Question

Integrate the following functions.
$\int \frac{1}{\sqrt{(x-a)(x-b)}}dx$

Answer 1

Let $I=\int \frac{1}{\sqrt{(x-a)(x-b)}}dx=\int \frac{1}{\sqrt{x^2-(a+b)x+ab}}dx$
$=\int \frac{1}{\sqrt{[x^2-(a+b)x+(\frac{a+b}{2}{)}^2-(\frac{a+b}{2}{)}^2+ab]}}dx$
$=\int \frac{1}{\sqrt{[x-(\frac{a+b}{2}){]}^2+ab-(\frac{a+b}{2}{)}^2}}dx=\int \frac{1}{\sqrt{[x-(\frac{a+b}{2}){]}^2+ab-(\frac{a^2+b^2+2ab}{4})}}dx$

$=\int \frac{1}{\sqrt{[x-(\frac{a+b}{2}){]}^2+\frac{4ab-a^2-b^2-2ab}{4}}}dx=\int \frac{1}{\sqrt{[x-(\frac{a+b}{2}){]}^2+\frac{2ab-a^2-b^2}{4}}}dx=\int \frac{1}{\sqrt{[x-(\frac{a+b}{2}){]}^2-(\frac{a^2+b^2-2ab}{4})}}dx=\int \frac{1}{\sqrt{(x-\frac{a+b}{2}{)}^2-(\frac{a-b}{2}{)}^2}}dxLetx-\frac{a+b}{2}=t\Rightarrow dx=dt\therefore I=\int \frac{1}{\sqrt{t^2-(\frac{a-b}{2}{)}^2}}dt=log|t+\sqrt{t^2-(\frac{a-b}{2}{)}^2}|+C[\because \int \frac{dx}{\sqrt{x^2-a^2}}=log|x+\sqrt{x^2-a^2}|]=log\left|\right(x-\frac{a+b}{2})+\sqrt{(x-\frac{a+b}{2}{)}^2-(\frac{a-b}{2}{)}^2}|+C(\because t=x-\frac{(a+b)}{2})=log\left|\right(x-\frac{a+b}{2})+\sqrt{(x-a)(x-b)}|+C$