Question

$\int \sqrt{\frac{x}{x-1}}dx,x\in (0,\pi /2)$ equals

• A. $\sqrt{x(x-1)}+\log (\sqrt{x}+\sqrt{x-1})+c$
• B. $\sqrt{x(x-1)}-\log (\sqrt{x}+\sqrt{x-1})+c$
• C. $\sqrt{x(x-1)}+\log (\sqrt{x}-\sqrt{x-1})+c$
• D. $\sqrt{x(x+1)}-\log (\sqrt{x}-\sqrt{x+1})+c$

Answer 1

The correct option is A $\sqrt{x(x-1)}+\log (\sqrt{x}+\sqrt{x-1})+c$

$x=se{c}^2\theta$

$dx=2se{c}^2\theta \cdot tan\theta \cdot d\theta$

$\int \sqrt{\frac{se{c}^2\theta }{ta{n}^2\theta }}\cdot 2se{c}^2\theta tan\theta d\theta$

$\int \frac{1}{sin\theta }\cdot 2se{c}^2\theta \cdot \frac{sin\theta }{cos\theta }\cdot d\theta$

$=2\int se{c}^3\theta \cdot d\theta \cdot$

$=2\int \left(sec\theta \right)\left(se{c}^2\theta \right)d\theta \cdot$

$=2[sec\theta tan\theta -\int (sec\theta tan\theta \cdot tan\theta )d\theta \cdot ]$

$2\int se{c}^3{\theta }^{d\theta }=2\left[sec\theta tan\theta \right]-2\int sec\theta (se{c}^2\theta -1)d\theta$

$2\int se{c}^3\theta d\theta =sec\theta tan\theta +\int sec\theta d\theta$

$2\int se{c}^3\theta d\theta =sec\theta tan\theta \to log|sec\theta +tan\theta |$

and$=\sqrt{x(x-1)}+log|\sqrt{x}+\sqrt{x-1}|+c$

$\int \frac{x}{\sqrt{x-1}}dx=\sqrt{x(x-1)}+log|\sqrt{x}+\sqrt{x-1}|+c\cdot$