Explanation

(a)
Let $I = \int_{0}^{\infty}{\log{\left( x + \frac{1}{x} \right)\frac{1}{1 + x^{2}}dx}}$
Put $x = \tan{\theta\ \ \Rightarrow \ dx = \sec^{2}{\theta\ d\theta}}$
$\therefore\ I = \int_{0}^{\pi/2}{\log{\left( \tan{\theta + \frac{1}{\tan\theta}} \right)\frac{\sec^{2}\theta}{\sec^{2}\theta}}}d\theta$
$\Rightarrow \ I = \int_{0}^{\pi/2}{\log{\frac{(1 + \tan^{2}{\theta)}}{\tan\theta}d\theta}}$
$\Rightarrow \ I = 2\int_{0}^{\pi/2}{\log{\sec{\theta\ d\theta - \int_{0}^{\pi/2}{\log{\tan{\theta\ d\theta}}}}}}$
$\Rightarrow \ I = 2\int_{0}^{\pi/2}{\log{\sec{\theta\ d\theta\ \ \ \left( \because\int_{0}^{\pi/2}{\log{\tan{\theta\ d\theta = 0}}} \right)}}}$
$\Rightarrow \ I = - 2\int_{0}^{\pi/2}{\log{\cos{\theta\ d\theta = - 2 \times \frac{- \pi}{2}}\log 2}}\ \ \ \left( \because\int_{0}^{\pi/2}{\log{\cos{\theta\ d\theta = - \frac{\pi}{2}\log 2}}} \right)$
$\therefore\ \ I = \pi\log 2$