Integrate the function: $\frac{1}{\sqrt{(x-1)(x-2)}}$

We need to evaluate the integral $I = \int \frac{1}{\sqrt{(x-1)(x-2)}} dx$.
First,expand the expression inside the square root:
$(x-1)(x-2) = x^2 - 3x + 2$.
Complete the square for the quadratic expression:
$x^2 - 3x + 2 = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4} = (x - \frac{3}{2})^2 - (\frac{1}{2})^2$.
Substitute this back into the integral:
$I = \int \frac{1}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}} dx$.
Let $t = x - \frac{3}{2}$,then $dt = dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{t^2 - a^2}} dt = \log |t + \sqrt{t^2 - a^2}| + C$:
$I = \log |(x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}| + C$.
Simplifying the expression inside the square root back to the original form:
$I = \log |(x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2}| + C$,where $C$ is an arbitrary constant.