Answer 1

Let I = `int (log "x")^2` dx

`= int (log "x")^2 * 1`dx

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

`= "x"(log "x")^2 * "x" - int 2 log "x" * 1/"x" * "x" * "dx"`

`= "x"(log "x")^2 - 2 int (log "x") * 1 * "dx"`

`= "x"(log "x")^2 - 2[log "x" int 1 * "dx" - int {"d"/"dx" (log "x") int 1 * "dx"}]`dx

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

`= "x"(log "x")^2 - 2["x" log "x" - int 1 * "dx"]`

`= "x"(log "x")^2 - 2("x" log "x" - "x")` + c

∴ I = x(log x)² - 2x log x - 2x + c