Explanation

(c)
$\because\ \sqrt{x + 2\sqrt{(x - 1)}} = \sqrt{\left( \sqrt{(x - 1)} \right)^{2} + 1^{2} + 2\sqrt{(x - 1)}}$
$= \sqrt{(x - 1)} + 1$
and $\sqrt{x - 2\sqrt{(x - 1)}} + \sqrt{\left( \sqrt{(x - 1)} \right)^{2} + 1^{2} - 2\sqrt{(x - 1)}}$
$= |\sqrt{(x - 1)} - 1|$
Then, $\int_{1}^{5}{(\sqrt{x + 2\sqrt{(x - 1})} + \sqrt{x - 2\sqrt{(x - 1})}})\ dx$
$= \int_{1}^{5}(\sqrt{(x - 1}) + 1)dx + \int_{1}^{5}{|\sqrt{(x - 1}}) - 1|dx$
$= \int_{1}^{5}(\sqrt{(x - 1}) + 1)dx + \int_{1}^{2}{(1 - \sqrt{(x - 1}}))\ dx + \int_{1}^{5}(\sqrt{(x - 1}) - 1)dx$
$= \int_{0}^{4}{\left( \sqrt{x} + 1 \right)dx + \int_{0}^{1}{\left( 1 - \sqrt{x} \right)dx +}}\int_{1}^{4}{\left( \sqrt{x} - 1 \right)dx}$
$= \left\lbrack \frac{2}{3}\left( x^{3/2} \right) + x \right\rbrack_{0}^{4} + \left\lbrack x - \frac{2}{3}x^{3/2} \right\rbrack_{0}^{1} + \left\lbrack \frac{2}{3}x^{3/2} - x \right\rbrack_{1}^{4}$
$= \left( \frac{16}{3} + 4 \right) + \left( 1 - \frac{2}{3} \right) + \left( \frac{16}{3} - 4 \right) + \left( \frac{2}{3} - 1 \right)$
$= \frac{32}{3}$