Answer
$$\int_{1}^{2} x \ln x d x
=2 \ln 2-\frac{3}{4} $$
Work Step by Step
Given $$\int_{1}^{2} x \ln x d x $$ Now, using integration by parts:
$$ u=\ln x \Rightarrow du= \frac{1}{x}dx $$ $$ dv=x dx \Rightarrow v= \frac{x^2}{2} $$ So, we get
\begin{aligned}
I&=\int_{1}^{2} x \ln x d x\\
&=uv|_{1}^{2}- \int_{1}^{2}vdu\\
&=\frac{x^{2} \ln x}{2}|_{1}^{2}-\frac{1}{2} \int_{1}^{2} \frac{x^{2}}{x} d x\\ &=\frac{x^{2} \ln x}{2}-\frac{1}{2} \int_{1}^{2} \frac{x^{2}}{x} d x\\
&=\frac{2^{2} \ln2}{2}-\frac{1^{2} \ln1}{2}-\frac{1}{2} \int_{1}^{2}xd x\\
&=\frac{2^{2} \ln2}{2}-\frac{1^{2} \ln1}{2}-\frac{x^2}{4} |_{1}^{2} \\
&=2 \ln 2-\frac{2^2}{4}+\frac{1^2}{4} \\
&=2 \ln 2-\frac{3}{4} \\
\end{aligned}
