Explanation

(d)
Case I If $x > 0,\ then\ |x| = x$
$\therefore\ \int_{}^{}{|x|\log{|x|dx = \int_{}^{}{x\log{x\ dx}}}}$
$= \log{x \bullet \frac{x^{2}}{2} - \int_{}^{}{\frac{1}{x} \bullet \frac{x^{2}}{2}dx}}$
$= \frac{x^{2}}{2} \bullet \log{x - \frac{x^{2}}{4} + c}$
$= - \frac{x^{2}}{2}\log{|x| - \frac{x^{2}}{4} + c}$
Case II If $x < 0$, then $|x| = - x$
$\therefore\ \int_{}^{}{|x|\log{|x|dx = - \int_{}^{}{x\log{( - x)dx}}}}$
$= - \left\{ \log{( - x) \bullet \frac{x^{2}}{2} - \frac{x^{2}}{4}} \right\} + c$
$= - \frac{x^{2}}{2}\log{|x| + \frac{x^{2}}{2} + c}$
On combining both cases, we get
$\int_{}^{}{|\ x\ |\log{|x|dx = \frac{1}{2}x|x|\log{|x| - \frac{1}{4}x|x| + c}}}$