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Question

Evaluate: 1xlogxlog(logx)dx

A
log(log(x))+c
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B
log(log(log(1x)))+c
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C
log(log(log(x)))+c
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D
log(log(1x))+c
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Solution

The correct option is B log(log(log(x)))+c
Now,
1xlogxlog(logx)dx
d{log(logx)}log(logx)dx
=log{log(logx)}+c [ Where c being integrating constant].

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