Solution
= Solution
Use <partial fraction decomposition>:
$$
\frac1{x(x-4)}=-\frac1{4x}+\frac1{4(x-4)}.
$$
Integrating the two logarithmic <derivatives> gives
$$
\boxed{\int\frac{dx}{x^2-4x}=\frac14\log\left|\frac{x-4}{x}\right|+C.}
$$
This holds on each interval avoiding the two <poles> $0$ and $4$, with an independent integration constant on each interval.