∫|x|log ⁡|x|dx is equal to (x≠0)

Case II If x<0, then |x|=−x∴∫|x|log ⁡|x|dx=∫xlog ⁡xdx=log ⁡x⋅x22−∫1x⋅x22dx=x22⋅log x−x24+C=+x22log⁡ |x|−x24+C Case II If x<0, then |x|=−x∴∫|x|log ⁡|x|dx=−∫xlog ⁡(−x)dx=−log ⁡(−x)⋅x22−x24+C=−x22log⁡ |x|+x24+COn combining both cases, we get∫|x|log⁡ |x|dx=12x|x|log ⁡|x|−14x|x|+C