Let $f(x)=\frac{x+2}{2x+3}$, $x>0$. If
$$\int \left( \frac{f(x)}{x^2} \right)^{1/2}dx=\frac{1}{\sqrt{2}}g \left(\frac{1+\sqrt{2f(x)}}{ 1-\sqrt{2f(x)}} \right) -\sqrt{\frac{2}{3}}h \left(\frac{\sqrt{3f(x)}+\sqrt{2}}{\sqrt{3f(x)}-\sqrt{2} } \right)+C$$
where $C$ is the constant of integration, then find $g(x)$ and $h(x)$.
I tried solving it like
$$\int \left( \frac{f(x)}{x^2} \right)^{1/2}dx=\int \sqrt{\frac{x+2}{2x^3+3x^2}}dx$$ but it leads us to nowhere.
Any hint will be of great help.
if you find this integral, you can
Hint: Let $$\sqrt{\dfrac{x+2}{2x+3}}=t,\Longrightarrow x=\dfrac{3t^2-2}{1-2t^2}$$
$$-\int \dfrac{1-2t^2}{3t^2-2}\cdot t\cdot\dfrac{2t}{(2t^2-1)^2}dt=\int\dfrac{2t^2}{(3t^2-2)(2t^2-1)}dt$$ and Note $$\dfrac{2t^2}{(3t^2-2)(2t^2-1)}=\dfrac{4}{3t^2-2}-\dfrac{2}{2t^2-1}$$
maybe as this result you can some calculation have your form