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

Answer

We know that, (x - a)(x - b) can be written as x2 - (a + b)x + ab.
Then, x2 - (a + b)x + ab = x2 - (a + b)x + $\frac{(a+b)^{2}}{4}-\frac{(a+b)^{2}}{4}+a b$ 
$\Rightarrow \left[\mathrm{x}-\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)\right]^{2}-\frac{(\mathrm{a}-\mathrm{b})^{2}}{4}$ 
$\Rightarrow \int \frac{1}{\sqrt{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})}} \mathrm{dx}=\int \frac{1}{\sqrt{\left\{\mathrm{x}-\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)\right\}^{2}-\frac{(\mathrm{a}-\mathrm{b})^{2}}{4}} \mathrm{d} \mathrm{x}} \mathrm{dx}$ 
Let $x-\left(\frac{a+b}{2}\right)=t$ 
$\Rightarrow$ dx = dt
$\Rightarrow \int \frac{1}{\sqrt{\left\{x-\left(\frac{a+b}{2}\right)\right\}^{2}-\frac{(a-b)^{2}}{4}}} d x=\int \frac{1}{\sqrt{t^{2}-\frac{(a-b)^{2}}{4}}}$ 
$= \log |t+\sqrt{t^{2}-\frac{(a-b)^{2}}{4}}|+C$ 
$=\log \left|\left\{\mathrm{x}-\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)\right\}+\sqrt{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})}\right|+\mathrm{C}$

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