If $f\left(x\right)=\frac{x+2}{2x+3}.$then $\int {\left(\frac{f\left(x\right)}{x^2}\right)}^{1/2}\mathrm{d}x$

is equal to $\frac{1}{\sqrt{2}}g\left(\frac{1+\sqrt{2f\left(x\right)}}{1-\sqrt{2f\left(x\right)}}\right)-\sqrt{\frac{2}{3}}h\left(\frac{\sqrt{3f\left(x\right)}+\sqrt{2}}{\sqrt{3f\left(x\right)}-\sqrt{2}}\right)$ 

$+C$ where 

Solution:

Putting $y^2=f\left(x\right)=\frac{x+2}{2x+3}$, we have 

$x=\frac{3y^2-2}{1-2y^2}$and $\mathrm{d}x=-\frac{2y}{{\left(1-2y^2\right)}^2}\mathrm{d}y.$

So $\begin{array}{l}=-\int y\cdot \frac{2y}{{\left(1-2y^2\right)}^2}\cdot \frac{1-2y^2}{3y^2-2}\mathrm{d}y\\ =2\int \frac{y^2}{\left(2y^2-1\right)\left(3y^2-2\right)}\mathrm{d}y\\ =-2\int \left[\frac{1}{2y^2-1}-\frac{2}{3y^2-2}\right]\mathrm{d}y\\ =\frac{1}{\sqrt{2}}\log \left|\frac{1+\sqrt{2}y}{1-\sqrt{2}y}\right|-\sqrt{\frac{2}{3}}\log \left|\frac{\sqrt{3}y+\sqrt{2}}{\sqrt{3}y-\sqrt{2}}\right|+C\end{array}$

Thus $g\left(x\right)=\log \left|x\right|\text{ and }h\left(x\right)=\log \left|x\right|$.