$I=\int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{{\left(3-2x\right)}^2}dx$

Let $\log \left(\frac{2x+3}{3-2x}\right)=t\Rightarrow \frac{2x+3}{3-2x}=e^t$

i.e. $\frac{\left(3-2x\right)2-\left(2x+3\right)\left(-2\right)}{{\left(3-2x\right)}^2}dx=e^{t}dt$

$\Rightarrow \frac{1}{{\left(3-2x\right)}^2}dx=\frac{1}{12}{e}^{t}dt$

So, $I=\frac{1}{24}\int e^t\cos tdt$

$=\frac{e^t}{24}\left(\cos t+\sin t\right)=\frac{\frac{2x+3}{3-2x}}{24}\left(\cos \log \left(\frac{2x+3}{3-2x}\right)+\sin \log \left(\frac{2x+3}{3-2x}\right)\right)$

i.e. $g\left(x\right)=\log \left(\frac{2x+3}{3-2x}\right)$, $f\left(x\right)=\frac{2x+3}{3-2x}$

$\Rightarrow g\left(x\right)=\log \left(f\left(x\right)\right)$

Hence, $g\left(1\right)=\log \left(f\left(1\right)\right)$