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Q.

∫1ee|ln x|dx=

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a

1−1e

b

1+1e

c

2  (1−1e)

d

2  (1+1e)

answer is C.

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Detailed Solution

∫1ee|logx|dx=∫1e1|logx|dx+∫1e|logx|dx                      =−∫1e1logx dx+∫1elogx dx                      =−(xlogx−x)1e1+(xlogx−x)1e                      =2−2e
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∫1ee|ln x|dx=