$Thedomainof\frac{{\tanh }^{-1}(2x-3)}{\sqrt{4-x^2}}is$

$\begin{array}{l}f\left(x\right)=\frac{{\mathrm{Tanh}}^{-1}(2x-3)}{\sqrt{4-x^2}}isdefined\\ iff-1<2x-3<1and4-x^2>0\\ \Rightarrow 2<2x<4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} and \hspace{0.17em}}{x}^2-4<0\\ \Rightarrow 1<x<2\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} and\hspace{0.17em}}(x+2)(x-2)<0 \Rightarrow -2<x<2\\ \\ x\in (1,2)\end{array}$