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

$$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}=\int_\gamma\frac{-y^2}{x^2+y^4}\,dx+\int_\gamma\frac{2xy}{x^2+y^4}\,dy$$
$$\gamma(t)=(t,2 t^2-2);\;t\in[-1,1];\;\gamma'(t)=(1, 4 t)$$
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
$$-\int_{-\infty}^{\infty} \frac{du}{1+u^2}=-\left[\arctan u\right]_{-\infty}^{\infty}=-\pi$$
