Explanation

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