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Q.

∫1+logx1+xlogx   dx  =

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a

log|1+xlogx|+c

b

log|xlogx|+c

c

log|1+logx|+c

d

1+xlogx+c

answer is A.

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Detailed Solution

∫1+logx1+xlogx   dx  = log|1+xlogx|+c   since ∫f'xfxdx=log fx                           ∫f|(x)f(x) dx
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