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Question
Find `int_ (log "x")^2 d"x"`
Solution
Let `I = int (log "x")^2 d"x"`
⇒ `I = int_ 1·(log "x")^2 d"x"`
⇒ `I = "x"·(log "x")^2 - int_ (2"x" log"x")/"x" d"x"`
⇒ `I = "x"·(log "x")^2 - I_1 + c_1` .....(i)
`I_1 = int_ 2·log "x"d"x"`
⇒ `I_1 = 2"x"· log"x"- 2 int_ "x"/"x" d"x"`
⇒ `I_1 = 2"x"·log "x" - 2"x" + c_2` .....(ii)
From (i) and (ii), we get
`I = "x"·(log "x")^2 - 2"x"·log "x"+ 2"x" + c_1 - c_2`
`I = "x"·(log "x")^2 - 2"x"·log "x"+ 2"x" + C` ...(where C = C1 - C2)
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