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प्रश्न
Evaluate the following.
`int [1/(log "x") - 1/(log "x")^2]` dx
उत्तर
Let I = `int [1/(log "x") - 1/(log "x")^2]` dx
Put log x = t
∴ x = et
∴ dx = et dt
∴ I = `int "e"^"t" [1/"t" - 1/"t"^2]` dt
Put f(t) = `1/"t"`
∴ f '(t) = `(-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
Notes
The answer in the textbook is incorrect.
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