Answer
$=\ln|x-2+\sqrt{x^{2}-4x}|+D $
Work Step by Step
$\displaystyle \int\frac{dx}{\sqrt{x^{2}-4x}}$ = ... expand to perfect square under the square root....
$=\displaystyle \int\frac{dx}{\sqrt{(x^{2}-4x+4)-4}}$
$=\displaystyle \int\frac{dx}{\sqrt{(x-2)^{2}-2^{2}}}$
We have a $\sqrt{u^{2}-a^{2}}^{ }$ form here, try the trigonometric substitution
$\left[\begin{array}{ll}
x-2=2\sec\theta & \\
dx=2\sec\theta\tan\theta d\theta &
\end{array}\right] $
$=\displaystyle \int\frac{2\sec\theta\tan\theta}{2\tan\theta}d\theta$ ... reduce 2 and tan ...
$=\displaystyle \int\sec\theta d\theta$
$=\ln|\sec\theta+\tan\theta|+C$
... bring back x
$\left[\begin{array}{llll}
\sec\theta = \frac{x-2}{2} & & & \\
\tan^{2}=\sec^{2}\theta-1 & \Rightarrow & \tan^{2}\theta & =(\frac{x-2}{2})^{2}-1\\
& & & =\frac{x^{2}-4x+4-4}{4}\\
& & \tan\theta & =\frac{\sqrt{x^{2}-4x}}{2}
\end{array}\right]$
$=\displaystyle \ln|\frac{x-2}{2}+\frac{\sqrt{x^{2}-4x}}{2}|+C$
$=\displaystyle \ln|\frac{1}{2}(x-2+\sqrt{x^{2}-4x})|+C$
$=\ln|x-2+\sqrt{x^{2}-4x}|-\ln 2+C$
$=\ln|x-2+\sqrt{x^{2}-4x}|+D,\qquad $, where $D=C-\ln 2$
