Explanation

(a)
Let $I = \int_{}^{}\frac{\log\sqrt{x}}{3x}\ dx$
Put $\sqrt{x} = t$
$\frac{1}{\sqrt{x}}dx = 2\ dt$
$\therefore\ I = \int_{}^{}\frac{2\log t}{3t}dt$
$= \frac{2}{3}\int_{}^{}\frac{\log t}{t}\ dt$
$= \frac{2}{3}\ .\ \frac{({\log{t)}}^{2}}{2} + c$
$= \frac{1}{3}\left( \log\sqrt{x} \right)^{2} + c$