If ${\int }_{\ln 2}^{\mathrm{x}} {\left({\mathrm{e}}^{\mathrm{x}}-1\right)}^{-1}\mathrm{dx}=\ln \left(\frac{3}{2}\right)$ then x =

${\int }_{{\ln }_2}^{\mathrm{x}} \frac{1}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}=\ln \left(\frac{3}{2}\right)\Rightarrow {\int }_{{\ln }_2}^{\mathrm{x}} \frac{{\mathrm{e}}^{-\mathrm{x}}}{1-{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}=\ln \left(\frac{3}{2}\right)$
$$\begin{gathered} \Rightarrow {\left[\ln \left|1-e^{-x}\right|\right]}_{\ln 2}^x=\ln \left(\frac{3}{2}\right) \\ \Rightarrow \ln \left|1-e^{-x}\right|-\ln \left(\frac{1}{2}\right)=\ln \left(\frac{3}{2}\right) \\ \Rightarrow \frac{e^x-1}{e^x}=\frac{3}{4} \\ \Rightarrow 4e^x-4=3e^x \\ \Rightarrow e^x=4\Rightarrow x=\ln 4 \end{gathered}$$