Explanation

(b)
Let $I = \int_{1}^{2}\frac{dx}{(x + 1)\sqrt{x^{2} - 1}}$
Put $x = \sec{\theta\ \Rightarrow \ dx = \sec{\theta\tan{\theta d\theta\ \ \ \ldots(i)}}}$
$\therefore I = \int_{0}^{\pi/3}\frac{\sec{\theta\tan\theta}}{(\sec{\theta + 1)\tan\theta}}\ d\theta$
$= \int_{0}^{\pi/3}{\frac{1}{1 + \cos\theta}d\theta = \int_{0}^{\pi/3}{\frac{\sec^{2}\frac{\theta}{2}}{2}d\theta}}$
$= \frac{1}{2}\left\lbrack \frac{\tan\frac{\theta}{2}}{\frac{1}{2}} \right\rbrack_{0}^{\frac{\pi}{3}} = \frac{1}{\sqrt{3}}$