Evaluate $$I=\int\_{0}^{2}\frac{(2x-2)dx}{2x-x^2}$$

I used two different methods to solve this. 

Method $1.$ Using the property that if $f(2a-x)=-f(x)$ Then $$\int\_{0}^{2a}f(x)dx=0$$

Now $$f(x)=\frac{(2x-2)}{2x-x^2}$$ So

$$f(2-x)=\frac{2(2-x)-2}{2(2-x)-(2-x)^2}=\frac{2-2x}{2x-x^2}=-f(x)$$

Hence $$I=0$$

Method $2.$ By partial fractions $$I=\int\_{0}^{2}\frac{-1}{x}-\frac{1}{x-2} dx$$

So

$$I=-\log |x| -\log |x-2| \vert\_{0}^{2}$$ which gives $$I=\infty$$

which is correct?