Consider the integral as attached, To determine the convergence or divergence of the integral, how many improper integrals must be analyzed? What must be true of each of these integrals for the given integral to converge? int_0^3 10/(x^2-2x)dx.

Step 1
Consider the integral ${\displaystyle {\int }_0^3\frac{10}{x^2-2x}dx}$.
To determine the convergence or divergence of the integral,and we need to analyze how many improper integrals are there.
and what must be true of each of these integrals for the given integral to converge.
Step 2
First we will see the dicontinuities of the function ${\displaystyle \frac{10}{x^2-2x}}$
${\displaystyle x^2-2x=0\ge x=0\text{or}x=2}$
We can write
${\displaystyle {\int }_0^3\frac{10}{x^2-2x}dx=\underset{a\to 0^{+}}{lim}{\int }_a^b\frac{10}{x^2-2x}dx+\underset{c\to 2^{-}}{lim}{\int }_b^c\frac{10}{x^2-2x}+\underset{d\to 2^{-}}{lim}{\int }_d^3\frac{10}{x^2-2x}}$
${\displaystyle Letc\in (0,2)}$
The integral must split in three improper integrals to contain one limit per integral. As 0 is the left integral limit we only need the right hand limit.
The dicontinuity at x=2 lies inside the interval (0,3) so we will consider both the limits.
Step 3
Now the limit must exists for all the three integrals to be convergent.
and each of the three integrals must be convergent for the integral ${\displaystyle {\int }_0^3\frac{10}{x^2-2x}}$ dx to be convergent.