Answer
$$-\frac{1}{4}$$
Work Step by Step
Integrate by parts:
\begin{align*}u&=\ln x\ \ \ \ \ \ dv=xdx\\
du&=\frac{ 1}{x}\ \ \ \ \ \ v=\frac{x^2}{2}\end{align*}
Then
\begin{align*}
\int_{0}^{1} x \ln x d x&=\lim _{b \rightarrow 0^{+}} \int_{b}^{1} x \ln x d x\\
&=\lim _{b \rightarrow 0^{+}}\left[\frac{x^{2}}{2} \ln x-\frac{x^{2}}{4}\right]_{b}^{1}\\
&=\lim _{b \rightarrow 0^{+}}\left[\left(\frac{1}{2} \ln 1-\frac{1}{4}\right)-\left(\frac{b^{2}}{2} \ln b-\frac{b^{2}}{4}\right)\right]\\
&=-\frac{1}{4}-\lim _{b \rightarrow 0^{+}} \frac{\ln b}{\left(\frac{2}{b^{2}}\right)}\\
&=-\frac{1}{4}-\lim _{b \rightarrow 0^{+}} \frac{\left(\frac{1}{b}\right)}{\left(-\frac{4}{b^{3}}\right)}\\
&=-\frac{1}{4}+\lim _{b \rightarrow 0^{+}}\left(\frac{b^{2}}{4}\right)\\
&=-\frac{1}{4}+0=-\frac{1}{4}
\end{align*}
