Question

The value of $\int \frac{dx}{(x-\beta )\sqrt{(x-\alpha )(\beta -x)}}$, is equal to

• A. $\frac{-1}{\beta -\alpha }\sqrt{\frac{x-\alpha }{\beta -x}}+C$
• B. $\frac{1}{\beta -\alpha }\sqrt{\frac{x-\alpha }{\beta -x}}+C$
• C. $\frac{2}{\alpha -\beta }\sqrt{\frac{x-\alpha }{\beta -x}}+C$
• D. none of these

Answer 1

The correct option is C $\frac{2}{\alpha -\beta }\sqrt{\frac{x-\alpha }{\beta -x}}+C$

Let $I=\int \frac{dx}{(x-\beta )\sqrt{(x-\alpha )(\beta -x)}}$
$=-\int \frac{dx}{(\beta -x{)}^2\sqrt{\frac{x-\alpha }{\beta -x}}}$
Put $\frac{x-\alpha }{\beta -x}=t$
$\Rightarrow \frac{\beta -\alpha }{(\beta -x{)}^2}dx=dt$
So, $I=-\frac{1}{\beta -\alpha }\int \frac{dt}{\sqrt{t}}$
$=\frac{2}{\alpha -\beta }{t}^{\frac{1}{2}}+C$
$I=\frac{2}{\alpha -\beta }\sqrt{\frac{x-\alpha }{\beta -x}}+C$