$$\int \frac{\ln{\left( \mathrm{e}\,x^{x+1} \right)} + \left[ \ln{ \left( x^{\sqrt{x}} \right) } \right]^2 }{1 + x\ln{ \left( x \right) } \ln{ \left( \mathrm{e}^x\,x^x \right) }}dx=f(x)+C $$ where $f(1)=0$, then $e^{e^{f(2)}-1}$ is equal to?
$\displaystyle \begin{align*} \frac{\ln{\left( \mathrm{e}\,x^{x+1} \right)} + \left[ \ln{ \left( x^{\sqrt{x}} \right) } \right]^2 }{1 + x\ln{ \left( x \right) } \ln{ \left( \mathrm{e}^x\,x^x \right) }} &= \frac{ \ln{\left( \mathrm{e} \right) } + \ln{ \left( x^{x+1} \right) } + \left[ \sqrt{x} \, \ln{ \left( x \right) } \right] ^2 }{ 1 + x\ln{ \left( x \right) } \left[ \ln{\left( \mathrm{e}^x \right) } + \ln{ \left( x^x \right) } \right] } \\ &= \frac{ 1 + \left( x + 1 \right) \ln{ \left( x \right) } + x \, \left[ \ln{ \left( x \right) } \right] ^2 }{ 1 + x \ln{ \left( x \right) } \left[ x + x\ln{ \left( x \right) } \right] } \end{align*}$, but could not solve it further.
Too long for a comment
I think that there is an error or a typo in the Question and the integral should be the following.
$$I=\int \frac{\ln{\left( \mathrm{e}\,x^{x+1} \right)} + \left[ \ln{ \left( x^{\sqrt{x}} \right) } \right]^2 }{1 + x\ln{ \left( x \right) } \ln{ \left( \mathrm{e}\,x^x \right) }}\mathrm dx$$
On simplification, we get
$$I=\int \frac{(1+x\ln(x)) (1+\ln(x))}{x^2 \ln^2(x) +x\ln(x)+1}\mathrm dx$$
Let $\space x\ln(x)=t \implies 1+\ln(x)=dt$
If we had our orignal integral then this nice substitution wouldn't have work because of that one extra $x$ in the denominator.
$$I=\int \frac{1+t}{t^2+t+1}\mathrm dt$$
$$\implies I=\frac12 \int \frac{2t+1}{t^2+t+1}+\frac{1}{t^2+t+1}\mathrm dt$$
$$I=\ln(t^2+t+1)+\frac{2}{\sqrt{3}}\arctan\bigg(\frac{2t+1}{\sqrt{3}}\bigg)+C$$
In the Question It is given that $I=\int F(x)\mathrm dx=f(x)+C$ and $f(1)=0$ but Here in my final answer $f(1)$ is clearly not $0$. I don't know what's the matter with the Question.
If we assume that the Question is Correct then we have to find the following integral to evaluate $f(2)$.
$$f(2)=\int_{1}^{2} \frac{(1+x\ln(x)) (1+\ln(x))}{x^2 \ln(x) (1+\ln(x))+1}$$
Using Desmos and wolfram Above Integral evaluates to $0.0315191338352$ which is $f(2)$ and is not recognised by inverse symbolic calculator as some simple logarithmic expression.Also note that in Question we have to find value of $e^{e^{f(2)}-1}$ so I think $f(2)-1$ should evaluate to something like $ln(x)$ then our answer would be of the form $e^{x}$.