Evaluation of $\int_{0}^{2}\frac{(2x-2)dx}{2x-x^2}$ 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?

The second gives the correct answer, but both solutions are, strictly speaking, incorrect: Since the integrand has singularities at both ends of the interval of integral, we define the value of the integral to be
$$\int_0^2 \frac{(2 x - 2)}{(2 x - x^2)} dx = \lim_{a \searrow 0} \int_a^c \frac{(2 x - 2)}{(2 x - x^2)} dx + \lim_{b \nearrow 2} \int_c^b \frac{(2 x - 2)}{(2 x - x^2)} dx$$
for some $c \in (a, b)$; it follows from elementary properties of definite integrals that this value does not depend on $c$.
For each $0 < a < b < 2$, the two integrals on the r.h.s. are proper, and they can be handled with the substitution $u = 2 x - x^2$, $du = 2 - 2 x$; we get
$$- \lim_{a \searrow 0} \int_{2a - a^2}^{2 c - c^2} \frac{du}{u} - \lim_{b \nearrow 2} \int_{2c - c^2}^{2b - b^2} \frac{du}{u} .$$ Setting $a' = 2 a - a^2$ and likewise defining $b', c'$ gives
$$- \lim_{a' \searrow 0} \int_{a'}^{c'} \frac{du}{u} - \lim_{b' \searrow 0} \int_{c'}^{b'} \frac{du}{u} .$$ The first limit is
$$\lim_{a' \searrow 0} \int_{a'}^{c'} \frac{du}{u} = \lim_{a' \searrow 0} \log u \vert_{a'}^{c'} = \lim_{a' \searrow 0} (\log c' - \log a') = +\infty .$$ So, the value of the integral itself does not exist, though we can denote its behavior in this case by $\int_0^2 \frac{(2 x - 2)}{(2 x - x^2)} dx = -\infty$.
