Answer 1

Let I = `int [1/(log "x") - 1/(log "x")^2]` dx

Put log x = t

∴ x = e^t

∴ dx = e^t dt

∴ I = `int "e"^"t" [1/"t" - 1/"t"^2]` dt

Put f(t) = `1/"t"`

∴ f '(x) = `(-1)/"t"^2`

∴ I = `int "e"^"t" ["f"("t") + "f" '("x")]` dt

`= "e"^"t"   "f"("t")` + c

∴ I = `"e"^"t" (1/"t") + "c" = "x"/(log "x")` + c