Question

The value of integral $\int \sqrt{\frac{x-2}{x-3}}dx$ is equal to
(Where $C$ is integration constant)

• A. $-\sqrt{(3-x)(2-x)}+\ln |\sqrt{3-x}+\sqrt{2-x}|+C$
• B. $-\frac{1}{2}\sqrt{(3-x)(2-x)}-\ln |\sqrt{3-x}+\sqrt{2-x}|+C$
• C. $-\frac{1}{2}\sqrt{(3-x)(2-x)}-\frac{1}{2}\ln |\sqrt{3-x}+\sqrt{2-x}|+C$
• D. $\frac{1}{2}\sqrt{(3-x)(2-x)}-\frac{1}{2}\ln |\sqrt{3-x}+\sqrt{2-x}|+C$

Answer 1

The correct option is A $-\sqrt{(3-x)(2-x)}+\ln |\sqrt{3-x}+\sqrt{2-x}|+C$

Let $I=\int \sqrt{\frac{x-2}{x-3}}dx$
Substituting, $x=2{\sec }^{2}y-3{\tan }^{2}y$
$\Rightarrow dx=-2{\sec }^{2}y\tan ydy$
and $-{\sec }^{2}y+3=x,\Rightarrow \sec y=\sqrt{3-x},\tan y=\sqrt{2-x}$
The integral becomes,
$I=\int \sqrt{\frac{-{\tan }^{2}y}{-{\sec }^{2}y}}\cdot (-2{\sec }^{2}y\tan y)dy=-2\int \sec y{\tan }^{2}ydy=2\int \underset{I}{\sin }y\cdot \underset{II}{\frac{-\sin y}{{\cos }^{3}y}}dy=2[\frac{\sin y}{-2{\cos }^{2}y}+\frac{1}{2}\int \frac{\cos y}{{\cos }^{2}y}dy]=-\sec y\tan y+\ln |\sec y+\tan y|+C=-\sqrt{(3-x)(2-x)}+\ln |\sqrt{3-x}+\sqrt{2-x}|+C$