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प्रश्न
`int(log(logx))/x "d"x`
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उत्तर
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
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