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Question
Fill in the Blank.
`int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" (log "x")^3` + c, then P = _______
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Solution
`int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" (log "x")^3` + c, then P = `underline(1/3)`
Explanation:
Let I = `int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" * (log "x")^3` + c
I = `int 1/"x"^3 [log "x"^"x"]^2 "dx" = int 1/"x"^3 ("x log x")^2 * "dx"`
`=int 1/"x"^3 * "x"^2 * (log "x")^2 "dx" = int 1/"x" (log "x")^2 * "dx"`
∴ Put log x = t
∴ `1/"x"` dx = dt
∴ I = `int "t"^2 * "dt"`
`= "t"^3/3 + "c"`
`= 1/3 (log "x")^3 + "c"`
∴ P = `1/3`
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