[tex]\int_1^e\dfrac{\ln x}{x}dx,~\text{if}~u=\ln x\to x=e^u\to dx=e^udu\\\\\text{So, finding the new limits:}\begin{cases}x=1\to u=\ln1\to u=0\\x=e\to u=\ln e\to u=1\end{cases}\\\\
\int_1^e\dfrac{\ln x}{x}dx=\int_0^1\dfrac{u}{e^u}e^udu=\int_0^1u\,du=\left[\dfrac{u^2}{2}\right]_0^1=\dfrac{1^2}{2}-\dfrac{0^2}{2}=\dfrac{1}{2}\\\\
\boxed{\int_1^e\dfrac{\ln x}{x}dx=\dfrac{1}{2}}[/tex]
