Explanation

(a)
$\because I = \int_{0}^{\infty}\frac{x\log x}{\left( 1 + x^{2} \right)^{2}}dx$
Put $x = \tan{\theta\ \Rightarrow dx = \sec^{2}{\theta\ d\theta}}$
$\therefore I = \int_{0}^{\pi/2}{\frac{\tan{\theta\log{(\tan{\theta)}}}}{\sec^{4}\theta}\sec^{2}\theta}d\theta$
$= \int_{0}^{\pi/2}{\frac{\sin{\theta\log{(\tan{\theta)}}}}{\cos{\theta \bullet \frac{1}{\cos^{2}\theta}}}d\theta}$
$= \int_{0}^{\pi/2}{\sin{\theta\cos{\theta\log{(\tan{\theta)\ d\theta}}}}}$
$= \frac{1}{2}\int_{0}^{\pi/2}{\sin{2\theta\log{(\tan{\theta)\ d\theta = 0}}\ \ \ \left( \because\int_{0}^{\pi/2}{\sin{\ 2\theta\log{\tan{\theta\ d\theta = 0}}}} \right)}}$