Answer
The given improper integral
$$
\int_{-1}^{1} \frac{dx}{x^{2}-2x}
$$
is divergent.
Work Step by Step
$$
\int_{-1}^{1} \frac{dx}{x^{2}-2x}
$$
Observe that the given integral is improper because $ f(x)= \frac{1}{x^{2}-2x} $ has the vertical asymptote $x=0 $. Since the infinite discontinuity occurs at the middle of the interval $ [-1,1] $ at $x = 0$, we must use part (c) of Definition 3 with $ c=0$:
$$
\int_{-1}^{1} \frac{dx}{x^{2}-2x}=\int_{-1}^{0} \frac{dx}{x^{2}-2x}+ \int_{0}^{1} \frac{dx}{x^{2}-2x}
$$
Now, decompose $\frac{1}{x^{2}-2x} $ into its partial fractions as follows,
$$
\frac{1}{x^{2}-2x} =\frac{1}{x\left(x-2\right)}=\frac{A} { x}+ \frac{B} { \left(x-2\right)}
$$
$\Rightarrow \quad 1=A(x-2)+Bx= (A+B)x -2A $ by comparison of the coefficients, we get $A=-\frac{1}{2}, B=\frac{1}{2}.$
Thus
$$
\frac{1}{x^{2}-2x} =\frac{-\frac{1}{2}} { x}+ \frac{\frac{1}{2}} { \left(x-2\right)}
$$
Now integrate the decomposed version.
$$ \begin{split}
\int_{0}^{1} \frac{dx}{x^{2}-2x} &= \int_{0}^{1} \left( \frac{-\frac{1}{2}} { x}+ \frac{\frac{1}{2}} { \left(x-2\right)}\right) dx \\
&= \lim _{t \rightarrow 0^{+}} \int_{t}^{1}\left(\frac{-\frac{1}{2}}{x}+\frac{\frac{1}{2}}{x-2}\right) d x \\
& = \lim _{t \rightarrow 0^{+}}\left[-\frac{1}{2} \ln |x|+\frac{1}{2} \ln |x-2|\right]_{t}^{1} \\
& = \lim _{t \rightarrow 0^{+}}\left[(0+0)-\left(-\frac{1}{2} \ln t+\frac{1}{2} \ln |t-2|\right)\right] \\
& = -\frac{1}{2} \ln 2+\frac{1}{2} \lim _{t \rightarrow 0^{+}} \ln t=-\infty
\end{split} $$
so the integral
$$
\int_{0}^{1} \frac{dx}{x^{2}-2x}
$$
is divergent.
Thus the given improper integral
$$
\int_{-1}^{1} \frac{dx}{x^{2}-2x}
$$
is divergent.
