Prove

$(x-1)(x-2)$ can be written as $x^{2}-3 x+2$.

Therefore,

$x^{2}-3 x+2$

$=x^{2}-3 x+\frac{9}{4}-\frac{9}{4}+2$

$=\left(x-\frac{3}{2}\right)^{2}-\frac{1}{4}$

$=\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{3}\right)^{2}$

$\therefore \int \frac{1}{\sqrt{(x-1)(x-2)}} d x=\int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x$

Let $x-\frac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x=\int \frac{1}{\sqrt{t^{2}-\left(\frac{1}{2}\right)^{2}}} d t$

$=\log \left|t+\sqrt{t^{2}-\left(\frac{1}{2}\right)^{2}}\right|+\mathrm{C}$

$=\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^{2}-3 x+2}\right|+\mathrm{C}$