line integral hard function to differentiate $$\int_\gamma \frac{(x^2+y^2-2)\,dx+(4y-x^2-y^2-2) \, dy}{x^2+y^2-2x-2y+2}$$

where $\gamma = 2\sin\left(\frac{\pi x}{2}\right)$ from $(2,0)$ to $(0,0)$.

I think it should be a shortcut to this problem that I cannot see , if that is not the case I will keep trying to simplify it . 

Thnaks in advance
Complete the squares in the denominator $(x-1)^2+(y-1)^2$ and change the variables $x-1\mapsto x$ and $y-1\mapsto y$. You get the vector field
\begin{align}
&amp;\left[\frac{x^2+2x+y^2+2y}{x^2+y^2},\frac{4y-x^2-2x-y^2-2y}{x^2+y^2}\right]=\\
&amp;=\left[1+\frac{2x}{x^2+y^2},-1+\frac{2y}{x^2+y^2}\right]-2\left[\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right].
\end{align}
The first term is conservative (with an easy potential), the second term has zero curl (easy to check).

P.S. Actually the second term has a potential too if you restrict you domain e.g. to the plain without the negative $y$-axis, but the potential is less obvious (easier to get in the polar coordinates).
Hint: Let $I$ denote the integral. Note that the curve can be parameterized using curve using

$$x(t) = t$$
$$y(t) = \sin\left(\frac{\pi x}{2} \right)$$

for $0 \leq t \leq 2$

Thus, separate your integral into two different integrals:

$$\int_\gamma \frac{x^2 + y^2 - 2}{x^2 + y^2 - 2x - 2y + 2} dx \qquad \mathrm{and} \qquad\int_\gamma \frac{4y - x^2 - y^2 - 2}{x^2 + y^2 - 2x - 2y + 2} dy$$

whose sum is equal to $I$.