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

$(x-a)(x-b)$ can be written as $x^{2}-(a+b) x+a b$.
Therefore,
$x^{2}-(a+b) x+a b = \left[x-\left(\frac{a+b}{2}\right)\right]^{2}-\frac{(a-b)^{2}}{4}$.
$\int \frac{1}{\sqrt{(x-a)(x-b)}} d x = \int \frac{1}{\sqrt{\left[x-\left(\frac{a+b}{2}\right)\right]^{2}-\frac{(a-b)^{2}}{4}}} d x$.
Let $x-\left(\frac{a+b}{2}\right)=t$,then $d x=d t$.
$\int \frac{1}{\sqrt{t^{2}-\left(\frac{a-b}{2}\right)^{2}}} d t = \log \left| t + \sqrt{t^{2}-\left(\frac{a-b}{2}\right)^{2}} \right| + C$.
Substituting back $t = x-\left(\frac{a+b}{2}\right)$,we get:
$\log \left| \left(x-\frac{a+b}{2}\right) + \sqrt{(x-a)(x-b)} \right| + C$.