∫log(logx)x dx - Mathematics and Statistics

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Sum

`int(log(logx))/x  "d"x`

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Solution

Let I = `int(log(logx))/x  "d"x`

Put log x = t

∴ `1/x  "d"x` = dt

∴ I = `int log "t"  "dt" = intlog"t"*1  "dt"`

= `log "t" int 1*"dt" - int ["d"/"dt"(log"t") int 1*"dt"]"dt"`

= `log "t"* "t" - int(1/"t" xx "t") "dt"`

= `"t"*log "t" - int "dt"`

= t log t − t + c

= t (log t − 1) + c

∴ I = logx [log (logx) − 1] + c

  Is there an error in this question or solution?
Chapter 2.3: Indefinite Integration - Short Answers I

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