Answer 1

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)