Answer
$$1-\frac{2}{e}$$
Work Step by Step
Given $$\int_{1}^{e} \frac{\ln x d x}{x^{2}}$$
Let
\begin{align*}
u&=\ln x\ \ \ \ \ \ \ \ dv= x^{-2}dx\\
du&= \frac{ 1}{x} dx\ \ \ \ \ \ \ \ v=-x^{-1}
\end{align*}
Then
\begin{align*}
\int_{1}^{e} \frac{\ln x d x}{x^{2}}&= -\frac{1}{x} \ln x\bigg|_{1}^{e} +\int_{1}^{e} \frac{1}{x^2} d x \\
&= -\frac{1}{x} \ln x\bigg|_{1}^{e}-\frac{1}{x}\bigg|_{1}^{e} \\
&=1-\frac{2}{e}
\end{align*}
