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Compute the work of the vector field $H: \mathbb{R^2} \setminus{(0,0}) \to \mathbb{R}$

$H(x,y)=\bigg(y^2-\frac{y}{x^2+y^2},1+2xy+\frac{x}{x^2+y^2}\bigg)$

in the path $g(t) = (1-t^2, t^2+t-1)$ with $t\in[-1,1]$

My attempt

So first I considered my vector field as a sum of 2 vector fields: $H = F + G$

$F(x,y)=\bigg(y^2,1+2xy\bigg)$

$G(x,y)=\bigg(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\bigg)$

The vector field $F$ is conservative with one of many potentials $A(x,y) = y^2x+y$ Then I worked out the work using the definition and fundamental theorem of calculus obtaining the value 2.

So no problems at this point.

But $G$ is not a conservative vector field (it doesn't have a potential, even though it's a closed field). How should I proceed? I tried the definition but we get to a very complicated integral... The path isn't closed so we can't apply Green's theorem... What should I do?

EDIT: After simplifying applying the definition

$\int_{-1}^{1} \frac{1+t^2}{2t^4+2t^3-3t^2+1} dt$

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    anyone?????????? – 2017-01-29
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    I'd like to see the integral you get calculating the work done on the particle. – 2017-01-29
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    The work done by the field G? Ok, 5 min and I will edit my question – 2017-01-29
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    The integral doesn't converge. The point $(0,0)$ it's the only one where, if passing over it, the work could diverge, but the path is away from it. I'd check the calculations. – 2017-01-29
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    So how should I proceed then? – 2017-01-29
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    I'm really not getting anywhere this integral is far too complicated there must be an easier solution PLEASE someone help me – 2017-01-29
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    I get this integral $\int_{-1}^1\frac{1+t^2}{2t^4+2t^3-3t^2-2t+2}$ – 2017-01-30

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