Explanation

(d)
On putting $y^{2} = f(x) = \frac{x + 2}{2x + 3}$, we have
$x = \frac{3y^{2} - 2}{1 - 2y^{2}}$ and $dx = - \frac{2y}{\left( 1 - 2y^{2} \right)^{2}}dy$
$\therefore\ \int_{}^{}{\left( f(x) \right)^{1/2}\frac{dx}{x} = - \int_{}^{}{y\frac{2y}{\left( 1 - 2y^{2} \right)^{2}} \bullet \frac{1 - 2y^{2}}{3y^{2} - 2}dy}}$
$= 2\int_{}^{}{\frac{y^{2}}{\left( 2y^{2} - 1 \right)(3y^{2} - 2)}dy}$
$= - 2\int_{}^{}{\left\lbrack \frac{1}{2y^{2} - 1} - \frac{2}{3y^{2} - 2} \right\rbrack dy}$
$= \frac{1}{\sqrt{2}}\log{\left| \frac{1 + \sqrt{2y}}{1 - \sqrt{2y}} \right| - \sqrt{\frac{2}{3}}\log{\left| \frac{\sqrt{3y} + \sqrt{2}}{\sqrt{3y} - \sqrt{2}} \right| + c}}$
Thus, $g(x) = \log{|x|}$ and $h(x) = \log{|x|}$