If  $\int \frac{\ln \left(e{x}^{x+1}\right)+{\left(\ln \left(x^{\sqrt{x}}\right)\right)}^2}{1+\left(x\ln x\right)\left(\ln \left(e^2{x}^x\right)\right)}dx=f\left(x\right)+C$, where  $f\left(1\right)=0$, then $e^{(e^{f\left(2\right)}-1)}$  is equal to:

Let  $I=\int \frac{\ln \left(e{x}^{x+1}\right)+{\left(\ln \left(x^{\sqrt{x}}\right)\right)}^2}{1+\left(x\ln x\right)\left(\ln \left(e^2{x}^x\right)\right)}dx$
 $$\begin{gathered} \left(\begin{array}{l}Use\left(1\right)\log m+\log n=\log mn\\ \left(2\right)\log a^m=m\log a\end{array}\right) \\ ={\int }_{ }^{ }\frac{\ln \left(e\right)+\ln \left(x^{x+1}\right)+{(\sqrt{x}.\ln \left(x\right))}^2}{1+(x.\ln \left(x\right)).(\ln e^2+\ln x^x)}dx \\ ={\int }_{ }^{ }\frac{1+(x+1)\ln \left(x\right)+x{\left(\ln \left(x\right)\right)}^2}{1+(x.\ln \left(x\right)).(2+x.\ln x)}dx \\ =\int \frac{1+\ln x+x\ln x+x{\ln }^{2}x}{1+2\left(x\ln x\right)+{\left(x\ln x\right)}^2}dx \\ I=\int \frac{(1+x\ln x)(1+\ln x)}{{(1+x\ln x)}^2}dx=\int \frac{1+\ln x}{(1+x\ln x)}dx(Use\int \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=\ln \left|f\left(x\right)\right|+c) \\ =\ln (1+x\ln x)+C \\ \therefore f\left(x\right)=\ln (1+x\ln x) \\ =f\left(2\right)=\ln |1+2\ln 2|\Rightarrow f\left(2\right)=\ln |1+\ln 4| \\ \Rightarrow e^{f\left(2\right)}=1+\ln 4\Rightarrow e^{f\left(2\right)}-1=\ln 4\Rightarrow e^{e^{f\left(2\right)}-1}=4 \end{gathered}$$