Explanation

(a)
Let $I = \int_{0}^{\infty}\frac{dx}{\left( x + \sqrt{x^{2} + 1} \right)^{3}}$
Put $x = \tan{\theta \Rightarrow dx = \sec^{2}{\theta\ d\theta}}$
$\therefore I = \int_{0}^{\pi/2}\frac{\sec^{2}{\theta d\theta}}{\left( \tan{\theta + \sqrt{\tan^{2}{\theta + 1}}} \right)^{3}}$
$= \int_{0}^{\pi/2}\frac{\sec^{2}{\theta d\theta}}{(\tan{\theta + \sec{\theta\left. \ \ \right)^{3}}}}$
$= \int_{0}^{\pi/2}{\frac{1/\cos^{2}\theta}{\left( \frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta} \right)^{3}}d\theta = \int_{0}^{\pi/2}{\frac{\cos\theta}{({\sin{\theta + 1)}}^{3}}d\theta}}$
$= \left\lbrack - \frac{1}{2(\sin{\theta + 1\left. \ \ \right)^{2}}} \right\rbrack_{0}^{\pi/2} = \frac{- 1}{8} + \frac{1}{2} = \frac{3}{8}\$