q_x=P_y$ it's potential vector field but the singularity is at origin, first i thought that i could evaluate it from $(-1,0)$ to $(1,0)$ along $x$ axis but the singularity is the problem, how should i proceed in this case?
Any suggestion would be great Thanks
 A: I will present two options to avoid doing the integral directly.
$\textbf{Option 1}$: Fundamental Theorem of Line Integrals
Since the vector field is conservative on any domain that doesn't contain the origin, we can find a potential function
$$f(x,y) = -\arctan\left(\frac{x}{y^2}\right)$$
There are many options for the potential function here, but this one is continuous on the lower half plane $y<0$. Being careful how we take limits, we approach the endpoints of the integral $(1,0)$ and $(-1,0)$ from below
$$I = \lim_{(x,y)\to(1,0^-)}-\arctan\left(\frac{x}{y^2}\right)+\lim_{(x,y)\to(-1,0^-)}\arctan\left(\frac{x}{y^2}\right) = -\frac{\pi}{2}-\frac{\pi}{2}=-\pi$$
$\textbf{Option 2}$: Green's Theorem/Partial Path Independence
As long we have two paths that (when taken together) do not enclose the origin, then we have path independence. This can be proven via Green's theorem. In this case we will automatically have path independence if we restrict our attention to paths only in the lower half plane. In this case consider the curve
$$x^2+y^4=1 \hspace{20 pt} y\leq 0$$
This curve is chosen because
(a) it simplifies the denominator to a constant and
(b) it contains the start and end points of the original curve.
We can parametrize this curve by
$$\begin{cases}x(t) = \cos t \\ y(t) = -\sqrt{-\sin t} \end{cases} \hspace{24 pt} t\in[-\pi,0] \hspace{18 pt}\implies \hspace{18 pt} \begin{cases}x'(t) = -\sin t \\ y'(t) = \frac{\cos t}{2\sqrt{-\sin t}} \end{cases}$$
Plugging into the integral gives us
$$\int_{-\pi}^0-\sin^2t-\cos^2t\:dt = -\pi$$
A: $$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}=\int_\gamma\frac{-y^2}{x^2+y^4}\,dx+\int_\gamma\frac{2xy}{x^2+y^4}\,dy$$
$$\gamma(t)=(t,2 t^2-2);\;t\in[-1,1];\;\gamma'(t)=(1, 4 t)$$
Substitute
$$\int_{-1}^1 -\frac{\left(2 t^2-2\right)^2}{\left(2 t^2-2\right)^4+t^2}\cdot 1\,dt+\int_{-1}^1 \frac{2 t \left(2 t^2-2\right)}{\left(2 t^2-2\right)^4+t^2}\cdot(4t)\,dt=$$
$$=\int_{-1}^1 \frac{4 \left(3 t^4-2 t^2-1\right)}{16 t^8-64 t^6+96 t^4-63 t^2+16}\,dt$$
Integrating is quite hard
denominator $16 t^8-64 t^6+96 t^4-63 t^2+16=(16 t^8-64 t^6+96 t^4-64 t^2+16)+t^2=16 \left(t^2-1\right)^4+t^2$
numerator can be factored $4 \left(3 t^4-2 t^2-1\right)=4\left(t^2-1\right) \left(3 t^2+1\right)$
and integral can be written as
$$\int_{-1}^1\frac{4\left(t^2-1\right) \left(3 t^2+1\right)}{16 \left(t^2-1\right)^4+t^2}\,dt$$
Now divide numerator and denominator by $16 \left(t^2-1\right)^4$
$$\int_{-1}^1\frac{\frac{4\left(t^2-1\right) \left(3 t^2+1\right)}{16 \left(t^2-1\right)^4}}{1+\frac{t^2}{16 \left(t^2-1\right)^4}}\,dt=\int_{-1}^1\frac{\frac{ 3 t^2+1}{4 \left(t^2-1\right)^3}}{1+\frac{t^2}{16 \left(t^2-1\right)^4}}\,dt$$
Now substitute $\frac{t}{4\left(t^2-1\right)^2}=u$
extremes become $t=-1\to u=-\infty;\;t=1\to u=+\infty$
while differentiating we get $\frac{3 t^2+1}{4 \left(t^2-1\right)^3}\,dt=-du$
Therefore the integral becomes
$$-\int_{-\infty}^{\infty} \frac{du}{1+u^2}=-\left[\arctan u\right]_{-\infty}^{\infty}=-\pi$$
A: You may also note that by changing variable $y\to{y'}=y^2$ you get:
$\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}=\int\frac{-y'dx+x\ dy'}{x^2+y'^2}$.
We know that the integrand is a potential vector field, so the integral depends only on the starting and finishing points and does not depend on the way - we just have to avoid the singularity at $(0,0)$.
The starting point for $(x,y')$ is $(-1,0)$ and finishing $(1,0)$. Now we can introduce a parametrization $x=\cos(t)$ and $y'=\sin(t)$, where $t$ is changing from $\pi$ to $0$ - to meet the requirements for the starting and finishing points.
After that the integration is trivial: $\int_{(-1.0)}^{(1,0)}\frac{-y'dx+x\ dy}{x^2+y'^2}=\int_{\pi}^{0}\frac{-\sin(s)(-\sin(s))ds+\cos(s)\cos(s) ds}{\cos^2(s)+\sin^2(s)}=\int_\pi^0ds=-\pi$
