Answer
$$\int_1^e\frac{1-\ln{x}}{x}\hspace{0.5mm}dx=\frac{1}{2}$$
Work Step by Step
Let $u=1-\ln{x}$
then $du=-\frac{1}{x}\hspace{1mm}dx$,
and therefore $-xdu=dx$.
Substituting this into our integral we get
$\int-\frac{u}{x}xdu=-\int u\hspace{0.5mm}du=-\frac{u^2}{2}+C=-\frac{(1-\ln{x})^2}{2}+C$
Plugging in our integration limits and ignoring our constant of integration we get
$-\frac{(1-\ln{x})^2}{2}\bigg\vert_1^e=-\frac{(1-\ln{e})^2}{2}-\left(-\frac{(1-\ln{1})^2}{2}\right)=\frac{(1-0)^2}{2}-\frac{(1-1)^2}{2}=\frac{1}{2}$
