$\int \left|\mathrm{x}\right|\log \left|\mathrm{x}\right|\mathrm{dx}\text{ is equal to }(\mathrm{x}\ne 0)$

Solution:

$\text{ Case II If }\mathrm{x}<0,\text{ then }\left|\mathrm{x}\right|=-\mathrm{x}$

$\begin{array}{l}\therefore \int \left|\mathrm{x}\right|\log \left|\mathrm{x}\right|\mathrm{dx}=\int \mathrm{xlog} \mathrm{xdx}\\ =\log \mathrm{x}\cdot \frac{{\mathrm{x}}^2}{2}-\int \frac{1}{\mathrm{x}}\cdot \frac{{\mathrm{x}}^2}{2}\mathrm{dx}\\ =\frac{{\mathrm{x}}^2}{2}\cdot \log \mathrm{x}-\frac{{\mathrm{x}}^2}{4}+\mathrm{C}=+\frac{{\mathrm{x}}^2}{2}\log \left|\mathrm{x}\right|-\frac{{\mathrm{x}}^2}{4}+\mathrm{C}\end{array}$

$\text{ Case II If }\mathrm{x}<0,\text{ then }\left|\mathrm{x}\right|=-\mathrm{x}$

$\begin{array}{l}\therefore \int \left|\mathrm{x}\right|\log \left|\mathrm{x}\right|\mathrm{dx}=-\int \mathrm{xlog} (-\mathrm{x})\mathrm{dx}\\ =-\left\{\log (-\mathrm{x})\cdot \frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^2}{4}\right\}+\mathrm{C}\\ =-\frac{{\mathrm{x}}^2}{2}\log \left|\mathrm{x}\right|+\frac{{\mathrm{x}}^2}{4}+\mathrm{C}\end{array}$

On combining both cases, we get

$\int \left|\mathrm{x}\right|\log \left|\mathrm{x}\right|\mathrm{dx}=\frac{1}{2}\mathrm{x}\left|\mathrm{x}\right|\log \left|\mathrm{x}\right|-\frac{1}{4}\mathrm{x}\left|\mathrm{x}\right|+\mathrm{C}$