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Question
Evaluate the following.
`int (log "x")/(1 + log "x")^2` dx
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Solution 1
Let I =`int (log "x")/(1 + log "x")^2` dx
Put log x = t
∴ x = et
∴ dx = et dt
∴ I = `int "t"/(1 + "t")^2 "e"^"t" "dt"`
`= int "e"^"t" [(("t" + 1) - 1)/(1 + "t")^2]` dt
`= int "e"^"t" [cancel("t + 1")/("1 + t")^cancel2 - 1/(1 + "t")^2]` dt
`= int "e"^"t" [1/(1 + "t") - 1/(1 + "t")^2]` dt
Put f(t) = `1/"1 + t"`
∴ f '(t) = `(-1)/(1 + "t")^2`
∴ `int "e"^"t" ["f"("t") + "f" '("t")]` dt
`= "e"^"t" "f"("t") + "c"`
`= "e"^"t" * 1/(1 + "t")` + c
∴ I = `"x"/(1 + log "x")` + c
Solution 2
Let I =`int (log x)/(1 + log x)^2` dx
Adding and subtracting 1 from the numerator,
I =`int ((1 + log x) - 1)/(1 + log x)^2` dx
I =`int cancel((1 + log x))/(1 + log x)^cancel2 - int 1/(1 + log x)^2` dx
I =`int 1/(1 + log x) - int 1/(1 + log x)^2` dx
I =`int (1 + log x)^(-1) - int 1/(1 + log x)^2` dx
I =`int (1 + log x)^(-1).1 - int 1/(1 + log x)^2` dx
Integration by Parts,
`∫"u"."v" "dx" = "u" ∫"v" "dx" − ∫ [ ∫"v" "dx". "du"/"dx"] "dx"`
`"I" = (1 + log x)^(-1) ∫1 "dx" − ∫ [∫1 "dx". "d"/"dx" (1 + log x)^(-1)] "dx" - int 1/(1 + log x)^2 "dx"`
`"I" = (1 + log x)^(-1). x − ∫[cancelx. (-1)/(cancelx(1 + log x)^2)] "dx" - int 1/(1 + log x)^2 "dx"`
`"I" = (1 + log x)^(-1). x + ∫cancel(1/(1 + log x)^2) - int cancel(1/(1 + log x)^2) + c`
I = `(1 + log x)^(-1). x + c`
I = `x/(1 + log x). + c`
Notes
The answer in the textbook is incorrect.
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