Answer
$$1$$
Work Step by Step
Integrate by parts with
\begin{align*}u&=\ln x\ \ \ \ \ \ dv=xdx\\
du&=\frac{ 1}{x}\ \ \ \ \ \ v=\frac{x^2}{2}\end{align*}
Then
\begin{align*}
\int_{0}^{1}(-\ln x) d x&=\lim _{b \rightarrow 0^{+}}\int_{b}^{1}(-\ln x) d x\\
&=\lim _{b \rightarrow 0^{+}}[x-x \ln x]_{b}^{1}\\
&=\lim _{b \rightarrow 0^{+}}[(1-1 \ln 1)-(b-b \ln b)]\\
&=1-0+\lim _{b \rightarrow 0^{+}} \frac{\ln b}{\left(\frac{1}{b}\right)}\\
&=1+\lim _{b \rightarrow 0^{+}} \frac{\left(\frac{1}{b}\right)}{\left(-\frac{1}{b^{2}}\right)}\\
&=1-\lim _{b \rightarrow 0^{+}} b =1
\end{align*}