Question

Solve:
$\int \frac{1}{x}\sqrt{\frac{x-1}{x+1}}dx$

Answer 1

$\begin{array}{c}I=\int \frac{1}{x}\sqrt{\frac{x-1}{x+1}}dx\\ I=\int \frac{1}{x}\sqrt{\frac{x-1}{x+1}\times \frac{x+1}{x+1}}dx\\ \int \frac{1}{x}\frac{\sqrt{x^2-1}}{x+1}dx\left(\begin{array}{c}x=\sec \theta \\ dx=\sec \theta .\tan \theta \end{array}\right)\\ \int \frac{1}{\sec \theta }.\frac{\sqrt{{\sec }^2\theta -1}}{\sec \theta +1}.\sec \theta .\tan \theta \left(\begin{array}{c}{\sec }^2\theta -{\tan }^2\theta =1\\ {\sec }^2\theta -1={\tan }^2\theta \end{array}\right)\\ \int \frac{\sqrt{{\tan }^2\theta }}{\sec \theta +1}.\tan \theta \\ \int \frac{{\tan }^2\theta }{\sec \theta +1}d\theta =\int \frac{{\sec }^2\theta -1}{\sec \theta +1}d\theta \\ I=\int \frac{\left(\sec \theta +1\right)\left(\sec \theta -1\right)}{\left(\sec \theta +1\right)}d\theta \\ \int \sec \theta d\theta -\int 1d\theta \\ \ln \left(\sec \theta +\tan \theta \right)-\theta +c\\ I=\ln \left(x+\sqrt{x^2-1}\right)-{\sec }^{-1}x+c.\end{array}$

Hence, this is the answer.