Question

For $x>0$, $\int \frac{\sqrt{x-1}}{x\sqrt{x+1}}dx$ is equal to (where $c$ is constant of integration)

• A. $\ln |x-\sqrt{x^2-1}|-{\tan }^{-1}x+c$
• B. $\ln |x+\sqrt{x^2-1}|-{\tan }^{-1}x+c$
• C. $\ln |x-\sqrt{x^2-1}|-{\sec }^{-1}x+c$
• D. $\ln |x+\sqrt{x^2-1}|-{\sec }^{-1}x+c$

Answer 1

The correct option is D $\ln |x+\sqrt{x^2-1}|-{\sec }^{-1}x+c$

$I=\int \frac{\sqrt{x-1}}{x\sqrt{x+1}}dx$
$=\int \frac{x-1}{x\sqrt{x^2-1}}dx$
$=\int \frac{dx}{\sqrt{x^2-1}}-\int \frac{dx}{x\sqrt{x^2-1}}$
$\because \int \frac{dx}{|x|\sqrt{x^2-1}}={\sec }^{-1}x+c$ and $x>0$ (given)
$=\ln |x+\sqrt{x^2-1}|-{\sec }^{-1}x+c$