Problem calculating line integral I have $\gamma=[0,1]\to\mathbb{R}^3$ defined by $\gamma(t)=(\cos(2\pi t), \sin (2\pi t), t^2-t)\;\forall t\in[0,1]$  and I'm asked to calculate $\displaystyle\int_{\gamma}\displaystyle\frac{2xy\mathrm{dx}-(x^2+z^2)\mathrm{dy}+2yz\mathrm{dz}}{(x^2+z^2)^2}$.
I have made an attempt to solve it but it seems that it leads to an integral hard to calculate (and very messy).
I thought that it would be a good idea make the following change of variables: $$x=\cos(2\pi t)\\y=\sin(2\pi t)\\z=t^2-t$$.
Then I'd have
$$\mathrm{dx} = -2\pi\sin(2\pi t) \mathrm{dt} \\ \mathrm{dy =-2\pi\cos(2\pi t)\mathrm{dt}} \\ \mathrm{dz} = 2t-1\;\mathrm{dt}$$.
Now, making the substitution returns a long integral:
$2xy\mathrm{dx}-(x^2+z^2)\mathrm{dy}+2yz\mathrm{dz} = [2\cos(2\pi t)\sin(2\pi t)(-2\pi\sin(2\pi t))dt]-(\cos^2(2\pi t) +t^4-2t^3+t^2)+2[\sin(2\pi t)(t^2-t)(2t-1)\mathrm{dt}]=-4\pi\sin^2(2\pi t)\cos(2\pi t)-\cos^2(2\pi t) -t^4+2t^3-t^2+4t^3\sin(2\pi t)-6t^3\sin(2\pi t)+2t\sin(2\pi t)\;\mathrm{dt} = [\sin(2\pi t)][-4\pi\sin(2\pi t)\cos(2\pi t)+\sin(2\pi t)+4t^3-6t^3+2t]+t^2(-t^2+2t-1)-1.$
And
$(x^2+z^2)^2= (\cos^2 (2\pi t)+t^2-2t^3+t^2)^2$
Which means I should calculate...
$$\int_0^1 \frac{[\sin(2\pi t)][-4\pi\sin(2\pi t)\cos(2\pi t)+\sin(2\pi t)+4t^3-6t^3+2t]+t^2(-t^2+2t-1)-1}{(\cos^2 (2\pi t)+t^2-2t^3+t^2)^2}dt$$.
Is that right?. How badly did I messed up?

(Expanding NotNotLogical's answer)
You got the $dy$ with the wrong sign. Correcting this would maybe lead to a simplification. At any rate, the original author of the problem had the following in mind: Analyzing the force field
$${\bf F}(x,y,z):=\left({2xy\over (x^2+z^2)^2},{-1\over x^2+z^2},{2yz\over (x^2+z^2)^2}\right)$$
one finds that ${\rm curl}\>{\bf F}\equiv{\bf 0}$. It follows that ${\bf F}$ is locally a gradient field, and indeed: The function
$$f(x,y,z):={-y\over x^2+z^2}$$
is well defined on $\gamma$ (check this!) and possesses the gradient
$$\nabla f(x,y,z)={\bf F}(x,y,z)\ .$$
It follows that the integral $J$ in question can be written as
$$J=\int_\gamma \nabla f({\bf x})\cdot d{\bf x}=f\bigl({\rm endpoint}(\gamma)\bigr)-f\bigl({\rm initial\ point}(\gamma)\bigr)=0\ .$$
