reverse the order of integration and evalute $\int_0^1 \int_{e^y}^e (x/\ln x)\,dx\,dy$ First step, I reversed the order of integration (not sure if correct) of...
$$\int_0^1\int_{e^y}^e \frac x {\ln x} \, dx\,dy$$
to...
$$\int_0^e \int_{\ln x}^1 \frac x {\ln x} \, dy \, dx$$
I then began to evaluate...
$$\int_0^e \Big[\frac{xy}{\ln x}\Big]_{y=\ln x}^{y=1} \, dx$$
Plugging in gives me...
$$\int_0^e \left(\frac x {\ln x} - x\right) \, dx$$
Now I am left with an expression that can only be solved using integration by parts. I tried using integration by parts but I keep getting stuck. Online resources are not helping me. Can someone show me how to evaluate the rest step-by-step?
Or if my steps are incorrect please tell me.

Note that we have
$$\begin{align}
\int_0^1\int_{e^y}^e \frac{x}{\log(x)}\,dx\,dy&=\int_1^e \frac{x}{\log(x)} \int_0^{\log(x)}(1) \,dy\,dx\\\\
&=\int_1^e x\,dx\\\\
&=\frac12(e^2-1)
\end{align}$$
