$\text{ If }\int \frac{\ln \left(e{x}^{x+1}\right)+{\left(\ln \left(x^{\sqrt{x}}\right)\right)}^2}{1+(x\ln x)\left(\ln \left(e^2{x}^x\right)\right)}dx=f\left(x\right)+\mathrm{C} \mathrm{Where} C \mathrm{is} \mathrm{arbitrary} \mathrm{Constant}, \mathrm{then} {\mathrm{e}}^{\left({\mathrm{e}}^{\mathrm{f}\left(2\right)}-1\right)} \mathrm{is} =$

$\begin{array}{l}\text{ Let }I=\int \frac{\ln \left(e{x}^{x+1}\right)+{\left(\ln \left(x^{\sqrt{x}}\right)\right)}^2}{1+(x\ln x)\left(\ln \left(e^2{x}^x\right)\right)}dx\\ \left(\begin{array}{c}\text{ Use }\left(1\right)\log m+\log n=\log mn\\ \left(2\right)\log a^m=m\log a\end{array}\right)\\ =\int \frac{\ln \left(e\right)+\ln \left(x^{x+1}\right)+(\sqrt{x}\cdot \ln (x){)}^2}{1+(x\cdot \ln (x\left)\right)\cdot \left(\ln e^2+\ln x^x\right)}dx\\ =\int \frac{1+(x+1)\ln \left(x\right)+x(\ln (x){)}^2}{1+(x\cdot \ln (x\left)\right)\cdot (2+x\cdot \ln x)}dx\\ =\int \frac{1+\ln x+x\ln x+x(\ln x{)}^2}{1+2(x\ln x)+(x\ln x{)}^2}dx\\ =\int \frac{(1+x\ln x)(1+\ln x)}{(1+x\ln x{)}^2}dx\end{array}$

$\begin{array}{l}=\int \frac{1+\ln x}{(1+x\ln x)}dx\left(\because \int \frac{f^{\mathrm{\prime }}\left(x\right)}{f\left(x\right)}=\ln \left|f\right(x\left)\right|+c\right)\\ =\ln (1+x\ln x)+C\\ \therefore f\left(x\right)=\ln (1+x\ln x)\\ \Rightarrow f\left(2\right)=\ln |1+2\ln 2|\\ \Rightarrow f\left(2\right)=\ln |1+\ln 4|\\ \Rightarrow e^{f\left(2\right)}=1+\ln 4\\ \Rightarrow e^{f\left(2\right)}-1=\ln 4\\ \Rightarrow e^{e^{f\left(2\right)}-1}=4\end{array}$

Therefore, the correct answer is (1).