Question

Evaluate the following.

`int (log "x")/(1 + log "x")^2` dx

Answer 1

Let I =`int (log "x")/(1 + log "x")^2` dx

Put log x = t

∴ x = e^t

∴ dx = e^t 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

Answer 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`