Explanation

(d)
Put $x = \tan{\theta \Rightarrow dx = \sec^{2}{\theta d\theta}}$
$\therefore\int_{0}^{\infty}{\frac{x\ dx}{(1 + x)(1 + x^{2})} = \int_{0}^{\pi/2}\frac{\tan{\theta\sec^{2}{\theta d\theta}}}{(1 + \tan{\theta)(1 + \tan^{2}{\theta)}}}}$
$= \int_{0}^{\pi/2}{\frac{\tan\theta}{1 + \tan\theta}d\theta}$
$= \int_{0}^{\pi/2}{\frac{\sin\theta}{\cos\theta + \sin\theta}d\theta}$
Let $I = \int_{0}^{\pi/2}{\frac{\sin\theta}{\cos{\theta + \sin\theta}}d\ \theta\ \ \ \ldots(i)}$
$\Rightarrow I = \int_{0}^{\pi/2}{\frac{\sin\left( \frac{\pi}{2} - \theta \right)}{\cos\left( \frac{\pi}{2} - \theta \right) + \sin\left( \frac{\pi}{2} - \theta \right)}d\theta}$
$= \int_{0}^{\pi/2}{\frac{\cos\theta}{\sin{\theta + \cos\theta}}d\theta}$ …(ii)
On adding Eqs. (i) and (ii), we get
$2I = \int_{0}^{\pi/2}{1d\theta = \lbrack\theta\rbrack_{0}^{\pi/2}} = \frac{\pi}{2}$
$\Rightarrow I = \frac{\pi}{4}$
$\therefore\ \int_{0}^{\infty}\frac{x\ dx}{(1 + x)(1 + x^{2})} = \frac{\pi}{4}$