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a
x22log |x|−x24+C
b
12x|x|log x+14x|x|+C
c
−x22log |x|+x24+C
d
12x|x|log |x|−14x|x|+C
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Correct option is D
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|+CSimilar Questions
The value of is equal to
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