∫|x|log ⁡|x|dx is equal to (x≠0)

|x|log |x|dx is equal to (x0)

  1. A

    x22log |x|x24+C

  2. B

    12x|x|log x+14x|x|+C

  3. C

    x22log |x|+x24+C

  4. D

    12x|x|log |x|14x|x|+C

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    Solution:

     Case II If x<0, then |x|=x

    |x|log |x|dx=xlog xdx=log xx221xx22dx=x22log xx24+C=+x22log |x|x24+C

     Case II If x<0, then |x|=x

    |x|log |x|dx=xlog (x)dx=log (x)x22x24+C=x22log |x|+x24+C

    On combining both cases, we get

    |x|log |x|dx=12x|x|log |x|14x|x|+C

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