求不定积分∫xln(x+1)dx

用分部积分公式
∫xln(x+1)dx=x^2ln(x+1)-∫[xln(x+1)+x^2/(x+1)]dx=x^2ln(x+1)-∫xln(x+1)dx-∫x^2/(x+1)dx……（1）
∫x^2/(x+1)dx=∫[(x+1)-2+1/(x+1)]dx=x^2/2-x+ln(x+1)+c
令∫xln(x+1)dx=y
由（1）式得y=x^2ln(x+1)-y-[x^2/2-x+ln(x+1)]
解出y=[x^2ln(x+1)]/2-[x^2/2-x+ln(x+1)]/2=-x^2/4+x/2-(x^2-1)[ln(x+1)]/2

用分布积分公式
∫uv'=uv-∫u'v 把x看成u ln(x+1)看成v
所以原式=(x*x/2)*ln(x+1)-(1/2)∫(x*x)/(x+1)dx
再看∫(x*x)/(x+1)dx=∫[(x+1)(x-1)+1]/(x+1)dx
=∫[(x-1)+1/(x+1)]dx
=∫(x-1)dx+∫1/(x+1)dx
=∫xdx-∫dx+∫1/(x+1)d(x+1)
=1/(2x*x)-x+ln|x+1|
把这个结果代入上式即可

∫xln(x+1)dx＝∫ln(x+1)d(1/2*x^2)＝1/2×x^2×ln(x+1)－1/2×∫x^2dln(x+1)＝1/2×x^2×ln(x+1)－1/2×∫x^2/(x+1)dx＝1/2×x^2×ln(x+1)－1/2×∫[x-1+1/(x+1)]dx＝1/2×x^2×ln(x+1)－1/2×[1/2×x^2－x＋ln(x+1...

∫xln(x+1)dx
=∫(x+1)ln(x+1)d(x+1)-∫ln(x+1)d(x+1)
=0.5(∫ln(x+1)d(x+1)^2-∫ln(x+1)d(x+1))
=0.5((x+1)^2ln(x+1)-∫(x+1)^2dln(x+1)-(x+1)ln(x+1)+∫(x+1)dln(x+1))
=0.5((x+1)^2ln(x+1)-∫(x+1)dx-(x+1)ln(x+1)+∫dx)
=0.5((x+1)^2ln(x+1)-0.5x^2-x-(x+1)ln(x+1)+x)
=0.5(x+1)^2ln(x+1)-0.25x^2-0.5(x+1)ln(x+1)+C