How do we evaluate this integral containing log

Question

If $f(x)=\frac{x}{1+(\ln x)(\ln x) \ldots \infty} \forall x \in[1, > \infty)$ then $\int_1^{2 e} f(x) d x$ equals is :

I have no idea on how to approach this type of integral. The only step I could write properly was:

$f(x)=\frac{x}{1+(\ln x)(\ln x) \ldots \infty} \Rightarrow f(x)=\lim _{n \rightarrow \infty} \frac{x}{1+(\ln x)^n}$

I tried thinking about where the graphs of log would give positive and how the powers of exponents would effect the limits. But I just can't get a knack on how to apply it here. Any help on how to proceed would be appreciated. Thanks.

Answer 1

When $x=1$ the $\ln x$ in the denominator is zero. At $x=e$ the function is $\frac e2$. When $x>e$ we just get $0$. When $1<x<e$ we have that $f(x)=x$. So the answer is $$\int_1^exdx=\frac{e^2-1}2$$We can numerically verify this answer on desmos by making $n$ bigger. This integral slowly converges by the way.

Answer 2

Hint:

Rewrite the integral as $$\int_1^{e}f(x)dx+\int_e^{2e}f(x)dx$$