Explanation

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