Evaluate for a double integral Evaluate $$\int_{0}^{1} \int_{e^y}^{e} \frac{x}{\ln x}\ dx\ dy$$

I'm new to double integral. Can anyone guide me for this question? Thanks in advance
Hint. This double integral is improper because the integrand is not bounded along the line $x=1$. Note that
$$\int_{y=0}^{1} \left(\int_{x=e^y}^{e} \frac{x}{\ln x} dx\right) dy
=\lim_{\epsilon\to 0} \int_{y=\epsilon}^{1} \left(\int_{x=e^y}^{e} \frac{x}{\ln x} dx\right) dy.$$
By changing the order of integration (Fubini's Theorem), we get
$$\int_{y=\epsilon}^{1} \left(\int_{x=e^y}^{e} \frac{x}{\ln x} dx\right) dy=
\int_{x=e^{\epsilon}}^{e} \left(\int_{y=\epsilon}^{\ln(x)} \frac{x}{\ln x} dy\right) dx=\int_{x=e^{\epsilon}}^{e}  \frac{x}{\ln x}(\ln(x)-\epsilon) dx.$$
Can you take it from here?

P.S. In order to avoid the limit you should use a version of Fubini's Theorem for improper integrals.
The idea of multiple integrals is to treat them as iterated integrals. For example, the integral provided can be rewritten as $$\int_{y=0}^{y=1}\left(\int_{x = e^y}^{x=e}\frac{x}{\ln x} dx\right)dy$$
It should be clear that we must evaluate the inner integral first, and the inner integral must resolve to a function of $y$. Assuming you could integrate the inner integral (let's call it $F(x)$), the double integral becomes $$\int_0^1(F(e)-F(e^y))dy$$ All that remains is to evaluate this final integral.