Integrate the function: 1 (x-a)(x-b) - Vidyadip

Question

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

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Answer

We know that, $(x - a)(x - b)$ can be written as $x^2 - (a + b)x + ab.$
Then, $ x^2 - (a + b)x + ab = x^2 - (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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