Answer
\[\ln\left|x-2+\sqrt{x^2-4x}\right|+C\]
Work Step by Step
Let \[I=\int\frac{dx}{\sqrt{x^2-4x}}\]
\[I=\int\frac{dx}{\sqrt{(x-2)^2-2^2}}\]
\[\left[\int\frac{1}{\sqrt{x^2-a^2}}dx=\ln\left|x+\sqrt{x^2-a^2}\right|\right]\]
\[I=\ln\left|x-2+\sqrt{(x-2)^2-2^2}\right|+C\]
Where $C$ is constant of integration
\[I=\ln\left|x-2+\sqrt{x^2-4x}\right|+C\]
Hence \[I=\ln\left|x-2+\sqrt{x^2-4x}\right|+C\]
