Explanation

(a)
Let $I = \int_{0}^{2}{\sqrt{\frac{2 + x}{2 - x}}dx = \int_{0}^{2}{\frac{2 + x}{\sqrt{4 - x^{2}}}dx}}$
$= \int_{0}^{2}{\frac{2}{\sqrt{4 - x^{2}}}dx + \int_{0}^{2}{\frac{x}{\sqrt{4 - x^{2}}}dx}}$
Put $x = 2$sin $\theta$ in the first integral and $4 - x^{2} = t\$in the second integral
$I = 2\int_{0}^{\pi/2}{\frac{2\cos{\theta d\theta}}{2\cos\theta} + \int_{4}^{0}{- \frac{dt}{2\sqrt{t}}}}$
$= 2\int_{0}^{\pi/2}{d\theta + \frac{1}{2}\int_{0}^{4}t^{- 1/2}}dt$
$= 2\lbrack\theta\rbrack_{0}^{\pi/2} + \frac{1}{2}\left\lbrack \frac{t^{1/2}}{1/2} \right\rbrack_{0}^{4}$
$= \pi + 2$