I can do the derivation but it is really long.
Just go here if you want the answer without the derivation.
Hello, Csou090490!
. . u .= .ln(x² + 2) . . . .dv .= .dx∫ ln[x² + 2] dx
Can this be done by integration by parts? . Yes!
. . . . . . . . .2x
. .du .= . -------- dx . . . .v .= .x
. . . . . . . x² + 2
. . . . . . . . . . . . . . . . . . . . . . .x²
We have: . x·ln(x² + 2) - 2 ∫ -------- dx
. . . . . . . . . . . . . . . . . . . . .x² + 2
. . . . . . . . . . . . . . . . . . . 2
Divide that fraction: .1 - --------
. . . . . . . . . . . . . . . . . .x² + 2
. . . and go for it!