问题标题：
呼呼~∫xln（x+1）dx
问题描述：
呼呼~
∫xln（x+1）dx
李金峰回答：
原式=1/2∫ln(x+1)dx²
=1/2x²ln(x+1)-1/2∫x²dln(x+1)
=[x²ln(x+1)]/2-1/2∫x²/(x+1)dx
单独求∫x²/(x+1)dx
=∫(x²-1+1)/(x+1)dx
=∫[(x²-1)/(x+1)+1/(x+1)]dx
=∫[(x-1)+1/(x+1)]dx
=∫(x-1)dx+∫dx/(x+1)
=x²/2-x+ln(x+1)+C
所以原式=[x²ln(x+1)]/2-x²/4+x/2-1/2*ln(x+1)+C
=[(x²-1)/2]ln(x+1)-x²/4+x/2+C
